Math 147 — Complex Analysis

Neil Donaldson

November 26, 2024

1 Complex Numbers

1.1 Deﬁnition and Basic Algebraic Properties

In the 1500s, Italian mathematician Rafael Bombelli posited a solution to the seemingly absurd equa-

tion x

= −1. By supposing that this solution behaved according to the usual rules of algebra,

Bombelli and others were able to describe the solutions of any quadratic equation. To some extent,

this was math for its own sake; Bombelli always considered his solutions to be entirely ‘ﬁctitious.’

For a modern deﬁnition, we start with the Cartesian plane R



(x, y) : x, y ∈ R



Deﬁnition 1.1. Given real numbers x, y, the complex number z = x + iy is the point with co-ordinates

(x, y) ∈ R

. Its real and imaginary parts are the individual co-ordinates

Re z = x, Im z = y

The complex numbers C comprise the real vector space R

with the extra

operation of complex multiplication: if z = x + iy and w = u + iv, deﬁne

(Vector) Addition: z + w := (x + u) + i(y + v)

Complex multiplication: zw := (xu − yv) + i(xv + yu)

When drawn with axes, the complex plane is known as the Argand

diagram and we refer, respectively, to the real and imaginary axes.

z + w

Since C = R

is a real vector space under addition, several properties are immediate:

Lemma 1.2 (Basic properties of complex addition).

Associativity: z

+ (z

+ z

) = (z

+ z

) + z

Commutativity: z + w = w + z (this is the parallelogram law illustrated in the picture)

(Real) scalar multiplication: ∀λ ∈ R, λ(x + iy) = λx + iλy

Additive inverse: −z = −(x + iy) = (−x) + i(−y) = −x −iy

Example 1.3. If z = 3 + 4i and w = 2 − 7i, then

z − w = z + (−w) =



3 −2





4 − (−7)



i = 1 + 11i

The natural distance measure from R

transfers to C (the complex numbers are a real metric space).

Deﬁnition 1.4. The modulus of a complex number z = x + iy is the

Euclidean distance of the point (x, y) from the origin:

+ y

In the picture, z = 1 +

√

3i has modulus

√

1 + 3 = 2.

0 1 2

z = 1 +

√

|z| = 2

Some natural inequalities follow straight from the picture in Deﬁnition 1.1.

Lemma 1.5 (Triangle inequalities). For all z, w ∈ C,

z + w

≤

and

z + w

≥



−



In the second (“extended”) inequality, we take the absolute value of the difference of the moduli. Un-

like in R, these inequalities relate to honest triangles! We can easily generalize the ﬁrst by induction,

+ z

+ ··· + z

≤

+ ··· +

The modulus may be used to describe various curves and regions in the plane.

Examples 1.6. 1.

= 3 describes the circle with radius 3 centered at

the origin. In Cartesian co-ordinates, this is x

+ y

= 9.

z − 2 −3i

≤ 2 describes the disk with radius 2 centered at 2 + 3i.

z − 2

= 4 describes an ellipse with foci 0 and 2. This is

more familiar after multiplying out (try it!):

+ y

(x −2)

+ y

= 4 =⇒ (x −2)

+ y

= ···

=⇒

(x −1)

= 1

4−4

−4i

Complex multiplication, division and the complex conjugate

Multiplication is what really distinguishes the complex numbers from R

, and lead to all the inter-

esting structure in this course. For starters, we instantly see that i is a solution to Bombelli’s absurd

equation:

= (0 + 1i)(0 + 1i) = (0 · 0 −1 ·1) + i(0 ·1 + 1 · 0) = −1

The upshot is that we can treat complex addition, subtraction and multiplication as if we are working

with linear polynomials

in the abstract variable i; simply replace i

with −1 when needed.

This is precisely the deﬁnition encountered in a Rings & Fields course: C is the factor ring of real polynomials modulo

the ideal



+ 1



Example 1.7. If z = 3 + 4i and w = 2 − 7i, then

zw = (3 + 4i)(2 − 7i) = 3 · 2 + 4i ·2 −3 ·7i −4i ·7i = 6 + 8i −21i −28i

= 6 + 8i −21i + 28 = 34 −13i

The basic algebraic properties of complex multiplication are straightforward, if tedious, to verify:

Lemma 1.8 (Basic properties of multiplication). For any complex numbers z

, z

Associativity: z

( z

) = (z

) z

Commutativity: z

= z

Distributivity: z

( z

+ z

) = z

+ z

To develop division, it is is helpful to introduce a new concept.

Deﬁnition 1.9. The (complex) conjugate of z = x + iy is z := x − iy (“z-bar”). Geometrically, z is

obtained by reﬂection in the real axis.

Observe that zz = (x + iy)(x −iy) = x

+ y

, from which:

Lemma 1.10. Every non-zero complex number z = x + iy has a unique multiplicative inverse

−1

x −iy

+ y

satisﬁes zz

−1

= 1

Proof. That zz

−1

= 1 is trivial. For uniqueness, suppose we also have zw = 1; now use associativity

and commutativity to conclude that

w = (z

−1

z)w = z

−1

( zw) = z

−1

Division is simply multiplication by an inverse.

Example 1.11. Given z = 3 + 4i and w = 2 − 7i, compute

2 −7i

3 + 4i

= wz

−1

(2 − 7i)(3 − 4i)

3 + 4i

2 ·3 −2 ·4i −7i ·3 + 7i ·4i

+ 4

6 −8i −21i + 28i

−22 − 29i

Alternatively, we could multiply numerator and denominator by the conjugate

of the denominator:

2 −7i

3 + 4i

2 −7i

3 + 4i

3 −4i

= ···

Compare this approach to elementary algebra where, for instance, we use the conjugate 5 +

√

3 of 5 −

√

3 to evaluate

the reciprocal

5−

√

(5−

√

3) (5+

√

This initial section should be revision, though some unfamiliarity with terminology is expected. The

goal at this stage is to be ﬂuent when computing with complex numbers. The exercises should feel

straightforward, particularly 1–6; if not, ask for help!

Exercises 1.1. 1. The real number 2 is also a complex number. What is it in co-ordinate form (x, y)?

What about the complex number i?

2. For any z ∈ C, prove that Re(iz) = −Im z and that Im(iz) = Re z.

3. (a) Check that both z = 2 + 3i and its conjugate z = 2 − 3i solve the quadratic equation

−4z + 13 = 0.

(b) More generally, suppose a, b, c ∈ R where ω := 4ac −b

> 0. Verify that z =

−b+i

√

and

its conjugate z both solve

the quadratic equation az

+ bz + c = 0.

4. Prove the commutativity of complex multiplication (Lemma 1.8) using Deﬁnition 1.1.

5. Evaluate the following in the form x + iy.

(a)

2 −i

3 −5i

(b) (1 + i)

−2

−(2 − 3i)

−2

6. Prove the following. Write z = x + iy rather than using the vector deﬁnition.

(a)

z = z (b) (z

−1

)

−1

= z (c) zw = z ·w

7. (a) Use the ﬁrst version of the triangle inequality (Lemma 1.5) to prove the second: for any

z, w ∈ C, we have

z + w

≥



−



(b) What relationship between z, w corresponds to equality here? Draw a picture!

8. Suppose that

≥ 2 and consider the polynomial P(z) = z

+ 3z −1.

(a) Use the triangle inequality to prove that



3z−1



≤

(b) Write

P(z)



+ 3z −1



1 +

3z−1



. Now use the extended triangle inequality

to prove that

P(z)

≥ 1.

(This shows that all zeros of P(z) lie inside the circle

< 2)

9. By considering the inequality (

−

)

≥ 0, prove that

√

≥

Re z

Im z

for any z ∈ C

10. Prove that the hyperbola x

−y

= 1 can be written in the form z

+ z

= 2.

11. Draw a picture of the ellipse satisfying the equation

z − 4i

= 6. Find the equation of the

curve in Cartesian coordinates:

(x−c)

(y−d)

= 1 where (c, d) is the center of the ellipse and

a, b are its semi-axes.

(Hint: write

z − 4i

= 6 −

, square both sides, cancel x

, y

terms and repeat. . . )

Since i

= −1, we may write i

√

ω =

√

−ω, whence the quadratic formula applies to all real quadratic equations!

1.2 The Exponential/Polar Form of a Complex Number

Deﬁnition 1.12. A complex number can be written in polar co-ordinates:

z = x + iy = r cos θ + ir sin θ = r(cos θ + i sin θ)

Plainly r =

is the modulus. We call the angle θ = arg z the argument of z.

The argument is multi-valued in the sense that arg z = θ + 2πn for any in-

teger n. For clarity, we distinguish the principal argument Arg z by insisting

that −π < Arg z ≤ π.

0 1 2

z = 1 +

√

|z| = 2

Arg z =

Note that 0 has no argument; it is the only such complex number.

Example 1.13. In the above picture, z = 1 +

√

3i has principal argument Arg z =

. You can write

the argument either as many different values, or as a set:

the following are all legitimate

arg z = {

+ 2πn : n ∈ Z} or arg z =

or arg z =

7π

If x = 0, is it almost trivial to compute the argument:

(

x = r cos θ

y = r sin θ

=⇒ tan θ =

???

=⇒ Arg z = tan

−1

This isn’t quite right. Since arctan has range (−

an addition or subtraction of π is required to ﬁnd the

correct value whenever z lies in the second or third

quadrants.

Example 1.14. Since z = −3 −3i lies in the third quad-

rant, we have

Arg z = tan

−1

−3

−π = −

3π

If we wanted the argument to be positive, we could

choose a non-principal argument arg z =

5π

Q1: tan

−1

Q2: tan

−1

+ π

Q3: tan

−1

−π Q4: tan

−1

−

The polar form allows us to deﬁne a crucial new function.

Deﬁnition 1.15. Given θ ∈ R, the complex exponential e

iθ

is deﬁned by Euler’s formula

iθ

:= cos θ + i sin θ

The polar form of a complex number is z = re

iθ

where r =

and θ = arg z (any argument! why?).

If w = u + iv is complex, then its exponential is deﬁned by e

:= e

= e

(cos v + i sin v).

This is merely the common mathematical fudge of replacing an equivalence class {

+ 2πn : n ∈ Z} by any of its

representatives, e.g.

7π

Euler’s formula provides a sensible deﬁnition of e

iθ

since it ﬁts with two common deﬁnitions of the

exponential in real analysis.

1. If k ∈ R, then e

kθ

is the solution to the initial value problem y

′

= ky with y(0) = 1. Assuming

that differentiation works when k = i, Euler’s formula also satisﬁes this criterion:

dθ

iθ

dθ

(cos θ + i sin θ) = −sin θ + i cos θ = i(cos θ + i sin θ) = ie

iθ

2. The real and imaginary parts of the Maclaurin series exp z =

∑

evaluated at z = iθ are,

respectively, the Maclaurin series of cos θ and sin θ.

A third reason is that the deﬁnition satisﬁes the usual exponential laws.

Lemma 1.16 (Exponential laws). Let z = re

iθ

and w = se

iψ

be written in polar form. Then:

1. zw = rse

i(θ+ψ)

. In particular,

and arg zw = arg z + arg w

i(θ−ψ)

3. z

= r

inθ

for all n ∈ Z

Note that the principal argument might not behave so nicely for products; the best we can say is that

Arg zw = Arg z + Arg w + 2πn for some n = 0, ±1

Proof. Part 1 follows from the multiple-angle formulæ for sine and cosine:

i(θ+ψ)

= cos(θ + ψ) + i sin(θ + ψ) = cos θ cos ψ −sin θ sin ψ + i(sin θ cos ψ + cos θ sin ψ)

= (cos θ + i sin θ)(cos ψ + i sin ψ) = e

iθ

iψ

Parts 2 and 3 are now straightforward.

Examples 1.17. 1. Given z = −7 + i and w = 3 + 4i, we compute the modulus and argument of zw

in two ways:

(a) Find the polar forms of z, w before applying the Lemma:

z =

i arg z

= 5

√

2 exp



i(π −tan

−1

)



, w =

i arg w

= 5 exp



i tan

−1



=⇒

= 25

√

2, arg zw = arg z + arg w = π − tan

−1

+ tan

−1

(b) Find zw = (−7 + i)(3 + 4i) = −25 − 25i then compute its polar form:

zw = 25

√

−

3πi

=⇒

= 25

√

2, Arg zw = −

3π

The ﬁrst approach is certainly uglier! It is usually better to use the second approach unless the

arguments of z, w are easily computable. In Exercise 7, we check that these values correspond.

2. We compute z

when z =

√

3 −i. First observe that z = 2e

−

πi

, from which

= 2

−

5πi

= 1024e

πi

= 512(1 +

√

3i)

3. The identity (e

iθ

)

= e

inθ

(n ∈ Z) is known as de Moivre’s formula. It is usually written

(cos θ + i sin θ)

= cos nθ + i sin nθ

Many trigonometric identities follow by taking real/imaginary parts. For instance, when n = 3,

cos 3θ + i sin 3θ = (cos θ + i sin θ)

= cos

θ + 3i cos

θ sin θ −3 cos θ sin

θ −i sin

=⇒ cos 3θ = cos

θ −3 cos θ sin

θ = 4 cos

θ −3 cos θ

As before, this shouldn’t be difﬁcult. This section is merely polar co-ordinates in a different language!

Exercises 1.2. 1. Use induction to prove that for any n ∈ N

≥2

iθ

···e

iθ

= e

i(θ

+θ

+···+θ

)

2. Find the principal argument of (1 + i)

2024

3. Prove that



iθ



= 1 and that e

iθ

= e

−iθ

4. (a) Show that if Re z > 0 and Re w > 0, then Arg zw = Arg z + Arg w.

(b) If z and w both lie in quadrant 2, explain why Arg zw = Arg z + Arg w −2π.

5. Prove that non-zero z, w ∈ C have the same modulus if and only if ∃p, q ∈ C such that z = pq

and w = pq.

6. Use de Moivre’s formula to establish the identity

cos 4θ = 8 cos

θ −8 cos

θ + 1

7. (a) Let α = tan

−1

and β = tan

−1

. Use right-triangles to show that

cos α =

, sin α =

, cos β =

√

, sin β =

√

Now use the cosine multiple-angle formula to check that α − β =

(This shows that arg zw =

5π

in Example 1.17(a))

(b) Generalize the approach in part (a): if 0 < β < α <

, prove the multiple-angle formula

tan(α − β) =

tan α − tan β

1 + tan α tan β

8. The polar form of a complex number is well-suited to describing circles. For instance the circle

centered at i with radius 3 may be described by z = i + 3e

iθ

where −π < θ ≤ π.

(a) Describe the circle centered at z

= 3 + 4i with radius 2.

(b) Show that the points z = re

iθ

for which r = 2a cos θ describe a circle.

(Hint: Multiply by r)

1.3 Roots of Complex Numbers

A na

ıve approach to taking roots in C is very messy.

Example 1.18. To ﬁnd c such that c

= −5 + 12i, we need to solve an equation:

−5 + 12i = c

= (x + iy)

= x

−y

+ 2ixy ⇐⇒

(

−y

= −5

xy = 6

Substituting y = 6x

−1

into the ﬁrst equation yields a quadratic in x

+ 5x

−36 = (x

−4)(x

+ 9)

We conclude that x = ±2, and obtain the square roots ±c = ±(2 + 3i).

The example is reassuring in that we obtain precisely two square roots. Unfortunately, extending the

method to cube, or higher, roots seems doomed! Instead we use the polar form. Suppose n ∈ N and

that c, z satisfy z = c

. In polar form

z = re

iθ

, c = se

iψ

=⇒ re

iθ

= s

inψ

By equating moduli and arguments, we see that

r = s

, nψ = θ + 2πk (∗)

where k is some integer. We’ll shortly put this together to obtain a proper deﬁnition, but we already

have enough for a calculation.

Example 1.19. We compute the ﬁfth roots of z = 2e

2πi

= −1 + i

√

In the above language, 2 = s

, whence

s =

√

2, ψ =



2π

+ 2πk



2π

(1 + 3k)

which results in the ﬁfth roots

√

2πi

√

8πi

√

14πi

√

20πi

√

−

10πi

√

26πi

√

−

4πi

There are precisely ﬁve ﬁfth roots: once k ≥ 5, the roots start re-

peating! The last two roots were written both with positive and

principal arguments; both have their advantages.

2π

√

Observe also how the ﬁfth roots form the vertices of a regular pentagon, equally spaced around the

circle of radius

√

2. Is it obvious to you why this is so?

Finally, note how essential the polar form was to this calculation. We could convert back to rectan-

gular form, but since cos

2π

and sin

2π

are unfriendly values, there is little beneﬁt.

Deﬁnition 1.20. Given a non zero complex number z = re

iθ

and a positive integer n, the n

roots of

z are the n distinct complex numbers

√

r exp

( θ + 2kπ)i

k = 0, 1, . . . , n −1

where

√

r is the real (positive!) n

root of r.

There are some conventions to follow. Let θ = Arg z be the principal argument of z:

(a) The principal n

root is written

√

z :=

√

iθ

(b) The set of n

roots is denoted z

:= {c

, . . . , c

n−1

Denote by ω

= e

2πi

a primitive n

root of unity (n

root of 1). Then the full set of n

roots of unity is

= {ω

: k = 0, . . . , n − 1} = {e

2πki

: k = 0, . . . , n − 1}

The n

roots of z may be written in terms of the principal root and the n

roots of unity

√

z 1

= {

√

zω

: k = 0, . . . , n − 1}

By the exponential laws (Lemma 1.16), multiplication by ω

= e

2πki

has the geometric effect of rotat-

ing counter-clockwise by

2πk

radians:

arg

√

zω

= arg

√

z + arg ω

= arg

√

z +

2πk

It follows that the n

roots of z = re

iθ

form the vertices of a regular n-gon spaced equally round the

circle of radius

√

r. Compare this with the previous example.

Examples 1.21. 1. As a sanity check, we compare what happens when n = 4 and z = r = 16 is real

and positive.

• The principal fourth root

√

16 = 2 is the usual positive real fourth root.

• Within the real numbers, we have two fourth roots: 16

= ±2.

• Within C, there are four fourth roots: 16

= {2, 2i, −2, −2i} where i = ω

= e

iπ

is a

primitive fourth root of unity.

2. We compute the fourth roots of z = 8

√

2( 1 + i).

First we write in polar form: z = 16e

πi

. Since Arg z =

, the principal fourth root is

√

2( 1 + i) = 2e

πi

To ﬁnd all fourth roots, simply multiply by the fourth roots of unity 1

= {1, i, −1, −i}:



√

2( 1 + i)



= {±2e

πi

, ±2ie

πi

} = {2e

πi

, 2e

9πi

, 2e

−

15πi

, 2e

−

7πi

}

Evaluating these in rectangular form is messy but doable (see Exercise 7). In practice it is better

to leave such expressions in polar form.

The language of n

roots is likely less familiar, and will take more getting used to, than the material

in the previous sections. With a little practice it is very easy, but ﬂuency will take some investment

of time.

Exercises 1.3. 1. Find the sixth roots of i in polar co-ordinates. Which is the principal root?

2. Find the square roots of −

√

3 + i and express them in rectangular co-ordinates.

(Hint: you may ﬁnd the fact that (

√

3 −1)

= 4 −2

√

3 useful)

3. Use the fact that the cube roots of unity are 1, ω

−1+

√

and ω

−1−

√

to evaluate the

cube roots of −27 in rectangular co-ordinates.

4. We previously found the fourth roots of 16. Use these to ﬁnd the fourth roots of −16. Hence

factorize the equation z

+ 16 = 0 as a product of two quadratic equations with real coefﬁcients.

5. If ω is an n

root of unity other than 1, prove that

n−1

∑

k=0

= 0.

(Hint: recall the geometric series formula)

6. (a) Suppose that a, b, c ∈ C with a = 0 and suppose that z satisﬁes the quadratic equation

+ bz + c = 0. Prove the quadratic formula:

z =

−b + (b

−4ac)

1/2

Note that (b

−4ac)

1/2

is the set of square roots of b

−4ac, so that this provides two solu-

tions whenever b

−4ac = 0.

(b) Find the roots of the equation iz

+ (1 + i)z + 3 = 0 in rectangular form.

7. Use the half-angle formula cos

(1 + cos α) to explicitly evaluate cos

and then cos

Hence ﬁnd an expression for the rectangular form of

√

2( 1 + i) = 2e

πi

using only square

roots.

8. Recall Example 1.18. Verify that the method in Deﬁnition 1.20 gives the same value for the

principal square root

√

−5 + 12i.

(You’ll need some trig identities. . . )

2 Holomorphic Functions

In this chapter we discuss functions of a complex variable and what it means for such to be differen-

tiable. This turns out to be more subtle and restrictive than in real analysis.

2.1 Functions of a Complex Variable

As in real analysis, it is common to express a function via a rule/formula.

Example 2.1. Deﬁne f : C → C by f (z) = z

−z. Evaluating is straightforward: e.g.

f (2 + i) = (2 + i)

−(2 + i) = 2

+ 3 ·2

i + 3 ·2i

+ i

−2 −i = 10i

The implied domain of a rule is the largest possible set D ⊆ C on which the rule is deﬁned.

Example 2.2. f (z) =

has implied domain D = C \ {±3i}. Complex functions can be repre-

sented in polar form by substituting z = re

iθ

f (z) =

2iθ

+ 9

It is also common to separate the real and imaginary parts by writing f ( z) = u(x, y) + iv(x, y) where

u, v : D → R. In this case,

f (z) =

(x + iy)

+ 9

−y

+ 9 + 2ixy

−y

+ 9

−y

+ 9)

+ 4x

+ i

−2xy

−y

+ 9)

+ 4x

These approaches may be combined by writing u, v as functions of r, θ.

The major initial challenge presented by functions of a complex variable is that of visualization. Since

C has two real dimensions, graphing a function f : C → C requires four real dimensions! Since we

cannot see four dimensions at once, any visualization we obtain will only be partial. To see this at

work, we consider a simple function in some detail.

Example 2.3. To help understand f ( z) = z

, we start by computing the various forms described

above:

f (z) = z

= x

−y

+ 2ixy

= r

2iθ

= r

cos 2θ + ir

sin 2θ

While we cannot graph the entire function (don’t even think about

a parabola!), we can visualize its real and imaginary parts

u(x, y) = x

−y

, v(x, y) = 2xy

as graphs of functions R

→ R. Both are saddle surfaces, which

may be analyzed using the standard tools of multivariable calcu-

lus. For instance, the level curves u = constant and v = constant are

hyperbolæ. You might ﬁnd this easier using the polar forms of u, v.

Level curves of u = x

−y

The polar form allows us to think about the what f does to the argument:

f (z) = r

2iθ

=⇒ arg f (z) = 2 arg z

Given a sector between arguments θ and ϕ, the function doubles this to the sector between 2θ and 2ϕ.

This can also be visualized pathwise. If z traces a path which once encircles the origin, then the path

traced by z

orbits the origin twice. The colored dots on the two paths correspond under z 7→ z

2θ

2φ

Similar behavior occurs with higher powers. For instance, z 7→ z

maps a single loop round the

origin to a triple loop. By contrast, the principal square root function

√

z =

√

iθ

halves the principal

argument: Arg

√

z ∈ (−

]. In the fractional-exponent language of Section 1.3,

z 7→ z

1/2

= {

√

iθ

, −

√

iθ

}

This is sometimes called a two-valued function—we’ll return to such objects in Chapter 3.

2D geometry with complex numbers

Complex functions easily describe the basic geometric transformations of the plane.

Translation If w is constant, the function f (z) = z + w translates the plane, shifting the origin to w.

Scaling If R > 0 is a real constant, the function f (z) = Rz scales/stretches the complex plane relative

to the origin.

Rotation Given ϕ ∈ R, the function f (z) = e

iϕ

z rotates the plane by ϕ radians counter-clockwise

around the origin. This is easy to see in polar form:

f (z) = f (re

iθ

) = e

iϕ

iθ

= re

i(θ+ϕ)

Reﬂection Complex conjugation f ( z) = z = x −iy reﬂects the plane in the horizontal axis.

By composing such functions, we can describe arbitrary rotations, reﬂections and scalings relative to

any points/lines.

Examples 2.4. 1. We construct rotation by

= 60° counter-clockwise about 2i in three stages:

(a) Translate 2i to the origin: z 7→ z −2i

(b) Rotate by

about the origin: z 7→ e

iπ

( z −2i)

iπ

( z −2i) + 2i

2. We may similarly produce any reﬂection. To reﬂect in the line through 0 and w with Arg w = ϕ;

(a) Rotate the plane by −ϕ: z 7→ e

−iϕ

(b) Reﬂect in the real axis: z 7→ e

−iϕ

z = e

iϕ

2iϕ

3. We compute the function that reﬂects across the line joining α = 2 + i and β = 4 + 3i.

Since β − α = 2 + 2i has argument ϕ =

, we combine reﬂection across the line making ϕ

through the origin with translation by α (translate by −α, reﬂect, translate back by α):

f (z) = e

2iϕ

( z −α) + α = e

iπ

( z −2 −i) + 2 + i = i(z −2 + i) + 2 + i

= iz + 1 − i

As a sanity check, you should verify explicitly that f (α) = α and f (β) = β: why?

Exercises 2.1. 1. For each function, describe its implied domain (page 11).

(a) f (z) =

4 + z

(b) f (z) =

z − 1

−1

+ z + 1

−1

2. Write the function in terms of its real and imaginary parts: f (z) = u(x, y) + iv(x, y).

(a) f (z) = z

−4z

+ 2 (b) f (z) =

1 − z

3. Write the function f (z) =

z in polar form.

4. Find an expression for the function which reﬂects across the vertical line through α = −1.

5. In Example 2.4.3, evaluate the function g(z) = e

2iϕ

( z − β) + β . Why are you not surprised by

the result?

6. Let ϕ = tan

−1

. Find the result (in rectangular co-ordinates) of rotating z = −2 + i counter-

clockwise by ϕ radians around the origin.

(Hint: consider a 3:4:5 triangle!)

7. Prove, using the expressions on page 12, that the composition of two reﬂections is a rotation,

and that the composition of a rotation and a reﬂection is a reﬂection.

2.2 Open sets, Limits and Continuity

This section is mostly for reference, since limits and continuity behave similarly to real analysis.

Indeed proofs are often identical, so we mostly omit them. If you’re only familiar with single-variable

real analysis, note particularly how the ﬁrst deﬁnition generalizes the notion of an open interval.

Deﬁnition 2.5 (Disks and Neighborhoods). Given δ > 0 and z

∈ C, the open disk (or δ-ball) centered

at z

∈ C with radius δ is the set

( z

) = {z ∈ C :

z − z

< δ}

Remove the central point z

for the corresponding punctured open disk:

{z ∈ C : 0 <

z − z

< δ}

Now let D ⊆ C be a subset.

An open neighborhood of z

• D is open if every point is interior: at every z

∈ D we may center some open disk B

( z

) ⊆ D:

∀z

∈ D, ∃δ > 0 such that

z − z

< δ =⇒ z ∈ D

• D is a neighborhood of z

if it contains some B

( z

). A neighborhood can, but need not, be open.

A punctured neighborhood instead contains a punctured disk.

• D is closed if its complement C \ D is open.

Example 2.6. The punctured plane D = C \ {2 + i} is an open set. For

instance, given z

∈ D, let δ =

−2 −i

, then B

( z

) ⊂ D.

(The

is superﬂuous, since the disk B

( z

) does not contain its boundary, it

just makes for a clearer picture.)

2 + i

Deﬁnition 2.7 (Sequences and Limits). Let (z

) = (z

, z

, . . .) be a sequence of complex numbers.

1. (z

) converges to z

∈ C, and we write lim

n→∞

= z

or simply z

→ z

, if

∀ϵ > 0, ∃N such that n > N =⇒

−z

< ϵ

2. (z

) is a Cauchy sequence if

∀ϵ > 0, ∃N such that m, n > N =⇒

−z

< ϵ

Theorem 2.8 (Useful Facts about Sequences). 1. (z

) is convergent/Cauchy if and only if its real

and imaginary parts are also, as follows from two straightforward inequalities:

−z

≤

− x

−y

≤ 2

−z

2. Cauchy completeness: (z

) converges in C if and only if it is Cauchy

3. Bolzano–Weierstraß: If (z

) is bounded, then it has a convergent subsequence.

4. D ⊆ C is closed if and only if every Cauchy sequence (z

) ⊆ D has its limit in D.

Deﬁnition 2.9 (Continuity). Let f : D → C and z

∈ D. We say that f is continuous at z

For all sequences (z

) ⊆ D with lim z

= z

we have lim f (z

) = f (z

)

f is continuous (on D) if it is continuous at all points z

∈ D.

Rather than using sequences, we’ll typically consider continuity via limits of functions.

Deﬁnition 2.10 (Limits of Functions). Let f : D → C, where D contains an open punctured neigh-

borhood of z

. We say that w

is the limit of f as z approaches z

, written lim

z→z

f (z) = w

, if

∀ϵ > 0, ∃δ > 0 such that 0 <

z − z

< δ =⇒

f (z) − w

< ϵ

Otherwise said, given an ϵ-ball centered at w

, there is a δ -ball B

( z

) such that f



( z

)



⊆ B

( w

Equivalently, ∀(z

) ⊆ D \{z

}, lim z

= z

=⇒ lim f (z

) = w

. This is almost continuity!

Theorem 2.11. Let z

be an interior point of D. Then f : D → C is continuous at z

if and only if

lim

z→z

f (z) = f (z

)

Example 2.12. Calculations tend to proceed similarly to real analysis. For instance, we show that

f (z) = z

is continuous (on C) by proving that lim

z→z

= z

for all z

Let z

∈ C and ϵ > 0 be given. Deﬁne δ = min{1,

1+2

}. By the triangle-inequality,

z − z

< δ =⇒

z + z

z − z

+ 2z

≤

z − z

+ 2

< δ + 2

≤ 1 + 2

=⇒



−z



z − z

z + z

< δ(1 + 2

) ≤ ϵ

The picture should help: given an ϵ-ball centered at w

= z

, we’ve described how to choose δ > 0

so that f maps the δ-ball centered at z

to a region inside the original ϵ-ball.

The picture illustrates the case when

3πi

√

( −1 + i), w

= −

, ϵ =

and δ = min

5/2

1+2·

= 1

Theorem 2.13 (Basic Facts about Limits of Functions). Suppose f , g : D → C are functions and

= x

+ iy

is a point satisfying the assumptions of Deﬁnition 2.10.

1. Limits are unique: If w

and

satisfy Deﬁnition 2.10, then

= w

2. If f ( z) = u(x, y) + iv(x, y) and w

= u

+ iv

, then

lim

z→z

f (z) = w

⇐⇒ lim

(x,y)→(x

)

u(x, y) = u

and lim

(x,y)→(x

)

v(x, y) = v

In particular lim

z→z

z = z

3. Suppose lim

z→z

f (z) = w

and lim

z→z

g(z) = w

(a) For any a, b ∈ C, lim

z→z



a f (z) + bg(z)



= aw

+ bw

(b) lim

z→z



f (z)g(z)



= w

z→z

f (z)

g(z)

, provided w

= 0 and g(z) = 0 on a punctured neighbourhood of z

(d) If h is a function such that lim

w→w

h(w) = w

, then lim

z→z

h( f (z)) = w

The basic limit laws should feel familiar and intuitive; indeed analogues of most should have been

proved in real analysis. Parts 2 & 3 have obvious corollaries for continuous functions. For instance:

Corollary 2.14. f : D ⊆ C → C is continuous if and only if its real and imaginary parts u, v : D → R

are also continuous.

Several pieces of part 3 follow from part 2 by considering real and imaginary parts and the limit laws

for functions R

→ R. For instance, write f (z) = u

+ iv

and g(z) = u

+ iv

, then

lim

z→z



f (z)g(z)



= lim

z→z



−v

+ i(u

+ v

)



= w

Examples 2.15. 1. lim

z→1+3i

( z

−iz) = (1 + 3i)

−i



1 + 3i



= 1 + 6i −9 − i(1 −3i) = −11 + 5i.

2. Every polynomial function is continuous on C.

3. Every rational function f (z) =

p(z)

q(z)

(where p, q are polynomials), is continuous on its implied

domain D = {z : q(z) = 0}.

4. The exponential function

exp(z) = e

= e

cos y + ie

sin y

is continuous on C, since the exponential, cosine and sine are continuous on R.

Limits and the Point at Inﬁnity

In (single-variable) real analysis there are two inﬁnities (±∞). In complex analysis, the convention is

to have only one: for instance, the sequences z

= n and w

= in both diverge to the same inﬁnity.

Deﬁnition 2.16. The extended complex plane (Riemann sphere) is the set C = C ∪{∞} where ∞ denotes

the point at inﬁnity. The concepts of neighborhood, openness and limit extend naturally:

1. A neighborhood of ∞ is any set containing an open disk at ∞, namely a subset of the form

{∞} ∪{z ∈ C :

> M}

2. lim

z→z

f (z) = ∞ means: ∀M > 0, ∃δ > 0 such that 0 <

z − z

< δ =⇒

f (z)

> M

3. lim

z→∞

f (z) = w

means: ∀ϵ > 0, ∃N > 0 such that

> N =⇒

f (z) − w

< ϵ

4. lim

z→∞

f (z) = ∞ means: ∀M > 0, ∃N > 0 such that

> N =⇒

f (z)

> M

These limits have the same interpretation as Deﬁnition 2.10: ‘z close to z

’ implies ‘ f (z) close to w

.’

The difﬁculty is merely about the meaning of ‘close to ∞’ when z

or w

is the point at inﬁnity.

Example 2.17. We verify that lim

z→−3i

z+3i

= ∞. Let M > 0 be given and deﬁne δ = min{1,

}. Then

0 <

z + 3i

< δ =⇒

△

≥

−

z + 3i

> 3 −δ ≥ 2 =⇒



z + 3i



≥ M

The relationship between limits, zero and inﬁnity feel like the dubious claims

∞

= 0 and

= ∞!

Theorem 2.18. Provided all limits make sense, Theorem 2.13 also applies to limits involving inﬁnity.

Moreover, we have the additional relationships:

1. lim

z→z

f (z) = ∞ ⇐⇒ lim

z→z

f (z)

= 0 2. lim

z→∞

f (z) = w

⇐⇒ lim

ζ→0





= w

3. lim

z→∞

f (z) = ∞ ⇐⇒ lim

ζ→0

f (1/ζ)

= 0

Sketch Proof. All six results are similar; we prove only the ⇒ direction of the second. Suppose ϵ > 0

is given. Then N exists according to Deﬁnition 2.16 (part 3). Let δ =

and ζ =

, then

0 <

< δ =⇒

> N =⇒



f (

) − w



f (z) − w

< ϵ

Example 2.19. Consider f (z) =

5iz+1

3z−2i

. Plainly

lim

z→

f (z)

= lim

z→

3z −2i

5iz + 1

= 0 =⇒ lim

z→

f (z) = ∞, lim

z→∞

f (z) =

5i + lim

3 −2i lim

Because of this, it is common to view f as a continuous bijection f : C → C (exercise):

f (z) =

5iz + 1

3z −2i



if z =

i, ∞



, f





= ∞, f (∞) =

Aside (non-examinable) The Riemann sphere is merely a fun diversion for us. Unless indicated

otherwise, all sets should be assumed to be subsets of the (ﬁnite) complex plane.

The Riemann sphere can be visualized as a sphere S

with ∞ playing the role of the north pole. The rest of

the sphere is identiﬁed bijectively with the equatorial

plane C via stereographic projection π : S

→ C. In the

picture, the image of P ∈ S

is the intersection π(P) of

C with the line through P and N = π

−1

( ∞).

Compactness, Path-Connectedness & Continuity

We ﬁnish with two further desirable properties of domains.

Deﬁnition 2.20. 1. A subset K ⊆ C is compact if it is closed and bounded.

2. K ⊆ C is (path-)connected

if any two points can be joined

by a path lying within K: more precisely, ∀p, q ∈ K,

∃z : [0, 1] → K continuous with z(0) = p and z(1) = q

z(t)

These concepts generalize notions of intervals from single-variable real analysis. In view of this, we

translate two familiar results, essentially the extreme and intermediate value theorems of real analysis.

Theorem 2.21. Suppose f : K → C is continuous.

1. If K is compact, so is the image f (K).

2. If K is path-connected, so is f (K).

Proof. 1. Consider the real-valued function

and apply the same argument from real analysis:

• Let M = sup{

f (z)

: z ∈ K}. Then ∃(z

) ⊆ K such that lim

f (z

)

= M.

• K is bounded; Bolzano–Weierstraß says (z

) has a convergent subsequence (z

• Let z

= lim z

. Since K is closed we have z

∈ K (Theorem 2.8.4), whence f (z

) is deﬁned.

• f continuous =⇒ lim f (z

) = f (z

); thus M = f ( z

) is ﬁnite and f (K) is bounded.

• The closure of f (K) is a short exercise.

2. This is also an exercise.

The upshot of this section is that continuity and limits are essentially the same in both real and com-

plex analysis and behave as you should expect. Differentiation, however, is another beast entirely. . .

Loosely speaking, a connected set consists of one ‘lump.’ If you’ve studied topology: almost all domains in these notes

will be open, so there is no beneﬁt to distinguishing between connectedness and path-connectedness.

Exercises 2.2. 1. As in Example 2.6, prove that the doubly-punctured plane D := C \ {1, −2i} is an

open set. State a function whose implied domain is this set.

2. Use the ϵ–δ deﬁnition (2.10) to prove the following.

(a) lim

z→z

z = z

(b) lim

z→0

= 0 (c) lim

z→2

z − i

2 −i

(d) lim

z→z

= z

3. Show that f (z) = (z/z)

equals 1 at all non-zero points on the real and imaginary axes, and −1

at all non-zero points on the line y = x. Explain why lim

z→0

f (z) doesn’t exist.

4. Prove part 3(c) of Theorem 2.13.

5. Suppose lim

z→z

f (z) = w

. Prove that lim

z→z

f (z)

6. Use Deﬁnition 2.16 to prove part of Theorem 2.18: lim

z→z

f (z) = ∞ =⇒ lim

z→z

f (z)

= 0.

7. Use Deﬁnition 2.16 to prove: (a) lim

z→2i

iz−1

z−2i

= ∞, (b) lim

z→∞

iz−1

z−2i

= i

8. (a) Show that f (z) =

5iz+1

3z−2i

deﬁnes a bijection of the Riemann sphere f : C → C.

(Hint: let w = f (z) and solve for z . . .)

(b) In general: Given α, β, γ, δ ∈ C, prove that f (z) =

αz+β

γz+δ

deﬁnes a bijection of the Riemann

sphere if and only if αδ − βγ = 0. How does this relate to the matrix



α β

γ δ



9. We complete the proof of Theorem 2.21. Suppose f : K ⊆ C → C is continuous.

(a) Let K be compact. Suppose (w

) ⊆ K is a sequence where



f (w

)



is convergent in

C. Explain why there exists a convergent subsequence (w

), and use it to show that

lim f (w

) ∈ f (K). Hence conclude that f (K) is closed.

(b) Suppose K is path-connected. If f (p), f (q) ∈ f (K), show that ∃w : [0, 1] → f (K) continu-

ous such that w(0) = f (p) and w(1) = f (q). Hence conclude that f (K) is path-connected.

10. (Hard) “Every open cover has a ﬁnite subcover,” is a crucial result in topology:

Theorem Suppose a compact K is a subset of a (possibly inﬁnite) union

of open

sets. Then there are ﬁnitely many U

(labelled WLOG) such that K ⊂ U

∪··· ∪U

Suppose z : [0, 1] → D ⊆ C is a path in an open domain D and deﬁne K = range z.

(a) Explain why K is compact.

(b) Prove that K may be covered by ﬁnitely many closed balls B

for which K ⊂ B

∪··· ∪ B

⊂ B

∪··· ∪ B

⊂ D.

(Hint: start by centering a ball at every point of K)

zontal/vertical segments) from z(0) to z(1).

z(0)

z(1)

(d) Show that we can ﬁt a ‘tube’ around K: ∃δ > 0 such that ∀z ∈ K,

z − w

≤ δ =⇒ w ∈ D.

(Hint: following part (b), let V =

and deﬁne δ := inf



z − v

: z ∈ K, v ∈ V



)

This is often taken as the deﬁnition of compactness in topology. Its equivalence to K being closed and bounded in C

(or any Euclidean space) is the famous Heine–Borel Theorem.

2.3 Derivatives & the Cauchy–Riemann Equations

Differentiation is where complex analysis starts to produce interesting and novel results. It won’t

appear so initially, since the deﬁnition of derivative is exactly as you’ve previously encountered.

Deﬁnition 2.22. Let f : D → C be a complex function and z

an interior point of D. We say that f is

differentiable at z

if the following limit exists:

lim

z→z

f (z) − f (z

)

z − z

We call this limit the derivative of f at z

and denote it f

′

( z

). If f is differentiable everywhere

on D

then the derivative is itself a function, written f

′

( z) or

d f

Example 2.23. The function f (z) = z

is everywhere differentiable,

lim

z→z

f (z) − f (z

)

z − z

= lim

z→z

−z

z − z

= lim

z→z

( z −z

)(z + z

)

z − z

= lim

z→z

( z + z

) = 2z

The second last equality follows since we only compute on the punctured disk 0 <

z − z

< δ when

taking limits (Deﬁnition 2.16).

The basic rules of differentiation are identical those in real analysis and can be proved similarly.

Theorem 2.24. Suppose f and g are differentiable (either at a point z

or as functions).

1. (Linearity) For any constants a, b ∈ C,



a f (z) + bg(z)



= a f

′

( z) + bg

′

( z)

2. (Power Law) For any n ∈ N

= nz

n−1

3. (Product rule)



f (z)g(z)



= f

′

( z)g(z) + f (z)g

′

( z)

4. (Quotient Rule) If g(z) = 0, then

f (z)

g(z)

′

( z)g(z) − f (z)g

′

( z)

[g(z)]

5. (Chain Rule) If h is differentiable at g(z

), then h ◦ g is differentiable at z

and

(h ◦ g)

′

( z

) = h

′



g(z

)



′

( z

)

We immediately see that all polynomials and rational functions are differentiable. Familiar examples

behave the same regardless of whether we are in R or C!

Example 2.25.

3(z

−2)

+ z

+ 1

[30z(z

−2)

+ 2z)](z

+ 1) −3z

[3(z

−2)

+ z

]

( z

+ 1)

Such holomorphic functions are the main topic of the course: we’ll consider them more properly in the next section.

Necessarily, D must be an open set if f is to be holomorphic (think about the deﬁnition of the limit!).

The Cauchy–Riemann Equations

Differentiation thus far seems uncontroversial; here is an example that might change your mind. . .

Example 2.26. Consider f (z) = z at a generic point z

= x

+ iy

, and the difference quotient

f (z) − f (z

)

z − z

x −iy − x

+ iy

x + iy − x

−iy

∆x −i∆y

∆x + i∆y

(where ∆x = x −x

and ∆y = y −y

)

For f

′

( z) to exist, we must obtain the same limit regardless of how

( ∆x, ∆y) → (0, 0) (equivalent to z → z

). There are two obvious ways

to take the limit:

Horizontally

f (z) − f (z

)

z − z

∆x

= 1

Vertically

f (z) − f (z

)

z − z

−i∆y

i∆y

= −1

∆y = 0

∆x = 0

The quotient plainly takes different values depending on how (∆x, ∆y) → (0, 0). We conclude that

′

( z

) does not exist, and that the function f (z) = z is nowhere differentiable!

The approach extends to general functions. We attempt to differentiate f (z) = u(x, y) + iv(x, y)

(written in real and imaginary parts) at a generic point z

= x

+ iy

lim

z→z

f (z) − f (z

)

z − z

= lim

(∆x,∆y)→(0,0)

u(x, y) − u(x

, y

) + i(v(x, y) −v(x

, y

)

∆x + i∆y

If this limit exists, the same result must be obtained when evaluating along paths approaching z

horizontally and vertically. This results in a key relationship between the partial derivatives of u, v.

Horizontally We have ∆y = 0 (y = y

), so the limit becomes

lim

∆x→0

u(x, y

) − u(x

, y

) + i(v(x, y

) − v(x

, y

)

∆x



∂u

∂x

+ i

∂v

∂x



, y

)

Vertically We have ∆x = 0 (x = x

), from which

lim

∆y→0

u(x

, y) −u(x

, y

) + i(v(x

, y) −v(x

, y

)

i∆y



∂u

∂y

+ i

∂v

∂y



, y

)

Theorem 2.27 (Cauchy–Riemann equations). Write f (z) = u + iv in its real and imaginary parts.

1. If f is complex-differentiable at z

, then u, v satisfy the Cauchy–Riemann equations at z

∂u

∂x

∂v

∂y

and

∂u

∂y

= −

∂v

∂x

(equivalently u

= v

and u

= −v

)

2. Conversely, if u, v have partial derivatives on a neighborhood of z

, which are continuous and

satisfy the Cauchy–Riemann equations at z

, then f is complex-differentiable at z

In either situation, we may write f

′

( z) = u

+ iv

= v

−iu

We proved part 1 above. For a sketch of the almost-converse (part 2), see Exercise 11.

Examples 2.28. 1. (Example 2.26) f (z) = z = x −iy has u(x, y) = x and v(x, y) = −y, whence

= 1 = −1 = v

and u

= 0 = v

Since u, v do not satisfy the Cauchy–Riemann equations, f fails to be differentiable anywhere.

2. f (z) =

+ y

has u =

+ y

and v = 0. Away from z

= 0, the Cauchy–Riemann

equations are

+ y

= 0 = v

, u

+ y

= 0 = −v

These equations are satisﬁed nowhere (u, v are not differentiable at z

= 0), whence f (z) is

nowhere differentiable.

3. f (z) = z

= x

−y

+ 2ixy has u(x, y) = x

−y

and v(x, y) = 2xy. We check

= 2x = v

, u

= −2y = −v

As expected, u, v satisfy the Cauchy–Riemann equations. Moreover,

′

( z) = 2z = 2x + 2iy = u

+ iv

= v

−iu

4. Consider f ( z) =

2 +

2 + x

+ y

2 + x

+ y

. We compute

2 − x

+ y

(2 + x

+ y

)

2 + x

−y

(2 + x

+ y

)

−2xy

(2 + x

+ y

)

−v

2xy

(2 + x

+ y

)

The Cauchy–Riemann equations are satisﬁed if and only if xy = 0 = x

− y

, which is if and

only if x = y = 0. We conclude that f is not differentiable at any non-zero z ∈ C.

Since the partial derivatives of u, v are continuous, the converse of the Cauchy–Riemann theo-

rem says that f is differentiable at z = 0: indeed

′

(0) = u

(0, 0) + iv

(0, 0) =

We can alternatively check that this straight from the deﬁnition of derivative:

′

(0) = lim

z→0

f (z) − f (0)

z − 0

= lim

z→0

2 +

The function f is therefore differentiable at precisely one point.

5. The complex exponential function is everywhere differentiable: indeed

f (z) = e

= e

cos y + ie

sin y

satisﬁes

= e

cos y = v

, u

= −e

sin x = −v

where these are certainly continuous on C. As expected,

′

( z) = u

+ iv

= e

cos y + ie

sin y = e

Exercises 2.3. 1. Use Theorem 2.24 to ﬁnd the derivatives of the following functions:

(a) f (z) =

+ 2z

(b) f (z) = (z

+ 2iz + 1)

(3z

−i)

(iz

+ 4)

2. Use the limit deﬁnition of the derivative to compute the derivative of the functions:

(a) f (z) = 3z

−iz

(b) f (z) =

3. Give a proof of the quotient rule, directly using the deﬁnition of the derivative.

4. Use the quotient rule to prove the power law for negative integer exponents:

∀n ∈ N,

−n

= −nz

−n−1

5. (L’Hˆopital’s rule) Suppose f (z

) = g(z

) = 0, and that f

′

( z

) and g

′

( z

) = 0 both exist. Use the

deﬁnition of the derivative to prove that

lim

z→z

f (z)

g(z)

′

( z

)

′

( z

)

6. Prove that f (z) = Re z and g(z) = Im z are not differentiable anywhere.

7. By writing z = re

iθ

in polar form, verify directly that f (z) =

+ y

is not differentiable

at z

= 0.

8. What, if anything, do the Cauchy–Riemann equations allow you to conclude for the following?

(a) f (z) = (z + i)

(b) f (z) =

z − i

−

(d) f (z) =



+ z



9. As in Example 2.28.5, prove that

= ke

for any complex constant k = a + ib.

10. Write a complex function f (z) = f (z, z) as a function of z and z. For example,

f (z) =

= zz

Noting that x =

( z + z) and y =

( z −z), use the chain rule

∂ f

∂z

∂ f

∂x

∂z

∂ f

∂y

∂z

to prove that f satisﬁes the Cauchy–Riemann equations if and only if

∂ f

∂z

= 0.

Hence give a quick proof that f (z) = zz

is not differentiable when z = 0.

11. Suppose f (z) = u + iv satisﬁes the Cauchy–Riemann equations at z

= x

+ iy

. Use the multi-

variable linear approximation

f (z) ≈ f (z

) + f

, y

) ∆x + f

, y

) ∆y

to prove that

f (z)−f (z

)

z−z

≈ u

, y

) + iv

, y

) whenever z = z

(The existence/continuity assumptions in Theorem 2.27, part 2, guarantee that this approxima-

tion becomes equality in the limit, though a rigorous proof requires more work)

2.4 Holomorphic and Harmonic Functions

We tend to be most interested in functions which are everywhere differentiable.

Deﬁnition 2.29. We say that f : D → C is:

• Holomorphic (or analytic

) on D if it is differentiable at all points z

∈ D (necessarily D is open!).

• Holomorphic at z

∈ D if it is differentiable (holomorphic) on some neighborhood of z

• Entire if it is holomorphic with domain D = C.

Examples 2.30. 1. The exponential function f (z) = e

is entire, as is every polynomial.

2. The rational function f (z) =

is holomorphic on its implied domain C \ {±2i}. Indeed, by

the quotient rule,

′

( z) =

i(z

+ 4) −2iz

( z

+ 4)

i(4 − z

)

( z

+ 4)

Our ﬁrst general result should seem very familiar.

Theorem 2.31. If f

′

( z) = 0 on an open, (path-)connected domain, then f (z) is constant.

The calculation in the proof should be compared to the corresponding argument from real analysis

which also relies on the mean value theorem.

Proof. Given p, q ∈ D = dom( f ), join them with a zig-zag path consisting of ﬁnitely many horizon-

tal/vertical segments (Exercise 2.2.10).

On each horizontal segment (x

≤ x ≤ x

, y constant), apply

the mean value theorem to the single-valued functions

x 7→ u(x, y) and x 7→ v(x, y)

We therefore obtain values

x, both in the interval (x

, x

which satisfy

u(x

, y) −u(x

, y)

− x

= u

(

x, y) and

v(x

, y) −v(x

, y)

− x

= v

(

x, y)

By assumption f

′

( z) = u

+ iv

= 0, whence both partial derivatives are everywhere zero. It follows

that f (z) = u + iv takes the same value at both endpoints of any horizontal segment.

The same holds along at the endpoints of any vertical segment; this time we utilize u

= v

= 0.

By stepping along the segments in this fashion, we conclude that f (q) = f (p). Since p, q were

arbitrary, f is constant on D.

We’ll use holomorphic/analytic interchangeably though the formal deﬁnition of analytic depends on power series. A

major part of the course involves showing that these deﬁnitions are equivalent.

Corollary 2.32. If f (z) is holomorphic with constant modulus k =

f (z)

, then f (z) is constant.

This is essentially trivial in the real case—think about why! The complex case needs a proof.

Proof. Plainly k

f (z)

= f (z) f (z) is constant. If k = 0, we are done. Otherwise, f (z) =

f (z)

holomorphic (quotient rule!). Write f (z) = u + iv, whence f (z) = u −iv, and consider the Cauchy–

Riemann equations for both:

= v

, u

= −v

, u

= −v

, u

= v

All partial derivatives are therefore zero: f

′

( z) = 0 and so f is constant.

The next result is worth stating now, since it provides a signiﬁcant contrast with the real case. We’ll

prove it later once we’ve developed contour integration.

Theorem 2.33. If f (z) = u + iv is holomorphic, then f is inﬁnitely differentiable. Otherwise said:

• f

(n)

( z) exists and is continuous for all n ∈ N.

• u and v have continuous partial derivatives of all orders.

In real analysis, functions which are everywhere differentiable need not even be twice differentiable,

let alone inﬁnitely so (Exercise 4).

Harmonic Functions The second partial derivatives of a holomorphic function satisfy a famous par-

tial differential equation:

∂

∂x

CR1

∂

∂x

= v

(∗)

= v

∂

∂y

CR2

= −

∂

∂y

= −u

Equality of the mixed partial derivatives (∗) follows because all derivatives are continuous (Theorem

2.33 and Clairaut’s Theorem). The same equation holds for v. We conclude:

Corollary 2.34. If f = u + iv is holomorphic, then u and v are harmonic functions; solutions to

Laplace’s equation

∇

u = u

+ u

= 0

Laplace’s equation is one of the most widely applied PDEs in mathematics and physics.

Example 2.35. f (z) =

x−iy

is holomorphic on C \{0}; its real and imaginary parts are therefore

harmonic away from the origin. Indeed,

+ u



∂

∂x

∂

∂y



+ y

∂

∂x

− x

+ y

)

−

∂

∂y

2xy

+ y

)

−2x(x

+ y

) − 4x(y

− x

)

+ y

)

−

2x( x

+ y

) − 8xy

+ y

)

= 0

Analytic Continuations We ﬁnish this introduction to holomorphic functions with another surpris-

ing result whose proof will have to wait (this time for the theory of residues).

Theorem 2.36. Suppose f and g are holomorphic functions on

an open connected domain E and assume that f (z) = g(z) on

some path contained in E.

Then f (z) = g(z) throughout E.

f = g =⇒ f = g

This is highly counter-intuitive; you need only know the values of a holomorphic function on a tiny

path to know the full function on its whole (connected) domain! This leads to a new concept.

Deﬁnition 2.37. Let D ⊆ E be open connected domains and

g : E → C holomorphic. Let f : D → C be the restriction of g

to D; that is f (z) = g(z) on D.

We call g the analytic continuation of f to E.

By Theorem 2.36, the analytic continuation of f to E is unique. There are, however, some subtleties,

which we explore a little in the next example:

• Given f analytic on D, an analytic continuation to E ⊇ D is not guaranteed to exist.

• The choice of extended domain E really matters.

Example 2.38. Let f (z) =

√

z =

√

iθ

be the principal square root whose domain D is the ﬁrst

quadrant. Exercise 10 veriﬁes that f is holomorphic on D.

The pictures describe two analytic continuations of f : in both cases the point w = e

−

3πi

= e

5πi

is used

for comparison.

1. Let G be the plane omitting the non-positive real axis

and deﬁne

g : G → C : z 7→

√

iθ

, θ = Arg z ∈ ( −π, π)

The codomain of g is the right half-plane. Observe that

g(w) = e

−

3πi

lies in the fourth quadrant.

2. Let H be the plane omitting the non-positive imaginary

axis and deﬁne

h : H → C : z 7→

√

iθ

, θ = arg z ∈ ( −

3π

)

The codomain of h is the upper-right half-plane. More-

over h(w) = e

5πi

= −g(w) lies in the second quadrant.

−

3π

G D

−

3π

g(w)

g(D)

5π

h(w)

h(D)

The two analytic continuations of f disagree on the intersection of their domains!

In fact the omissions chosen for G, H are necessary: there is no analytic continuation of f to the

punctured plane C \ {0}, or indeed to any domain in which it is possible to loop completely around

the origin. We shall return to this topic in Chapter 3. . .

Exercises 2.4. 1. Suppose g

′

( z) = h

′

( z) on an open connected domain D. Prove that h(z) = g(z) + c

for some constant c ∈ C.

Equivalently: if g(z) and h(z) are anti-derivatives of f ( z), then g(z) − h(z) is constant

2. Verify that u = e

cos ny and v = e

sin ny are harmonic functions for any n ∈ Z.

3. Prove or disprove: if u, v : D → R are harmonic functions on an open set D ⊆ R

, then

f (z) := u(x, y) + iv(x, y) is holomorphic.

4. Deﬁne f : R → R by

f (x) = x

(

if x ≥ 0

−x

if x < 0

Show that f is differentiable, but not twice so (an impossible property for complex functions).

5. Suppose f ( z) = u + iv and write z = x + iy = re

iθ

in polar form.

(a) Use the chain rule applied to the polar co-ordinate relations x = r cos θ, y = r sin θ to

compute the partial derivatives u

, u

, v

in terms of u

, u

, v

(b) Deduce the polar form of the Cauchy–Riemann equations:

= v

= −rv

, f

′

( z) = e

−iθ

( u

+ iv

) =

−i

( u

+ iv

)

6. Prove the polar form of Laplace’s equation:

+ ru

+ u

θθ

= 0

7. Show that u = r

cos nθ is a harmonic function for any n ∈ N: ﬁnd two ways to show that this

is true!

8. Suppose f : C → C is an entire function such that f (iy) = −iy

whenever z = iy lies on the

imaginary axis. What is the value f (2)? Explain your answer.

(Hint: Think about analytic continuations)

9. The function f (z) =

is holomorphic on C \{0}. Explain why there is no analytic continuation

of f that is holomorphic at z = 0.

10. (a) Use Exercise 5 to prove that f (z) =

√

z =

√

iθ

is holomorphic on the ﬁrst quadrant and

ﬁnd f

′

( z). Moreover, explain why the functions g, h in Example 2.38 are holomorphic and

therefore both analytic continuations of f .

(b) Prove that there exists no analytic continuation of g to any set larger than G.

(Hint: suppose the extended domain contains −r ∈ R

−

. Now use the fact that an analytic contin-

uation must be continuous at −r . . .)

3 Elementary Functions

We’ve already considered polynomials, rational functions and, to some extent, n

roots and the

exponential. We now develop the logarithmic and trigonometric functions.

3.1 Exponential and Logarithmic Functions

The exponential function exp : C → C : z 7→ e

was deﬁned earlier using Euler’s formula

exp(z) = e

:= e

cos y + ie

sin y (∗)

For reference, we collect some basic properties: parts 1–3 are identical to real analysis.

Lemma 3.1. Throughout let z, w ∈ C.

1. The exponential function is entire, with derivative

= e

2. e

= 0

3. (Exponential laws) e

z+w

= e

, e

z−w

and (e

)

= e

(whenever n ∈ Z).

4. e

is periodic with period 2πi. Moreover,

= e

⇐⇒ z −w = 2πin for some n ∈ Z

Sketch Proof. 1. This is Example 2.28.5: check Cauchy–Riemann and compute

= u

+ iv

2. This follows trivially from (∗): e

> 0 while cos y and sin y are never both zero.

3. Apply the multiple-angle formulæ for cosine and sine. For the third law, induct with z = w.

4. Certainly e

w+2πin

= e

by the periodicity of sine and cosine. Now suppose e

= e

where

z = x + iy and w = u + iv. By considering the modulus and argument, we see that

= e

=⇒

(

= e

y = v + 2πin for some n ∈ Z

We conclude that x = u and so z −w = i(y − v) = 2πin.

Example 3.2. Find all z ∈ C such that e

= 5(−1 + i).

Following part 4 of the Lemma, write z = x + iy and take the polar form of 5(−1 + i):

= 5(−1 + i) ⇐⇒ e

= 5

√

3πi

= e

ln(5

√

3πi

⇐⇒

(

x = ln(5

√

y =

3π

+ 2πn for some n ∈ Z

⇐⇒ z = ln(5

√

2) +



3π

+ 2πn



i for some n ∈ Z

We see that there are inﬁnitely many suitable z!

Duplicate Notation Warning! When n ∈ N, the expression e

could mean two things. For instance

might mean either:

1. The set of cube roots of e, namely {

√

2πi

√

−

2πi

};

2. The real value

√

e ∈ R

Since e

is such a common function, we default to the second meaning: if you mean the set of n

roots, say so! Remember you can always write exp(z) for the function if necessary.

The periodicity of the exponential leads to the more subtle notion of the complex logarithm.

Deﬁnition 3.3. Let z = re

iθ

be a non-zero complex number with principal argument θ = Arg z. The

principal logarithm of z is the value

Log z := ln r + iθ = ln

+ i Arg z

where ln is the usual (real!) natural logarithm. The logarithm of z is any (and all) of the values

log z = ln

+ i arg z = ln r + i(θ + 2πn) : n ∈ Z

Examples 3.4. 1. Since −4 = 4e

πi

, we see that

Log(−4) = ln 4 + πi and log(−4) = ln 4 + (1 + 2n)πi

2. Again write in polar form to compute:

Log(

√

3 −i) = Log(2e

−

πi

) = ln 2 −

πi

and log(

√

3 −i) = ln 2 −

πi

+ 2πni

These examples involve solving equations of the form e

= z: writing z = re

iθ

= e

ln r+iθ

as above,

and appealing to part 4 of Lemma 3.1, we instantly see that

= z ⇐⇒ w = log z

Read this carefully, remembering that the logarithm is multi-valued and the exponential periodic:

log z

= z and log e

= w + 2πni where n ∈ Z

For reference, we summarize some of the basic properties of the principal logarithm function. All

parts should be clear from Deﬁnition 3.3.

Lemma 3.5. Throughout, z and w are complex numbers with z = 0, and n ∈ Z.

• Log : C \ {0} → {w ∈ C : −π < Im w ≤ π} is a bijection with inverse exp.

• Log e

= w + 2πni where n ∈ Z is chosen such that Im(Log e

) = Im w + 2πn ∈ (−π, π].

• If z ∈ R

, then Log z = ln z is the usual natural logarithm.

This is similar to how arg z means either the set {Arg z + 2πni} or some particular value from this set, dependent on

context. We’ll more formally discuss such multi-valued functions in Section 3.2.

The Logarithm Laws Just as the standard rules for exponentiation (Lemma 3.1 part 3) apply to the

complex exponential, the log laws also translate. However, the multi-valued nature of the logarithm

adds some subtlety.

Suppose non-zero z, w are given: since

and arg zw = arg z + arg w, we conclude that

log zw = ln(

) + i(arg z + arg w)

= ln

+ i arg z + ln

+ i arg w

= log z + log w

Be very careful, for this expression is not an identity of functions. What it really means is that the

following two sets are identical:

log z + log w =



α + β : α ∈ log z, β ∈ log w





+ i Arg z + 2πki +

+ i Arg w + 2πmi : k, m ∈ Z



log zw =



+ i Arg(zw) + 2πni : n ∈ Z



Unless you are sure you won’t make a mistake, it is safer to write

log zw = log z + log w + 2πni for some n ∈ Z

Given its restricted range, we can be more precise for the principal logarithm:

Log zw = Log z + Log w + 2πni for some n = 0, −1, 1

Example 3.6. Let z = −

√

3 + i = 2e

5πi

and w =

√

2( 1 + i) = 2e

πi

. Then, for some k, m, n,

log z = ln 2 +

5πi

+ 2πki, log w = ln 2 +

πi

+ 2πmi

log zw = log(4e

5πi

πi

) = log(4e

13πi

) = ln 4 +

13πi

+ 2πni

We can choose particular logarithms satisfying log zw = log z + log w provided n = k + m. For

principal logarithms, we don’t get to make a choice:

Log z = ln 2 +

5πi

, Log w = ln 2 +

πi

Log zw = Log(4e

−

11π

) = Log(4e

−

11π

) = ln 4 −

11πi

= ln 4 +

5πi

πi

−2πi = Log z + Log w − 2πi

We can similarly demonstrate the second log law, with exactly the same caveat:

log

= log z −log w

As before, principal logarithms might require a correction term of ±πi.

We can try to translate the ﬁnal log law (log z

= n log z, n ∈ N), though even more care is needed!

Example 3.7. Let z = −

√

3 + i = 2e

5πi

and compute what is meant by the set 2 log z:

2 log z =





ln 2 +

5πi

+ 2πmi



: m ∈ Z





ln 4 +

5πi

+ 4πmi : m ∈ Z



This has only half the terms of the set

log z



log( 4e

10πi

) = ln 4 +

5πi

+ 2πki : k ∈ Z



It is therefore safer to state that log z

= 2 log z.

Since the principal logarithm is a function rather than a set, we can be more precise: for any n ∈ N,

Log z

= n Log z + 2πki for some integer k with

≤

Example 3.8. Let z = e

−

13πi

and consider z

. We see that

Log z

= Log e

−13πi

= Log e

πi

= iπ, 16 Log z = −13πi =⇒ Log z

= 16 Log z + 14πi

In this case

= 7 ≤

Exercises 3.1. 1. Compute (be careful with (c)!):

(a) exp(3 −

i) (b) Log(ie) (c) log(3 −4i) (d) Log[(−1 + i)

]

2. (a) If e

is real, show that Im z = nπ for some integer n.

(b) If e

is imaginary, what restriction is placed on z?

3. Show in two ways that the function f ( z) = exp(z

) is entire, and ﬁnd its derivative.

4. Prove, for any z ∈ C, that



exp(z

)



≤ exp

. What must z satisfy if this is to be equality?

5. Find Re e

in terms of x and y. Why is Re e

harmonic on any domain not containing the origin?

6. Show that Log i

= 3 Log i.

7. Show that Re(log(z −1)) =

ln[(x −1)

+ y

] whenever z = 1.

8. Prove the above boxed formula for Log z

9. The square roots of i are

√

i = e

πi

and −

√

i = e

−

3πi

(a) Compute Log

√

i and Log(−

√

i) and check that Log

√

i =

Log i.

(b) Show that the set of all logarithms of all square roots of i is

log i



n +



πi where n ∈ Z

Hence deduce that log i

log i as sets.

3.2 Multi-valued Functions, Branch Cuts and the Power Function

Recall that each log z represents a set of complex numbers; as such, the complex logarithm is termed

a multi-valued function. We have previously encountered others of this ilk:

• The argument of a complex number is any of the values arg z = Arg z + 2πn where n ∈ Z. The

complex logarithm is merely a modiﬁcation of this: log z = ln

+ i arg z.

• The n

root of z is the set of values z

= {

√

zω

: k = 0, . . . , n −1} where ω

= e

2πi

is an n

root of unity and

√

z the principal n

root.

It is an abuse of language to refer to a multi-valued function, since any function should assign exactly

one output to each element of its domain. While this problem can be ﬁxed using equivalence classes,

another approach is simpler to visualize.

Deﬁnition 3.9. A branch of a multi-valued function f is a single-valued function F on a domain D

which is holomorphic (differentiable) on D and such that each F(z) is one of the values of f ( z).

Let D = C \ ℓ where ℓ is a line or curve in C. If F : D → C is a branch of f , we call ℓ a branch cut. A

branch point is any point common to all possible branch cuts.

Branches of the Logarithm The principal branch of the logarithm is a

slightly restricted version of the principal logarithm

Log z = ln r + iθ where θ = Arg z ∈ (−π, π)

The branch cut is the non-positive real axis. To check holomorphicity,

we verify the Cauchy–Riemann equations:

= r

∂

∂r

ln r = 1 =

∂

∂θ

θ = v

, u

= 0 = −rv

The partial derivatives are certainly continuous, whence log z is holo-

morphic with derivative

log z = e

−iθ

( u

+ iv

) =

−iθ

More generally, for any angle α we could take a branch cut ℓ to be the

line with argument α, which deﬁnes a new branch of the logarithm:

log z = ln r + iθ where θ ∈ (α −2π, α)

In this description, the principal branch corresponds to α = π. Note

that choosing α = −π results in the same branch cut, but a different

branch:

Principal Branch

α-Branch

log z = ln r + iθ where θ ∈ (−3π, −π)

We use the polar form: ru

= v

, rv

= −u

, f

′

(z) = e

−iθ

+ iv

) (see Exercise 2.4.5).

More esoteric branch cuts are possible, such as the pictured ‘squiggle.’

At issue is the fact that traversing a counter-clockwise loop around the

origin increases the value of a logarithm by 2πi. It is therefore impossi-

ble for a branch to be continuous (let along holomorphic) on any domain

containing such a loop; to make log z single-valued, a branch cut must

‘cut’ any such path, and thus connect the two branch points 0 and ∞.

Clarity is crucial here: when you write log z, do you mean a set, a partic-

ular element of that set, or a branch? Certain expressions may be true or

false depending on the meaning.

Example 3.10. Consider z =

√

(1 + i) = e

πi

. For the principal branch, we have

Log z

= Log e

πi

= 2 Log z

For the branch with α =

(that is, arg z ∈ (

−5π

)), we have

= e

πi

= e

−

3πi

=⇒ log z

= −

3πi

= 2 log z

Recall also (Example 3.7), that as sets, log z

= 2 log z.

For individual branches of the logarithm, speciﬁc versions of the logarithm laws (page 30) are avail-

able, though they are not worth trying to remember; just carefully think out the possibilities when

calculating.

General Exponential Functions The logarithm can be used to create exponential functions for any

non-zero complex base c. Choose a value log c and deﬁne

:= e

z log c

Provided c is not a non-positive real number, the standard is to use the principal logarithm. Regard-

less of your choice, log c is constant and the exponential function is holomorphic everywhere:

= c

log c

Example 3.11. Let c = i = e

πi

and use the principal logarithm to deﬁne

:= e

z Log i

= exp



πiz



= exp



−

y + i



= e

−

πy

cos

x + i sin

It is simple to check the Cauchy–Riemann equations and see that i

is entire (holomorphic on C).

If we instead took a different branch of the logarithm with arg i = −

3πi

, then

= exp



−3πiz



= e

3πy



cos

3π

x −i sin

3π



This is still entire, though it is a completely different function! Note that both choices of i

agree

whenever z is an integer, and for z =

they produce the two distinct square roots of i!

Power Functions Following a similar approach, for any non-zero z and complex number c we may

deﬁne the (typically multi-valued) function

:= e

c log z

In this case, restricting to the principal branch of the logarithm gives an unambiguous function.

Deﬁnition 3.12. The principal value of z

is the function

P. V. z

:= e

c Log z

whose domain is that of the logarithm (C excluding the non-positive real axis).

Example 3.13. Using the principal branch of the logarithm (Θ = Arg z), we obtain

P. V. z

= exp



(ln r + iΘ)



= exp



√

r +

iΘ



√

iΘ

√

precisely the principal cube-root of z as deﬁned previously.

If we chose a different branch with arg z = θ, then

= e

log z

= exp



(ln r + iθ)



√

r exp



( Θ + (θ − Θ))



√

i(θ−Θ)

Since, for any z, the difference in the arguments θ − Θ = 2πn is a multiple of 2π, this expression

really does return one of the cube-roots of z.

Lemma 3.14. Choose a branch of the logarithm so that z

= e

c log z

is single-valued. Then z

holomorphic on the same domain as the logarithm; moreover

= cz

c−1

Proof. Since log z is holomorphic, simply use the chain rule:

c log z

= e

c log z

( c log z) = e

c log z

= ce

c log z

−log z

= ce

(c−1) log z

= cz

c−1

Example 3.15. If the principal branch of the logarithm is used, then

( zw)

= exp(c Log(zw)) = exp(c Log z + c Log w + 2πcni) = z

2πcni

for some n ∈ {0, ±1}. We do not expect simple exponent rules such as (ab)

= a

to hold in complex

analysis! Note, however, that this does work in the case where c is an integer.

As an example, again using principal values, if z = w = e

3πi

, then zw = e

3πi

= e

−

πi

, whence

P. V.(zw)

= exp



5i ·

−πi



= e

5π

P. V. z

= P. V. w

= exp



5i ·

3πi



= e

−

15π

=⇒ (zw)

= e

5π

= e

−

15π

10π

= z

2π·5i·ni

with n = −1

Exercises 3.2. 1. Show that the function f (z) = Log(z −i) is holomorphic everywhere except on the

portion x ≤ 0 of the line y = 1.

2. Show that the function f (z) =

Log(z + 4) is holomorphic everywhere except at the points

√

2(1−i)

and on the portion x ≤ −4 of the real axis.

3. Show that the set z

as deﬁned earlier in the course coincides with the set z

:= exp



log z



deﬁned in this section.

4. Show that (1 + i)

= exp



−

+ 2nπ



exp



ln 2



where n ∈ Z.

5. Find the principal values of the following:

(a) i

(b) (1 −i)

√

3 + i)

1+4πi

6. Suppose c, c

, c

and z are complex numbers where z = 0. If all the powers involved are princi-

pal values, show that,

(a) z

= z

(b) (z

)

= z

for any n ∈ N.

7. The power function z

is usually multi-valued. However, if c = m is an integer, prove that z

single-valued: i.e. it is independent of the branch of logarithm used in its deﬁnition.

8. Check the claim at the bottom of Example 3.11: if m ∈ Z, then i

is the same value for the two

deﬁnitions of i

9. Continuing the previous question, suppose c = 0 and deﬁne c

= e

z log c

where any choice of

the branch of the logarithm is made.

(a) Let m ∈ Z. Prove that c

produces the same value, regardless of the branch of logarithm

used to deﬁne log c.

(b) If z =

, show that c

really is an m

root of c. If the principal branch of the logarithm is

used, show that c

is the principal m

root of c. For every m

root of c, show that there

exists a branch of the logarithm for which c

equals the given m

root.

3.3 Trigonometric and Inverse Trigonometric Functions

A sensible deﬁnition of the basic trigonometric functions comes simply by modifying Euler’s for-

mula. For instance, if y ∈ R, then

+ e

−iy

= cos y + i sin y + cos y −i sin y = 2 cos y

This motivates the primary deﬁnition.

Deﬁnition 3.16. For any z ∈ C we deﬁne

cos z =

+ e

−iz

sin z =

−e

−iz

Example 3.17. cos(

+ i) =



iπ

−1

+ e

−iπ





−1

√

(1 + i) +

√

(1 − i)



e+e

−1

√

−

e−e

−1

√

Theorem 3.18. Sine and cosine are entire with the same derivatives are their real counterparts

sin z = cos z

cos z = −sin z

The usual identities, including double and multiple-angle formulæ, are satisﬁed: for instance

sin

z + cos

z = 1, cos(z + w) = cos z cos w −sin z sin w, cos 2z = 2 cos

z − 1, etc.

In particular, sin z = cos(z −

) and cos z = sin(z +

). Sine and cosine are also 2π-periodic and

have exactly the same zeros as their real versions:

sin z = 0 ⇐⇒ z = nπ, cos z = 0 ⇐⇒ z =

+ nπ where n ∈ Z

The upshot of the Theorem is that sine and cosine behave exactly as you’d expect.

The proofs are straightforward applications of properties of the exponential function. For instance;

sin z =

−e

−iz

(ie

+ ie

−iz

) = cos z

and,

sin z = 0 ⇐⇒ e

= e

−iz

⇐⇒ e

2iz

= 1 ⇐⇒ e

−2y

(cos 2x + i sin 2x) = 1 ⇐⇒ z = πn

The remaining trigonometric functions are deﬁned in the expected way: e.g.,

tan z =

sin z

cos z

−e

−iz

i(e

+ e

−iz

)

whenever z =

+ nπ

All trigonometric functions are holomorphic wherever deﬁned and have the usual expressions for

their derivatives: e.g., by the quotient rule,

tan z =

sin z

cos z

cos

z + sin

cos

= sec

Aside: Hyperbolic Functions By considering real and imaginary parts,

sin z =

( e

ix−y

−e

−ix+y

) =

( e

−y

cos x + ie

−y

sin x −e

cos x + ie

sin x)

( e

+ e

−y

) sin x +

( e

−e

−y

) cos x = sin x cosh y + i cos x sinh y

cos z = cos x cosh y −i sin x sinh y

where cosh y =

( e

+ e

−y

) and sinh y =

( e

− e

−y

) are the usual (real) hyperbolic functions. We

could have written Example 3.17 this way

cos



+ i



e + e

−1

√

−i

e − e

−1

√

= cos

cosh 1 − i sin

sinh 1

Hyperbolic functions are a convenient short-cut, but never necessary; use or ignore as you like. All

their properties can be derived from their relationship to exponential and trigonometric functions:

cosh z =

+ e

−z

= cos iz, sinh z =

−e

−z

= −i sin iz

For instance

sinh z = −i

sin iz = −i

cos iz = cosh z and

cosh z = sinh z

cosh

z − sinh

z = cos

(iz) + sin

(iz) = 1

Inverse Trigonometric Functions As with logarithms, inverting trigonometric functions results in

multi-valued functions.

Example 3.19. We ﬁnd an expression for cos

−1

z and compute its derivative.

w = cos

−1

z ⇐⇒ z = cos w ⇐⇒ 2z = e

+ e

−iw

⇐⇒ (e

)

−2ze

+ 1 = 0

which is quadratic in e

. Apply the quadratic formula to see that

2z + (4z

−4)

= z + i(1 − z

)

⇐⇒ cos

−1

z = w = −i log

z + i(1 − z

)

A branch of cos

−1

requires us to choose branches of both the square-root and the logarithm.

Inverse cosine is holomorphic since it is a composition of holomorphic functions. By the chain rule,

cos

−1

z =

−i

z + i(1 − z

)

z + i(1 − z

)

−i

z + i(1 − z

)

1 −

(1 − z

)

−i

z + i(1 − z

)

(1 − z

)

−iz

(1 − z

)

−1

(1 − z

)

If we ﬁx a branch of the square-root (and logarithm) so that cos

−1

z is single-valued, this is necessarily

the same branch that appears in the expression of the derivative.

Expressions such as these are not worth memorizing; they are not difﬁcult to derive when needed

using the method in Example 3.19. Here is the complete list for the three basic functions.

Theorem 3.20. The inverse sine, cosine and tangent functions are given by the expressions

sin

−1

z = −i log

iz + (1 − z

)

cos

−1

z = −i log

z + i(1 − z

)

tan

−1

z =

log

i + z

i −z

Once branches of the square-root (sine/cosine only) and logarithm are chosen, these are holomorphic

on their domains and have familiar derivatives:

sin

−1

z =

(1 − z

)

cos

−1

z =

−1

(1 − z

)

tan

−1

z =

1 + z

The branches of the square-root in the derivatives of inverse sine and cosine are identical to those

used in the deﬁnitions of the original functions.

Examples 3.21. 1. To evaluate sin

−1

√

as a complex number, we could follow the approach of

Example 3.19 (solve

−e

−iz

√

directly), or use the formula in Theorem 3.20:

sin

−1

√

= −i log

√

1 −

= −i log

i ±1

√

Now evaluate the logarithms separately:

−i log

i + 1

√

= −i log e

πi

= −i



πi

+ 2πni



+ 2πn

−i log

i −1

√

= −i log e

3πi

= −i



3πi

+ 2πni



3π

+ 2πn

The set of values sin

−1

√

generated by all branches of the square-root and logarithm is pre-

cisely the set we’d have found working entirely within R!

2. We can also evaluate inverse sines that would have no meaning in R. For instance,

sin

−1

7 = −i log[7i ±

√

−48] = −i log(7 ±4

√

3)i = −i log(7 ±4

√

3)e

πi

= −i



ln( 7 ±4

√

3) +

πi

+ 2πni



= −i ln(7 ± 4

√

3) +

+ 2πn

Note that 7 > 4

√

3, so we are always taking natural log of a positive real number!

3. Compute tan

−1

(i −2

√

3) . First compute the required fraction in polar form:

i + (i − 2

√

i −(i − 2

√

2i −2

√

= −1 +

√

5πi

It follows that

tan

−1

(i −2

√

3) =



√

5π

i −2πni



= −

5π

√

+ πn : n ∈ Z

Choosing the principal value of the logarithm (n = 0) yields −

5π

√

Exercises 3.3. 1. Find the real and imaginary parts of sin i, cos(1 + i) and tan(2i ln 5 +

2. As a sanity check, if w = −i log

z + (z

−1)

, compute cos w =

( e

+ e

−iw

) directly and

verify that you obtain z, irrespective of which branches are chosen.

3. Using the real and imaginary parts of sin z, directly verify that the Cauchy–Riemann equations

are satisﬁed.

4. Prove the following double/multiple-angle formulæ using the deﬁnitions in this section:

(a) cos 2z = 2 cos

z − 1

(b) sin(z −w) = sin z cos w −cos z sin w

tan z + tan z

1 −tan z tan w

5. Find all the values of tan

−1

(1 + i).

6. Solve the equation cos z =

√

2 for z.

7. Recall Exercise 3: check explicitly that tan w = i −2

√

3 when w = −

5π

√

(Hint: use tan w =

2iw

−1

i(e

2iw

+1)

. Why is this true?)

8. Suppose z > 1 is real. Prove that Re sin

−1

z =

+ 2πn is independent of z. What is Im sin

−1

9. If the same branch of square-root is chosen in each case, prove that sin

−1

z + cos

−1

z is constant.

10. Derive the expressions for tan

−1

z and its derivative in Theorem 3.20.

11. (a) Given cosh z = cos(−iz), ﬁnd an expression in terms of the complex logarithm for

cosh

−1

(b) Using your answer to part (a), or otherwise, ﬁnd all solutions to the equation cosh z =

√

−1

4 Integration

At ﬁrst glance, integration might appear straightforward; surely we can write the following?

dz =



−1) = 21

In real analysis such a statement reﬂects the equivalence of two distinct concepts:

Anti-derivatives

is an anti-derivative of z

Deﬁnite Integrals The ‘area’ under the curve y = z

on an interval [0, z] equals

This amazing and important equivalence is suitably named the fundamental theorem of calculus. But

does it hold in complex analysis? While anti-derivatives make immediate sense, deﬁnite integrals are

more delicate. For instance, what should we mean by

3+i

dz ?

A Riemann sum construction requires us to partition some curve joining 3 + i and 4i; but which curve?

Does it matter? These questions lead us to revisit contour/path integrals from multi-variable calculus.

4.1 Functions of a Real Variable and Contour Integrals

We start by considering complex-valued functions of a real variable w : I → C where I ⊆ R is an

interval; derivatives and deﬁnite integrals are built from those of their real and imaginary parts:

′

( t) = u

′

( t) + iv

′

( t),

w(t) dt =

u(t) dt + i

v(t) dt (a, b are real or ±∞)

Examples 4.1. 1. If w(t) = 5t

+ it, then

′

( t) = 10t + i,

w(t) dt =

dt + i

t dt =

2. If w(t) = t

+ e

= t

+ cos t + i sin t, then

′

( t) = 2t −sin t + i cos t,

2π

w(t) dt =

+ sin t −i cos t



2π

We see the natural extension of the (real) fundamental theorem here:

+ sin t − i cos t is

plainly an anti-derivative of w(t). This requires no proof since it applies separately to the real

and imaginary parts of w(t).

Basic rules of integration are easily veriﬁed by separately considering real and imaginary parts.

Lemma 4.2. • Linearity: if k ∈ C, then

kw(t) = kw

′

( t) and

kw(t) dt = k

w(t) dt

• Product rule:

w(t)z(t) = w

′

( t)z(t) + w(t)z

′

( t)

• Chain rule: If s( t) is a real function then



s(t)



= w

′



s(t)



′

( t)

Complex substitutions are more subtle, so we provide a proof.

Lemma 4.3 (Complex Chain Rule). Suppose w(t) = F



z(t)



where

• z(t) = x(t) + iy(t) is differentiable at t, and,

• F(z) = u(x, y) + iv(x, y) is holomorphic at z(t)

Then w is differentiable at t, and w

′

( t) = F

′



z(t)



′

( t).

If z

′

( t) is integrable on [a, b] and F holomorphic on z([a, b]), we can put this in integral form

′



z(t)



′

( t) dt = F



z(b)



− F



z( a)



Proof. Apply the multi-variable chain rule from real calculus and the Cauchy–Riemann equations:

+ i

∂u

∂x

∂u

∂y

+ i



∂v

∂x

∂v

∂y



= (u

+ iv

)

+ i(v

−iu

)

= (u

+ iv

)



+ i



(Cauchy–Riemann)

= F

′



z(t)



′

( t)

Examples 4.4. 1. Let w(t) = e

t−it

= F



z(t)



where F(z) = e

and z(t) = t −it

. Then

′

( t) = e

t−it

( t −it

) = e

t−it

(1 − 2it)

Compare with the method of Example 4.1 which gives the same result, if more slowly

′

( t) =

( e

cos t

−ie

sin t

) = e

(cos t

−2t sin t

) − ie

(sin t

+ 2t cos t

)

2. Since w(t) = F



z(t)



(1 − t

+ it)

has w

′

( t) = (i − 2t)(1 −t

+ it)

, we see that

(i −2t)(1 − t

+ it)

dt =

(1 − t

+ it)



−1

= −

This would be horriﬁc if we instead had to multiply out to ﬁnd the real and imaginary parts!

3. Sometimes the real and imaginary part approach is simply not tenable,

3it

1 + it

dt = (1 + it

)

3/2



= (1 + i)

3/2

−1 =

√

3πi

−1

We evaluate using the principal square root since 1 + it

lies in the ﬁrst quadrant.

While most of the basic rules of real calculus translate to complex-valued functions of a real variable,

not everything goes through. Be particularly careful of existence results such as the mean value

theorem which apply perfectly well to real and imaginary parts, but not to the whole. . .

Contours and Contour Integrals

We want to integrate complex functions along curves. But what sort of curves?

Deﬁnition 4.5. A smooth arc is an oriented curve C in the complex plane for which there exists a

regular parametrization; a differentiable function z : [a, b] → C such that:

1. z([a, b]) = C where z(a) is the start of the curve and z(b) is the end.

2. z

′

( t) is continuous

on [a, b] and non-zero on (a, b).

A contour is a piecewise smooth arc C consisting of ﬁnitely many smooth arcs joined end-to-end. A

parametrization z(t) of C is therefore continuous with piecewise continuous derivative.

If we reverse the orientation of a contour C, the resulting contour is labelled −C.

Additionally, we say that a contour is:

• Closed if it starts and ends at the same point, z(a) = z(b);

• Simple if it does not cross itself (though endpoints can be the same, z(b) = z(a));

• Positively oriented if it is simple, closed and traversed counter-clockwise.

Examples 4.6. Here are three contours with explicit parametrizations:

0 1

−1 1

−i

−1 1

−i

Simple, piecewise Closed, non-simple Positively oriented

z =

(

t + it

t ∈ [0, 1]

2 − t + it t ∈ [1, 2]

z = cos t + i sin 2t, t ∈ [0, 2π] z = e

, t ∈ [0, 2π]

Deﬁnition 4.7. Let C be a contour parametrized by z : [a, b] → C and suppose that f (z) is a complex

function deﬁned on the range of z. The contour integral of f (z) along C is

f =

f (z) dz :=



z(t)



′

( t) dt

This is often written

f if C is positively oriented (simple and closed).

We plainly require, and assume from now on by default, that the function f



z(t)



′

( t) is integrable

on the real interval [a, b]; indeed we typically assume that this expression is piecewise continuous.

Since

′

(t)

is bounded on [a, b], the arc-length

′

(t)

dt is ﬁnite. This extends to any contour.

Examples 4.8. We evaluate several contour integrals.

1. For the contour C

parametrized by z( t) = t + it

, t ∈ [0, 1],

we compute

z dz =

z(t)z

′

( t) dt =

( t −it

)(1 + 2it) dt

t + 2t

+ it

dt =

= 1 +

2. For the contour C

with z(t) = e

with t ∈ [0, π],

dz =

dt = πi

−1 1

+ 1 dz =

( e

2it

+ 1)ie

dt =

3it



( e

3πi

−1) + e

πi

−1 = −

3. We compute the same integrals as the previous example, but over the lower semi-circle C

parametrized by z(t) = e

−it

with t ∈ [0, π]. This time

dz =

−ie

−it

dt = −πi

+ 1 dz =

( e

−2it

+ 1)(−ie

−it

) dt =

−i

−3i

−3it

−i

−it



= −

The sign of one integral changed but the other did not! We’ll return to this problem shortly. . .

Before considering more examples we develop some of the basic properties of contour integrals.

Several are immediate from our earlier discussion, for instance linearity:



a f (z) + bg(z)



dz = a

f (z) dz + b

g(z) dz

Of more importance are the following:

Theorem 4.9 (Basic rules for contour integrals). Suppose C is a contour parametrized by z(t).

1. If C = C

∪C

, where C

and C

are contours such that the end of the

ﬁrst is the start of the second, then

f (z) dz =

f (z) dz +

f (z) dz

f is independent of (orientation-preserving) parametrization.

3. Reversing orientation changes the sign of the integral:

−C

f = −

f .

Example 4.10. By the previous example and part 3 of the Theorem, if C is the unit circle centered at

the origin, then

dz = 2πi. This can also be veriﬁed directly.

Proof. Part 1 follows from the well-known property

of real integrals. Armed with this,

it is enough to check the other parts for a single smooth arc C.

2. Suppose z : [a, b] → C and w : [α, β] → C are parametrizations of C. Then w



s(t)



= z(t) for

some continuously differentiable s with positive derivative. Now compute,

′

( t) = w



s(t)



′

( t), s(a) = α, s(b) = β, from which,



z(t)



′

( t) dt =



w(s(t))



′

( s(t))

dt =

s(b)

s(a)



w(s)



′

( s) ds



w(s)



′

( s) ds

3. This is almost identical, except that −C requires a reparametrization w(s) with s

′

< 0, s(a) = β

and s(b) = α. The upshot is that the limits ﬂip on the ﬁnal integral:

f =



z(t)



′

( t) dt =



w(s)



′

( s) ds = −



w(s)



′

( s) ds = −

−C

Contour Integrals of multi-valued functions If a function is multi-valued, we must specify a branch

before integrating. It is acceptable to have the contour start and/or ﬁnish on the branch cut.

Examples 4.11. 1. Compute

1/2

dz over the unit circle using the principal square root.

Since the question speciﬁes the principal branch, we must

must start and ﬁnish the contour at z = −1. Parametrize via

z(t) = e

where t ∈ (−π, π). Since

z(t) z

′

( t) = e

= ie

3it

is continuous on [−π, π], we compute

1/2

dz =

−π

3it

dt =

3it



−π

( e

3πi

−e

−

3πi

) = −

2. Find

dz over the unit semi-circle shown using P. V. z

Let z(t) = e

where t ∈ [0, π]. Since

′

= e

i Log z

′

= exp

(

i(it)

)

= ie

(i−1)t

is continuous when 0 ≤ t ≤ π, we have

1−1 0

dz =

(i−1)t

dt =

i −1



(i−1)π

−1



i −1



1 + e

−π



Warning! Things are more difﬁcult if we allow a contour to cross a branch cut, since we’d then have

to work with multiple branches simultaneously. We won’t do this.

We extend the integrand continuously: recall that if g(t) is uniformly continuous on a bounded open interval (a, b), then

it has a continuous extension to the closed interval [a, b] and we can therefore deﬁne

g(t) dt.

Exercises 4.1. 1. Evaluate the derivatives and integrals:

(a)

sin t + i

√

(b)

(i + t

)

(c)

πit

(d)

π/2

(1 + e

)

dt (e)

(i −t)

dt (f)

(i −1) cos



(1 + i)t



2. Show that if m, n ∈ Z, then

2π

imt

−int

dt = 2πδ

where δ

= 1 when m = n and 0 otherwise. Do this two ways:

(a) Using the exponential law and the chain rule/substitution.

(b) By multiplying out and working with real and imaginary parts.

3. Use the chain rule to evaluate the integral

(1+i)t

dt.

Hence ﬁnd both

cos t dt and

sin t dt without using integration by parts.

4. Prove the product rule for functions of a real variable:

( wz) = w

′

z + wz

′

(Hint: let w(t) = u(t) + iv(t), z(t) = x(t) + iy(t) and multiply out. . . )

5. Check that the mean value theorem fails for the function w(t) =

√

t + it

on the interval [0, 1].

That is, there is no ξ ∈ (0, 1) for which w

′

( ξ) =

w(1)−w(0)

1−0

6. Justify the integral form of the complex chain rule by considering the real and imaginary parts

of F. What facts from real calculus are you using?

7. Evaluate each contour integral

f (z) dz by explicitly parametrizing C:

(a) f (z) = z

; C is the straight line from z = 1 to z = i.

(b) f (z) = z; C consists of the straight lines joining z = 1 to 1 + i to −1 + i to −1.

clockwise from −3i to 3i.

(d) f (z) = exp z where C is the straight line from z = i to z = 3 − i.

8. Explicitly check that

z dz =

− A

) along the straight line joining A and B.

(Hint: the line can be parametrized by z(t) = (1 − t)A + tB where t ∈ [0, 1])

9. Let n ∈ Z and let C

be the positively oriented circle centered at z

with radius R > 0. Explicitly

parametrize this circle to show that

( z −z

)

n−1

dz =

(

2πi if n = 0,

0 otherwise

10. Suppose z : [a, b] → C is a regular parametrization of a smooth arc C. Then the arc-length of the

curve is the integral of the speed of the parametrization:

L =



′

( t)



(a) Compute the arc-length of the circle of radius R centered at the origin.

(b) Compute the arc-length of the simple piecewise curve in Example 4.6.

(This requires a tough substitution; perhaps look up a table of integrals. . . )

not change the arc-length.

(d) Let s(t) =

′

( τ)

dτ be the arc-length as a function of t ∈ [a, b]. Consider a new

parametrization w(s) = z(t(s)), where t(s) is the inverse function of s(t).

Prove that



= 1.

(This proves that every smooth arc has a unit-speed parametrization)

The last three questions elaborate a little on the approach in Examples 4.11.

11. (a) Compute the integral of the principal value of z

1/3

around the positively oriented unit

circle starting and ﬁnishing at z = −1.

(b) Now consider the branch z

1/3

= exp(

log z) where arg z ∈ (0, 2π). Integrate this around

the positively oriented unit circle starting and ﬁnishing at z = 1. What do you observe?

12. Compute

1/2

dz where we take the α-branch z

1/2

= exp(

log z) with arg z ∈ (α, α + 2π)

and the unit circle C traced from angle α to α + 2π.

13. Let ϵ > 0 be small and suppose that C

is the circular arc of radius

1 centered at the origin, traversed counter-clockwise from angle

−π + ϵ to π − ϵ .

By parametrizing C

, explicitly evaluate

√

z dz and verify that

lim

ϵ→0

√

z dz = −

is the value obtained previously for

√

z dz.

π − ǫ

4.2 Path-independence, the Fundamental Theorem & Integral Estimation

We start by revisiting an earlier example.

Example (4.8, parts 2 & 3). Let F(z) =

+ z and observe that F

′

( z) = z

+ 1. If z : [a, b] → C

parametrizes a smooth arc C such that z(a) = 1 and z(b) = −1, then Lemma 4.3 shows that

+ 1 dz =



z(t)

+ 1



′

( t) dt =

′



z(t)



′

( t) dz =



z(t)



= F



z(b)



− F



z( a)



+ z



−1

= −

−

= −

The contour integral is independent of the choice of arc C, depending only on its endpoints.

Deﬁnition 4.12. Suppose C is a contour in some path-connected D ⊆ C

and that f : D → C. We say that

f is path-independent if its value

depends only on the endpoints of C and not otherwise on the contour.

Otherwise said,

f =

f for any contour

C with the same endpoints.

When

f is path-independent (and only in such cases), it is permissible to write a contour integral

with explicit limits: to write

f is to assert that the integral evaluates identically along any contour

with endpoints z

, z

Before tackling the main result, we tidy up a connection between path independence and closed

curves. You should have seen exactly this result in multi-variable calculus.

Lemma 4.13. Let f deﬁned on a domain D ⊆ C be given. Every contour integral

f (z) dz over a

contour in D is path-independent if and only if

f (z) dz = 0 round every closed contour.

Proof. (⇒) Assume path-independence for all integrals

f and let C be a given closed contour in D.

Choose any points z

, z

∈ C and decompose C = C

∪C

into two contours:

from z

to z

and back again. Then,

f (z) dz =

f (z) dz +

f (z) dz (Theorem 4.9, part 1)

f (z) dz −

−C

f (z) dz (Theorem 4.9, part 3)

Since C

and −C

share the same endpoints, path-independence tells us that

the integrals are equal, whence

f = 0.

(⇐) Conversely, suppose

f (z) dz = 0 round any closed contour in D, and let C

, −C

be given

contours in D with the same endpoints z

, z

Plainly C = C

∪ C

is a closed contour, and the previous calculation demonstrates path-

independence:

f (z) dz =

−C

f (z) dz.

In the above example, the existence of an anti-derivative F(z) =

+ z of f (z) = z

+ 1 demonstrated

path-independence, thus facilitating easy computation of the integral. This should seem familiar. . .

Theorem 4.14 (Fundamental Theorem). Let f be continuous on an open domain. Then,

f has an anti-derivative ⇐⇒ all contour integrals

f are path-independent

In such situations,

f (z) dz = F(z

) − F(z

) where F(z) is any anti-derivative of f (z).

The openness and continuity assumptions are necessary and will be used in the proof. In multi-

variable calculus, the (⇒) direction is often known as the Fundamental Theorem of Line Integrals.

Examples 4.15. 1. The function f (z) = (z + i)

has anti-derivative F(z) =

( z + i)

. It follows that

1−2i

( z + i)



1−2i

(1 − i)

−i

( −4 −1) = −

regardless of the contour used to travel from z = 0 to 1 − 2i.

2. The principal root f (z) =

√

z has anti-derivative F(z) =

3/2

(

√

(also using the principal branch!). If C is any contour from z

= 1 to

= i which doesn’t cross the branch cut, then

√

z dz =

(

√





(

√

−(

√





3πi

−1



No!

3. Since f (z) = sin z has an anti-derivative F(z) = −cos z, valid everywhere, we see that

sin z dz = 0 round any closed curve C.

4. Around the unit circle

z dz =

−π

−it

dt = 2πi = 0. Thus f (z) = z has no anti-derivative

on any domain including the unit circle!

5. (Example 4.10, cont) Around the unit circle,

dz = 2πi is non-zero. By the fundamental

theorem, f (z) =

cannot have an anti-derivative on any domain containing said circle. But

this is obvious from our discussion in Section 3.2: If an anti-derivative F(z) existed, then

F(z) =

log z =⇒ F(z) = log z + c (Theorem 2.31)

for some branch of the logarithm, which contradicts the fact that no branch can be made single-

valued on a path encircling the origin.

We can, however, use the fundamental theorem to evaluate

over a contour within the domain of a single branch. In the pic-

ture, given the contour C, we choose the branch cut shown and the

logarithm with −

5π

< arg z <

dz = log(1 + i) − log i = log

√

πi

−log e

−

3πi

= ln

√

2 +

πi

−

3πi

(ln 4 + 7πi)

1 + i

Proving the Fundamental Theorem The forward direction is straightforward, though note the im-

portance of a contour consisting of only ﬁnitely many smooth arcs.

Proof. (⇒) Suppose f has an anti-derivative F. Given a contour C parametrized by a piecewise

smooth function z(t), write C = C

∪···∪C

where each C

is a smooth arc between endpoints

k−1

= z(t

k−1

) and z

= z(t

). By Theorem 4.9 and Lemma 4.3, we see that

f =

∑

k=1

f =

k−1

′



z(t)



′

( t) dt =

∑

k=1



F(z

) − F(z

k−1

)



= F(z

) − F(z

)

The converse is similar to a standard proof of part I of the fundamental theorem from real analysis:

f continuous =⇒

f (t) dt = f (x)

There are, however, a couple of extra subtleties: we must choose how to travel from one endpoint of

an integral to the other (path-independence says it doesn’t matter how); we also need to bound the

modulus of a contour integral. This last isn’t as trivial as it may seem. . .

(⇐) We prove when the domain D is connected.

Fix z

∈ D and deﬁne

F(z) :=

f (ζ) dζ where the integral is taken along any curve (path-

independence). We need to show that lim

w→z

F(w)−F(z)

w−z

= f (z) on D.

Fix z ∈ D and let ϵ > 0 be given. Since f is continuous and D open,

∃δ > 0 such that

ζ − z

< δ =⇒ ζ ∈ D and

f (ζ) − f (z)

Let w ∈ D be such that 0 <

w − z

< δ. Evaluating along any curve

joining z, w (path-independence again), we obtain

F(w) − F(z) =

f (ζ) dζ

For simplicity, we choose the straight line segment from z to w. The proof is almost complete:



F(w) − F(z)

w − z

− f (z)



w − z

f (ζ) − f (z) dζ



w − z



f (ζ) − f (z) dζ



≤

w − z

< ϵ

The second last inequality should give you pause. At ﬁrst glance it appears as if we argued



f (ζ) − f (z) dζ



≤

f (ζ) − f (z)

dζ ≤

dζ =

w − z

This is ﬁne in real analysis (if z ≤ w), but is utter nonsense in complex-land: the middle term

f (ζ) − f (z)

dζ =





ζ(t)



− f (z)



′

( t) dt

need not be a real number! We had better tidy this up. . .

If not, repeat the argument choosing a new z

for each connected component (‘lump’) of D.

Theorem 4.16 (Integral Estimation). 1. Suppose w : [a, b] → C is piecewise continuous. Then



w(t) dt



≤

w(t)

2. Suppose C is a contour with length L, and let f be piecewise continuous on C. Then

f (z)

bounded by some M ≥ 0 on C, and



f (z) dz



≤ ML

Part 2 justiﬁes the suspect inequality in the proof of the Fundamental Theorem: f (ζ) − f (z) is

bounded by M =

and the path of integration is the straight line of length L =

w − z

Proof. 1. Let

w(t) dt = re

iθ

. Since θ is constant, r =

−iθ

w(t) dt is real. In particular,

r =



−iθ

w(t)



+ i Im



−iθ

w(t)



dt =



−iθ

w(t)



Appealing to Re z ≤

, we see that



w(t) dt



= r =

Re(e

−iθ

w(t)) dt ≤



−iθ

w(t)



dt =

w(t)

2. Parametrize the contour integral by z : [a, b] → C. Since f (z(t)) is piecewise continuous on a

closed bounded interval, it is bounded and thus M exists. But now,



f (z) dz





z(t)



′

( t) dt



≤





z(t)



′

( t)



dt (part 1)

≤



′

( t)



dt = ML

While an explicit computation of arc-length is usually impractical, it is straightforward for circular

or straight-line arcs. Indeed the estimation of integrals around circles is crucial for later sections.

Examples 4.17. 1. On the straight line C joining z = 4 to 4i, we see that

√

2 =

2( 1 + i)

≤

≤ 4 =⇒

z + 1

≤

+ 1 ≤ 5

By the extended triangle inequality,



+ 4



≥



−4



≥



√

−4



= 60

Since C has length 4

√

2, it follows that



z + 1

+ 4



≤

√

0 2 4

2. For the same function f (z) =

z+1

, consider the circle C

with radius R >

√

4 centered at the

origin. On C

, we have

> 4; the triangle inequality tells us that



z + 1

+ 4



≤

+ 1

−4

R + 1

−4

=⇒



z + 1

+ 4



≤

2πR(R + 1)

−4

In particular, this approaches zero as R → ∞.

3. Let C

be the circle with radius r < 2 centered at z = 1. Then

z − 1

= r and

z + 1

2 − (1 −z)

≥ 2 −r

from which



−1



≤

2πr

(2 − r)r

2π

2 −r

−−→

r→0

1−1

In Exercise 11, it will be veriﬁed that

−1

dz = iπ for any r < 2.

Examples 2 & 3 are typical of the calculations dominating later sections.

Exercises 4.2. 1. Evaluate each contour integral

f (z) dz using the fundamental theorem:

(a) f (z) = z

where C is the straight line from z = 1 to z = i.

(b) f (z) =

where C is the pair of straight lines from z = 1 to −1 −i to −i.

, where C is the straight line from the origin to z = i

√

π.

(d) f (z) =

1 + z

where C is the straight line from z = 1 to 2 + i.

(e) f (z) =

√

where C is any path z(t) with Re z > 0 joining z = 1 + i and z = 4.

(f) f (z) = P. V. z

−1−2i

along the quarter circle z(t) = e

where 0 ≤ t ≤

2. Let n ∈ N

. Prove that for every contour C from z

to z

dz =

n + 1



n+1

−z

n+1



3. If C is a closed curve not containing z

, and n ∈ Z \ {0}, prove that

( z −z

)

n−1

dz = 0.

4. Let f (z) = z

1/3

be the branch where arg z ∈ (−

3π

). Evaluate the

integral

f (z) dz where C

is the curve shown in the picture.

5. Evaluate

1+i

Log z dz where the curve C lies in the upper half-plane.

(Hint: use integration by parts)

−1 0 1

6. Evaluate

−1

2−i

dz where z

2−i

is the principal branch, and the integral is over any contour

which, apart from its endpoints, lies above the real axis.

7. Let C be the arc of the circle

= 2 from z = 2 to 2i. Without evaluating the integral, show that

(a)



z + 4

−1



≤

6π

(b)



−1



≤

8. If C is the straight line joining the origin to 1 + i, show that



2iz



≤ 4

9. If C is the boundary of the triangle with vertices 0, 3i and −4, prove that



( e

−

z) dz



≤ 60

(Hint: show that

−z

≤ e

+ y

. . . )

10. Let C

be the circle of radius R > 1 centered at the origin. Prove that



Log z



< 2π



π + ln R



and thus conclude that lim

R→∞

Log z

dz = 0.

11. (Hard) We continue and expand Example 4.17.3. Let f ( z) =

−1

(a) Explain why F(z) =



Log( 1 − z) − Log(1 + z)



is an anti-

derivative of f (z) on the domain D = C \ ℓ where ℓ consists of

the two segments of the real axis where

≥ 1.

Use F(z) to evaluate

1+2i

f (z) dz along any curve in D.

−1 1

(b) Show that

f (z) dz = iπ where C

is a circle of radius r < 2 cen-

tered at z = 1.

(Hint: parametrize z(t) = 1 + re

where −π < t < π and use F(z))

f (z) dz along a curve C

passing to the right of z = 1 as

in the picture. Compare your answer with parts (a) and (b).

(Hint: you need a different branch of the log for one of the terms in F( z))

4.3 The Cauchy–Goursat Theorem and Cauchy’s Integral Formula

In this section we start to extend the fundamental theorem with the goal of more fully understanding

and evaluating holomorphic functions. We ﬁrst require another piece of topology.

Deﬁnition 4.18. A simply-connected region D is a path-connected set such

that any simple closed curve C ⊆ D may be shrunk to a point without

leaving D. Otherwise said, the inside of any such C also lies in D.

A non-simply connected region has holes: e.g., the set D pictured at the

bottom of the page.

Theorem 4.19 (Cauchy–Goursat, version 1). Suppose C is a closed contour in a simply-connected

region D. If f is holomorphic on D, then

f (z) dz = 0.

A basic proof invokes Green’s Theorem (from multi-variable calculus) after assuming that the real and

imaginary parts of f have continuous partial derivatives and that C is simple. A proof without these

assumption is signiﬁcantly longer and more challenging.

Lemma 4.20 (Green’s Theorem). Suppose D is a closed bounded simply-connected domain with

boundary C and that P, Q : D → R have continuous partial derivatives. Then

P dx + Q dy =

∂Q

∂x

−

∂P

∂y

Sketch Proof of Cauchy–Goursat. Suppose f ( z) = u + iv has continuous partial derivatives and that C

is simple (assume WLOG that C is positively oriented). Parametrize C by z(t), then

f (z) dz =

(

z(t)

)

′

( t) dt =

(

u(z(t)) + iv(z(t))

)



′

( t) + iy

′

( t)



′

−vy

′

dt + i

′

+ uy

′

u dx − v dy + i

v dx + u dy (deﬁnition of line integral in R

)

−v

−u

dA + i

−v

dA (Green’s Theorem)

Both double integrals are zero by the Cauchy–Riemann equations.

To extend Cauchy–Goursat, we perform a sneaky trick. From the

interior of a simple closed contour C, remove the regions inside

simple closed non-intersecting contours C

, . . . , C

. Orient these

clockwise so that the interior region D lies to all contours’ left.

By cutting D as indicated and traversing each cut twice in op-

posite directions, C, C

, . . . , C

and the cuts may be joined to

form a new simple closed contour. We may now apply Cauchy–

Goursat. . .

Corollary 4.21 (Cauchy–Goursat, version 2). Suppose C is a simple closed contour, oriented

counter-clockwise. Let C

, . . . , C

be non-intersecting simple closed contours in the interior of C,

oriented clockwise. If f (z) is holomorphic on the region between and including C and the interior

boundaries C

, . . . , C

, then

f (z) dz +

∑

j=1

f (z) dz = 0

The power of this result lies in how it allows us to compare integrals around different contours.

Corollary 4.22. Suppose C

, C

are non-intersecting positively oriented simple closed contours. If f

is holomorphic on the region between and including the curves, then

f (z) dz =

f (z) dz

Examples 4.23. 1. (Example 4.17.3, cont.) The function

f (z) =

−1

is holomorphic on and between any two circles C

centered at z = 1

with radius r < 2. Cauchy–Goursat thus conﬁrms part of the conclu-

sion of Exercise 4.2.11, that

−1

dz is independent of the radius r.

We will shortly develop a better method to evaluate this integral.

2. We compute the integral of f (z) =

around any simple closed contour C staying away from

the origin.

• If the origin is outside C, then f is holomorphic on and inside C. By Cauchy–Goursat, we

conclude that

dz = 0.

• If the origin is inside C, then there is some minimum

distance d of C to the origin. Choose any circle C

with

radius r < d centered at the origin. Since f is holomor-

phic on the region D between and including C and C

we conclude that

dz =

2π

rie

dθ = 2πi

More generally, if C is a positively-oriented simple closed contour not containing z

, then

z − z

dz =

(

2πi if z

lies inside C

0 if z

lies outside C

This example generalizes to perhaps the most powerful result in complex analysis. . .

Theorem 4.24 (Cauchy’s Integral Formula). Suppose f is holomorphic everywhere on and inside a

simple closed contour C. If z

is any point inside C, then

f (z

) =

2πi

f (z)

z − z

More generally, f is inﬁnitely differentiable at z

with n

derivative

(n)

( z

) =

2πi

f (z)

( z −z

)

n+1

Example 4.25. If f (z) is a polynomial, we can check the integral formula explicitly by appealing to

the Cauchy–Goursat Theorem and Exercise 4.1.9: if C is a simple closed contour encircling z

, then

( z −z

)

n−1

dz =

(

2πi if n = 0

0 otherwise

For instance, if f (z) = 3z

+ 2 and z

= 0, then

2πi

+ 2

dz =

2πi

dz +

2πi

dz = 6 + 0 = f

′′

(0)

Proof of the basic Integral formula

. Let ϵ > 0 be given. Denote by D the open region interior to C.

Since f is holomorphic it is also continuous. Combining these:

∃δ > 0 such that

z − z

< δ =⇒ z ∈ D and

f (z) − f (z

)

Draw a circle C

of radius r < δ centered at z

. This lies entirely in D. Since

f (z)

z−z

is holomorphic between and on C

and C, Corollary 4.22 tells us that

f (z)

z − z

dz =

f (z)

z − z

We need only bound an integral to complete the proof:



2πi

f (z)

z − z

dz − f (z

)



2πi

f (z)

z − z

dz −

f (z

)

2πi

z − z



2πi

f (z) − f (z

)

z − z



Thm

≤

4.16

2π

·2πr < ϵ

since

z − z

= r and

f (z) − f (z

)

on C

, which has circumference 2πr.

Informally, it appears as if the general formula follows by repeated differentiation (induction):

(n+1)

) =

(n)

)

???

2πi

∂

∂z

f (z)

(z −z

)

n+1

dz =

(n + 1)!

2πi

f (z)

(z −z

)

n+2

One cannot blindly bring a derivative inside an integral like this, so a formal proof is required; Exercise 13 offers a sketch.

Examples 4.26. We can use the integral formula to evaluate certain integrals that would be difﬁcult

if not impossible to evaluate by parametrization.

1. The integral formula makes Example 4.17.3 and its extensions very easy: apply the formula to

f (z) =

2πi

z+1

around (say) the circle with radius 1 centered at z = 1 to see that

−1

dz =

2πi

( z + 1)(z − 1)

dz = f (1) = iπ

Indeed

−1

dz = iπ around any simple closed curve encircling z = 1 but not z = −1.

2. If C is the circle centered at z = i with radius 1, then f (z) =

3z sin z

z+i

is holomorphic on and inside

C, whence

3z sin z

+ 1

dz =

3z sin z

( z + i)(z − i)

dz = 2πi f (i) = 2πi

3i sin i

= 3πi sin i

( e

−1

−e

)

Contrast this with parametrizing z(t) = i + e

and attempting to evaluate directly!

2π

3(i + e

) sin(i + e

)ie

(i + e

)

+ 1

dt . . .

3. Let C be a simple closed contour staying away from z

= 4. Since f (z) =

is entire,

+ 7

( z −4)

dz =

(

0 if z

= 4 is outside C

2πi f (4) = 110πie

−4

if z

= 4 is inside C

4. Let C be the circle of radius 2 centered at z = 1 + i. Then g(z) =

+1)

(z+i)

(z−i)

holomorphic on and inside C, except at z = i. We conclude that

g(z) dz =

( z + i)

( z −i)

dz =

2πi



( z + i)

−3

= 12πi(2i)

−5

3π

We ﬁnish with a previously mentioned corollary of the integral formula.

Corollary 4.27 (Theorem 2.33). Holomorphic functions are inﬁnitely differentiable with all deriva-

tives holomorphic. In particular, their real and imaginary parts have continuous partial derivatives

of all orders.

Proof. If f (z) is holomorphic at w, then it is holomorphic on an open set D containing w. We may

therefore choose a circle C

centered at w lying inside D.

The integral formula with n = 2 says that the second derivative f

′′

( z

) exists at every point z

inside

. Otherwise said, the derivative f

′

( z) is holomorphic at w.

Now induct to see the f

′′

( z) is holomorphic at w, etc.

Finally, note that if f has an anti-derivative F, then F is necessarily holomorphic and so, by the

Corollary, is f itself. Combining our results yields the following summary.

Theorem (Summary). Suppose f is continuous on an open domain D and that curves C lie in D.

all

f (z) dz = 0

Lemma 4.13



f holomorphic on D

Cauchy–Goursat

(if D simply-connected)

all

f (z) dz path-independent

Fundamental Thm

f has an anti-derivative on D

Cauchy Integral Formula

Exercises 4.3. 1. Apply the Cauchy–Goursat Theorem to show that

f (z) dz = 0 when the contour

C is the unit circle

= 1.

(a) f (z) =

z + 3

(b) f (z) = ze

−z

2. Let C

be the square with sides the lines x = ±1, y = ±1, and C

the circle

= 4. Explain why

+ 1

dz =

+ 1

3. Let C be the square with sides x = 0, 1 and y = 0, 1. Evaluate the integral

z−a

dz when:

(a) a is exterior to the square;

(b) a is interior to the square.

4. Let C be the positively-oriented boundary of the half-disk 0 ≤ r ≤ 1, 0 ≤ θ ≤ π and deﬁne

f (z) =

√

z =

√

iθ/2

and f (0) = 0 using the branch of z

1/2

with −

< θ <

3π

. Prove that

f (z) dz = 0

by evaluating three contour integrals: over the semicircle, and over two segments of the real

axis joining 0 to ±1. Why does Cauchy–Goursat not apply here?

5. If C is a positively-oriented simple closed contour, prove that the area enclosed by C is

z dz

(Hint: Mirror the sketch proof of Cauchy–Goursat, even though z isn’t holomorphic. . . )

6. Let C denote the boundary of the square with sides x = ±2, y = ±2. Evaluate the following:

(a)

−

πz

z − i

(b)

+ e

−z

z(z

+ 10)

dz (c)

z dz

3z + i

(d)

sec(z/2)

( z −1 −i)

7. Evaluate the integral

g(z) dz around the circle of radius 3 centered at z = i when:

(a) g(z) =

+ 9

(b) g(z) =

( z

+ 9)

8. Prove that if f is holomorphic on and inside a simple closed contour C and z

is not on C, then

′

( z)

z − z

dz =

f (z)

( z −z

)

9. Properly justify the claim at the end of Example 4.26.1, that

−1

dz = iπ

when C is any simple closed curve encircling z = 1 but not z = −1. What happens if instead C

encircles z = −1 but not z = 1?

10. Suppose we have a polynomial p(z) =

∑

k=0

( z − z

)

centered at z

. Use Cauchy’s integral

formula to prove that a

(k)

)

is the usual Taylor coefﬁcient.

11. Let C be the unit circle z = e

iθ

where −π < θ ≤ π and suppose a ∈ R is constant. By ﬁrst

evaluating

−1

dz, prove that

a cos θ

cos(a sin θ) dθ = π

12. (a) Suppose that f (z) is continuous on and inside a simple closed contour C. Prove that the

function g(z) deﬁned by

g(z

) :=

2πi

f (z)

z − z

is holomorphic at every point z

inside C and that

′

( z

) :=

2πi

f (z)

( z −z

)

(Hint: consider the formula in part (b) of the next question, after replacing the ﬁrst two f ’s by g)

(b) If f (z) = x(1 − x)(1 − y) and C is the square with vertices 0, 1, 1 + i, i compute g(z

13. We prove the 1

derivative version of Cauchy’s integral formula. As previously, let δ > 0 be

such that

w − z

< δ =⇒ w ∈ D.

(a) If

∆z

< δ and z ∈ C explain why

z − z

≥ δ

ii.

z − z

−∆z

> 0

(b) Use the basic integral formula on C to evaluate f (z

+ ∆z) − f (z

) and thus prove that



f (z

+ ∆z) − f (z

)

∆z

−

2πi

f (z)

( z −z

)



≤

2π(δ − ∆z)δ

∆z

where M is an upper bound for

f (z)

on C, and L is the length of C. Hence conclude that

′

( z

) =

2πi

f (z)

(z−z

)

dz.

If you want a challenge, try to do the same thing for higher derivatives!

4.4 Liouville’s Theorem and The Maximum Modulus Principle

In this section we deduce several powerful corollaries of Cauchy’s integral formula. The ﬁrst is a

straightforward combination of the integral formula with integral estimation (Theorem 4.16).

Lemma 4.28 (Cauchy’s Inequality). If f is holomorphic on and inside the circle C of radius R cen-

tered at z

, and

f (z)

≤ M on C, then



(n)

( z

)



2π



f (z)

( z −z

)

n+1



≤

n!M

This seems innocuous, but it has surprising applications. If f is entire and bounded (

f (z)

≤ M for

all z), then Cauchy’s inequality applies for any circle centered at any z

∀z

∈ C, R > 0,



′

( z

)



≤

=⇒ ∀z

, f

′

( z

) = 0

Thm

=⇒

2.31

f (z) constant

Theorem 4.29 (Liouville). Every bounded, entire function is constant.

Liouville’s Theorem is the key ingredient in perhaps the simplest proof of a very famous result.

Corollary 4.30 (Fund. Thm. of Algebra). Every non-constant polynomial has a root in C.

Proof. WLOG suppose a non-constant polynomial p(z) = z

∑

n−1

k=0

is monic. We assume, for

contradiction, that p(z) = 0 has no solutions. It follows that is

p(z)

is entire. Our strategy is to

establish that

p(z)

is also bounded; since p(z) is non-constant, this contradicts Liouville’s Theorem.

In fact it is enough to bound

p(z)

for all large z: that is, we want to ﬁnd some R > 0 such that

> R =⇒

p(z)

bounded

This is since

p(z)

, being continuous, is automatically bounded on any disk

≤ R (compactness,

Theorem 2.21). So how do we ﬁnd a suitable R? Everything follows from the triangle inequality,

> R =⇒

p(z)



1 +

n−1

∑

k=0

n−k



≥



1 −



n−1

∑

k=0

n−k



To complete the proof, we need only choose R such that



∑

n−1

k=0

n−k



≤

∑

n−1

k=0

n−k

≤

, for then

> R =⇒

p(z)

≥

=⇒

p(z)

≤

is bounded. It is sufﬁcient to force each term in the above sum to be ≤

, as may be accomplished

(see Exercise 8) by deﬁning

R := max

(2n

)

n−k

: 0 ≤ k < n

The proof used the fact that a continuous function f on a compact (closed, bounded) domain K is

bounded (Theorem 2.21). As you should recall from real analysis, the least upper bound is achieved:

∃z

∈ K such that

f (z

)

= sup



f (z)

: z ∈ K



For holomorphic functions, a more restrictive and surprising result holds: the maximum modulus of a

non-constant such function on a compact domain is always and only achieved at an edge point.

Example 4.31. Let f (z) = e

on the unit disk

≤ 1. Then

f (z)

= e

, whence the maximum

modulus

f (1)

= e occurs at z = 1, on the edge of the disk.

The theorem establishing this important observation is usually stated a little differently.

Theorem 4.32 (Maximum Modulus Principle). Suppose f is holomorphic and non-constant on a

connected, open domain D. Then

f (z)

has no maximum value on D.

Holomorphicity is critical: see, for instance, Exercise 5.

Proof. We start by demonstrating a special case.

Fix δ > 0, assume f is holomorphic on and inside B

:= {z :

z − z

≤ δ} and

that

f (z)

attains its maximum at z

. We prove that f (z) is constant on B

Let r ≤ δ and apply the integral formula to the circle C

(z(t) = z

+ re

f (z

)

2π



f (z)

z − z



2π



2π

f (z

+ re

)

ire



2π



2π

f (z

+ re

) dt



≤

2π



f (z

+ re

)



dt (Theorem 4.16)

≤

2π

f (z

)

dt =

f (z

)

(

f (z

)

is the maximum of

f (z)

)

Both inequalities are therefore equalities, the second of which now becomes

2π

f (z

)

−



f (z

+ re

)



dt = 0

Since the integrand is non-negative and continuous, it must be zero. But then

f (z)

f (z

)

on all

such circles C

, whence

f (z)

is constant on B

. By Corollary 2.32, f (z) is also constant on B

Now suppose f is holomorphic on some connected open set D and that

f (z)

attains its maximum

at z

. Let w ∈ D be any other point and join z

to w by a smooth arc C.

Let δ be the tube-radius from Exercise 2.2.10. We may there-

fore cover C with ﬁnitely many closed disks B

, B

, . . . , B

⊂ D

of radius δ, indexed so that each center z

lies in one of the

previous disks (at most

length( C)

disks are needed).

By the special case, f (z) is constant on B

. Since z

∈ C ∩ B

we see that

f (z

)

f (z

)

is also the maximum modulus on

D. The special case now says that f (z) is constant on B

∪ B

Continuing in this fashion (essentially induction), we conclude that f (z) = f (z

) is constant on

∪··· ∪ B

. In particular f ( w) = f (z

). Since w was arbitrary, f (z) is constant on D.

Example 4.33. Let f (z) = 2z

+ i on the upper semi-disk with radius 1. There are two boundaries:

• y = 0: plainly

f (z)

√

+ 1 is maximal at z = ±1.

• r = 1: write f (z) = 2e

2iθ

+ i from which

f (z)

(2 cos 2θ)

+ (2 sin 2θ + 1)

√

5 + 4 sin 2θ

which is maximal when θ =

with



f (e

iπ

)



= 3.

−1 0 1

√

The color indicates the value of

f (z)

, and the arrows its direction of increase on the boundary.

Exercises 4.4. 1. (a) Suppose f is entire and that

f (z)

≤ c

for some constant c ∈ R

. Prove that

f (z) = kz where k ∈ C satisﬁes

≤ c.

(b) What can you say about f if it is entire and there exists some linear polynomial cz + d with

c = 0 such that

f (z)

≤

cz + d

for all z ∈ C?

2. If f (z) = u + iv is entire and u(x, y) is bounded above, apply Liouville’s Theorem to exp( f (z))

to prove that u is constant.

3. Let f (z) be a non-zero holomorphic function on a closed bounded domain. By considering

g(z) =

f (z)

, show that the minimum value of

f (z)

also occurs on the boundary.

4. Find the maximum and minimum values of



+ 4i



on the unit disk

≤ 1.

5. Consider f (z) = exp(−

) deﬁned on the unit disk

≤ 1. What is its maximum modulus,

and where is it found? Why doesn’t this contradict the maximum modulus principle?

6. (a) On the rectangle 0 ≤ x ≤ π, 0 ≤ y ≤ 1, show that

sin z

is maximal at the point z =

+ i.

(Hint: ﬁrst show that

sin z

= sin

x + sinh

(b) Prove that there exists a sequence (z

) ⊂ C such that lim sin z

= ∞.

7. Revisit the standard method from multivariable calculus (compute ∇

f (z)

) to check that the

maximum value of



+ i



in Example 4.33 really does occur on the boundary.

8. Complete the proof of the fundamental theorem of algebra:

(a) Verify that R := max



(2n

)

n−k

: 0 ≤ k < n



is positive.

(b) Prove that

> R =⇒ ∀k,

n−k

9. (a) Prove the factor theorem: if p(z

) = 0, then p(z) = (z −z

) q(z) for some polynomial q(z).

(Hint: This follows easily from the division algorithm: if deg p ≥ deg g, then there exist unique

polynomials q(z), r(z) for which

p(z) = g(z)q(z) + r(z) and deg r < deg g

For a challenge, prove some version of this sufﬁcient for our needs. . . )

(b) Prove that a degree n ≥ 1 polynomial p(z) factors uniquely over C: up to order, there exist

unique a, z

, . . . , z

∈ C such that

p(z) = a(z − z

) ···(z − z

)

5 Power Series, Taylor’s Theorem & Analytic Functions

The theory of series in complex analysis differs signiﬁcantly from the real situation, particularly with

regard to two concepts.

• Taylor’s Theorem: Holomorphic functions equal their Taylor series. This is false in real analy-

sis where differentiable functions need not have, nor equal, a Taylor series.

• Laurent Series (chapter 6): series with negative powers (z

−1

+ 3z

−2

+ ···) are a powerful tool.

We start be reviewing the basics of both inﬁnite and power series: as we saw for sequences in Section

2.2, this is essentially identical to the real situation.

5.1 A Brief Review of Power Series

Deﬁnition 5.1 (Inﬁnite Series). The n

partial sum of a sequence (z

)

∞

n=0

is the complex number

∑

k=0

= z

+ ··· + z

The (inﬁnite) series

∑

:= lim s

is said to converge/diverge if the sequence (s

) converges/diverges.

The series converges absolutely if

∑

converges.

The series converges conditionally if it converges but not absolutely.

By convention, the initial term of the sequence/series is z

. This isn’t required: it could be z

, etc.

Theorem 5.2 (Basic Series Facts). Let

∑

and

∑

be series of complex numbers.

1. If z

= x

+ iy

, then

∑

converges if and only if

∑

and

∑

both converge, in which case

∑

+ i

∑

2. If a, b ∈ C, and

∑

and

∑

converge, then

∑

+ bw

converges, in which case

∑

+ bw

= a

∑

+ b

∑

3. (n

term/divergence test) If

∑

converges, then lim z

= 0.

4. The (real!) comparison, ratio and root tests apply to the series

∑

5. Absolute convergence implies convergence; moreover

∑

≤

∑

Proof. 1. This is Theorem 2.8, part 1.

2, 3. These follow from 1 and the corresponding results for the real series

∑

4. This requires no proof:

∑

is a series of non-negative real numbers, so the real version apply!

5. Since

≤

, the (real) comparison test says that

∑

and

∑

are absolutely con-

vergent and thus convergent. By part 1,

∑

converges. Finally, apply the triangle inequality



∑

k=0



≤

∑

k=0

≤

∞

∑

n=0

and take limits as m → ∞.

Deﬁnition 5.3 (Power Series & Analyticity). A power series centered at z

is a function of the form

p(z) =

∞

∑

n=0

( z −z

)

and the coefﬁcients a

are constants)

A function f : D → C is analytic if every z

∈ D has a neighborhood on which f (z) equals a power

series centered at z

. That is,

∀z

∈ D, ∃δ > 0, (a

) such that

z − z

< δ =⇒ f (z) =

∞

∑

n=0

( z −z

)

To be analytic at a point z

is to be analytic on some neighborhood of z

Example 5.4 (Geometric series). By the n

term test, the power series

∑

diverges when

≥ 1.

Otherwise, inside the unit circle

< 1 we have z

n+1

→ 0, from which

−zs

= 1 −z

n+1

=⇒ s

1 − z

n+1

1 − z

=⇒

∞

∑

n=0

= lim s

1 − z

In fact f ( z) =

1−z

is analytic on its whole domain C \ {1}: the substitution z 7→

z−z

1−z

= 1) to see

that

1−z

equals a power series centered at z

1 − z

1 −

z−z

1−z

1 − z

∞

∑

n=0



z − z

1 − z



whenever

z − z

1 − z

Regardless of the center z

, observe how the geometric series converges on a disk (radius

1 − z

This behavior in fact happens for every power series, analogous to the interval/radius of convergence

discussion from real analysis.

Theorem 5.5 (Radius of Convergence). Suppose p(z) =

∑

( z −z

)

1. If p(z) converges at z

= z

, then it is absolutely convergent at every

every z satisfying

z − z

−z

2. Deﬁne the radius of convergence R

:= sup



z − z

: p(z) converges



Then p(z) converges absolutely whenever

z − z

< R

, and diverges

whenever

z − z

> R

Proof. 1. By the n

term test, the sequence



( z

−z

)



converges (to 0); it is therefore bounded

by some M ∈ R

. But then

z − z

−z



z − z

−z



≤ Mr

where r =

z − z

−z

< 1

Since

∑

converges, the comparison test says that

∑

z − z

converges.

2. If

z − z

< R

, then ∃z

such that

z − z

−z

≤ R

and p(z

) converges; now apply

part (a). The divergence condition is an exercise.

A power series therefore has a disk of convergence. As in real analysis, we have to test convergence

separately on the boundary circle

z − z

= R

; a key technique for this is Abel’s Test (Exercise 4).

Note particularly the two extreme cases:

• If R

= ∞, the series is absolutely convergent on C.

• If R

= 0, the series converges only when z = z

As in real analysis, we could also compute R

= lim inf

−1/n

, though for us this will mostly

be redundant since the condition observed in Exercise 1 (proved generally in Section 5.3) is often

obvious.

Exercises 5.1. 1. Sketch the disks of convergence with centers z

= −1, 1 + i, 3 − 2i for the function

f (z) =

1−z

(Example 5.4). Complete the following observation:

1 − z

is the distance from z

to the nearest point at which f (z) is

2. By mimicking Example 5.4, ﬁnd a power series centered at z

= i which equals the function

g(z) =

1+iz

. What is its radius of convergence?

3. Revisit the proof of Theorem 5.5.

(a) Complete part 2: If

z − z

> R

, prove that p(z) =

∑

( z −z

)

diverges.

(b) In part 1, explain why we couldn’t simply use the comparison test to say

z − z

−z

=⇒

∑

z − z

converges

(Hint: Think carefully about the hypothesis!)

4. (Abel’s Test) In real analysis, the alternating series test was often used to decide convergence at

the endpoints of an interval of convergence. Here is a generalization to the complex situation.

Consider the power series

∑

where (a

) is a real sequence such that

≥ 0, a

n+1

≤ a

, lim

n→∞

= 0

(a) Write s

( z) =

∑

k=0

for the partial sum and prove that

(1 − z)s

( z) = a

− a

n+1

∑

k=1

− a

k−1

) z

(b) Prove Abel’s Test:

∑

converges on the closed disk

≤ 1, except perhaps when z = 1.

(Hint: Show that

∑

− a

k−1

) z

converges absolutely by comparison with a telescoping series)

∑

(all z ∈ C at which the series converges).

ii. Prove that the real series

∑

cos nθ

converges except when θ is divisible by 2π. For what

values of θ does the series

∑

sin nθ

converge?

(Hint: Use part (i). . . )

(d) Find all values of z for which the series

∑

1+i

(n+i)(4+3i)

( z −1 + 2i)

converges and sketch the

disk of convergence.

(Hint: Let w =

z−1+2i

4+3i

and think about real and imaginary parts)

5.2 Taylor Series and Taylor’s Theorem

The overarching goal of the next two sections is the establishment of a key result:

A function is holomorphic if and only if it is analytic

Otherwise said, f (z) is differentiable on an open domain D if and only if for each z

∈ D there is

some neighborhood of z

on which it equals a power series f (z) =

∑

( z −z

)

If a holomorphic function is to equal a power series, it is natural ask which one? The answer revisits a

familiar deﬁnition and leads to a startling difference between the real and complex cases.

Deﬁnition 5.6. If f (z) is inﬁnitely differentiable at z

, then its Taylor series is the power series

∞

∑

n=0

( z −z

)

∞

∑

n=0

(n)

( z

)

( z −z

)

The Taylor coefﬁcients are a

(n)

)

. A Maclaurin series is a Taylor series centered at z

= 0.

Example (5.4, cont.). On the disk

< 1, the function f (z) =

1−z

has

(n)

( z) =

(1 − z)

n+1

=⇒

∞

∑

n=0

(n)

(0)

∞

∑

n=0

= f (z)

The Maclaurin series is precisely the geometric power series representation

of f (z) on the open disk

< 1!

In complex analysis, this situation is completely general. . .

−1 1

−i

Theorem 5.7 (Taylor’s Theorem). Suppose f (z) is holomorphic on a disk

z − z

< R. Then,

f (z) =

∞

∑

n=0

(n)

( z

)

( z −z

)

whenever

z − z

< R

In comparison to real analysis, this is a very strong statement: in the real case, Taylor’s Theorem is

usually stated with one of several awkward remainder terms, and there exist inﬁnitely differentiable

functions which do not equal their Taylor series (see Exercise 4).

Plainly R cannot be larger than the radius of convergence R

of the Taylor series. If f is entire, then

the result holds for all positive R and the series has inﬁnite radius of convergence.

Examples 5.8. Familiar Maclaurin series are identical to real analysis:

∞

∑

n=0

sin z =

∞

∑

n=0

( −1)

(2n + 1)!

2n+1

cos z =

∞

∑

n=0

( −1)

(2n)!

Since e

, sin z and cos z are entire, we may take R = ∞ in Taylor’s Theorem: each function equals its

Maclaurin series everywhere on C.

Before seeing the proof of Taylor’s Theorem, we apply it to quickly deduce half of our key result.

If f : D → C is holomorphic, then, for every z

∈ D, there exists some disk

z − z

< R on which f

is holomorphic. By Taylor’s Theorem, f (z) equals its Taylor series on this disk. Otherwise said:

Corollary 5.9. Every holomorphic function is analytic.

We’ll obtain the converse in the next section (Corollary 5.15).

Why is Taylor’s Theorem so much more speciﬁc in complex analysis? The reason is that we have can

apply a powerful tool unavailable in real analysis: Cauchy’s integral formula.

Proof of Taylor’s Theorem. By relabelling

f (z) = f (z − z

), it is enough to prove when z

= 0, that is

for Maclaurin series. We therefore suppose f (z) is holomorphic when

< R.

Let w be given where

< R. Apply the geometric series formula (Exam-

ple 5.4), to see that if z = 0, then

n−1

∑

k=0





1 −





z − w

=⇒

z − w

n−1

∑

k=0

k+1

z − w





Let C

be a circle centered at the origin with radius satisfying

< r < R.

Since 0 and w both lie inside C

, we may apply Cauchy’s integral formula

twice, using the previous expression to substitute.

f (w) =

2πi

f (z)

z − w

dz (Cauchy for C

around w)

n−1

∑

k=0

2πi

f (z)

k+1

dz +

2πi

f (z)

( z −w)

dz (subsititute for

z−w

)

n−1

∑

k=0

(k)

(0)

2πi

f (z)

( z −w)

dz (Cauchy for C

around 0)

To ﬁnish the proof, we need to control the ﬁnal integral. Since f is holomorphic on C

, it is bounded

by some M > 0. Moreover, for z ∈ C

we have

z − w

≥

−

= r −

. It follows that



f (w) −

n−1

∑

k=0

(k)

(0)



2π



f (z)

( z −w)



≤

M · 2πr

2πr

(r −

)

r −





Since

< r, this last converges to zero. Otherwise said

f (w) = lim

n→∞

n−1

∑

k=0

(k)

(0)

∞

∑

n=0

(n)

(0)

so that f (w) equals its Maclaurin series whenever

< R.

Example 5.10. The principal logarithm f (z) = Log z is holomorphic on the open disk

z − i

< 1.

Whenever n ≥ 1, we have

(n)

(i) =

( −1)

n−1

( n −1)!



z=i

= −i

( n −1)!

By Taylor’s Theorem, f (z) equals its Taylor series on the disk:

Log z = Log i −

∞

∑

n=1

( z −i)

πi

−

∞

∑

n=1

(iz + 1)

−1 0 1

In case you’re feeling skeptical, convergence inside the disk (

z − i

< 1) can be veriﬁed directly using

the comparison test:

z − i

= r < 1 =⇒

z − i

≤ r

=⇒

∑

( z −i)

converges absolutely

At z = 0, we recognize the divergent harmonic series

∑

, so the radius of convergence of the Taylor

series is in fact R

= 1. Exercise 3 shows that the series converges everywhere else on the boundary

circle

z − i

= 1.

Exercises 5.2. 1. (a) Compute the Maclaurin series of cos z directly from the deﬁnition.

(b) Evaluate the Taylor series of sin z about z

and conﬁrm that it equals your answer to

part (a) when z is replaced with z −

= i.

2. Consider f (z) =

. For any z

= 0, ﬁnd the Taylor series of f (z) about z

. What is its disk of

convergence?

3. Use Abel’s test (Exercise 5.1.4) to verify that the Taylor series for Log z centered at z

= i (Ex-

ample 5.10) converges everywhere on the boundary circle

z − i

= 1, except when z = 0.

4. Consider the function

f (z) =

(

−1/z

if z = 0

0 if z = 0

When z ∈ R this provides the classic example of an inﬁnitely differentiable function whose

Maclaurin series (being identically zero) does not equal the original function except at the ori-

gin. When z ∈ C, explain why f (z) does not contradict Taylor’s Theorem.

5.3 Uniform Convergence: Continuity, Integrability and Differentiability

As in real analysis, we’d like to establish some useful facts:

Uniqueness of representation Equal power series have identical coefﬁcients. Otherwise said, on a

given disk

z − z

< R, a function can only equal at most one power series

∑

( z − z

)

(its

Taylor series).

Computability Power series are continuous, differentiable and integrable inside their disk of con-

vergence. They can moreover be differentiated and integrated term-by-term like polynomials.

The arguments are intertwined. Since the ﬁrst results are identical to the real case, we will be brief,

postponing all examples until the end. The critical ingredient is uniform convergence.

Deﬁnition 5.11. Suppose f (z) =

∑

( z − z

)

is a power series with n

partial sum s

( z) and

remainder ρ

( z) = f (z) −s

( z). We say that the series converges uniformly on a domain D if

∀ϵ > 0, ∃N such that n > N, z ∈ D =⇒

( z)

< ϵ

Uniformity means that N = N(ϵ) is independent of the location z ∈ D. If N = N(ϵ, z) were permitted

to depend on z, we’d refer to the convergence as pointwise.

Theorem 5.12. Let R

be the radius of convergence of a power series centered at z

. If R

< R

, then

the series converges uniformly on the closed disk

z − z

≤ R

This ﬁrst argument also applies in real analysis.

Proof. As preparation, suppose z

satisﬁes

−z

= R

. Since R

< R

, the series converges abso-

lutely at z

(Theorem 5.5). Denote by σ

the n

remainder of this absolutely convergent series:

∞

∑

k=n+1

−z

∞

∑

k=n+1

Now suppose ϵ > 0 is given. Since the above series converges, the

-term test says that lim

n→∞

= 0: that is,

∃N such that n > N =⇒ σ

< ϵ

By the comparison test, if z satisﬁes

z − z

≤ R

, then

( z)



∞

∑

k=n+1

( z −z

)



≤

∞

∑

k=n+1

z − z

≤

∞

∑

k=n+1

= σ

< ϵ

Since N depends only on ϵ (not on z), we conclude that the convergence is uniform.

That R

is strictly less than the radius of convergence R

is important. In Exercise 9, we’ll see that a

power series need not converge uniformly on the full open disk of convergence

z − z

< R

It looks as if N might also depend on the choice of z

in the ﬁrst line. However, any suitable z

produces the same

value R

−z

and thus the same sequence (σ

). It is from the convergence of this sequence that we get N.

Theorem 5.13 (Continuity). Suppose f (z) =

∑

( z −z

)

has radius of convergence R

. Then f (z)

is continuous on the open disk of convergence

z − z

< R

This is identical to the famous

-proof encountered in real analysis.

Proof. Fix w and R

such that

w − z

< R

. Let ϵ > 0 be given. Observe:

• Uniform convergence whenever

z − z

≤ R

(Theorem 5.12):

∃N such that n > N =⇒

( z)

and

( w)

• Openness and continuity (s

is a polynomial!): for any n > N,

∃δ > 0 such that

z − w

< δ =⇒

(

z − z

< R

( z) −s

( w)

Put everything together to see that f (z) is continuous at w: for any n > N,

z − w

< δ =⇒

f (z) − f (w)

f (z) − s

( z) + s

( z) −s

( w) + s

( w) − f (w)

≤

( z)

( z) −s

( w)

< ϵ

The discussion now differs from the real approach via our employ of contour integrals. The remaining

results follow from a general version of term-by-term integration.

Theorem 5.14. Suppose f (z) =

∑

( z −z

)

has radius of convergence R

Let g(z) be continuous on some contour C in the open disk of convergence

z − z

< R

. We may then integrate term-by-term:

g(z) f (z) dz =

∞

∑

n=0

g(z)(z − z

)

Proof. The integral

g(z) f (z) dz exists since f , g are continuous on C. Since C is a compact set:

• C lies in some closed disk

z − z

≤ R

< R

on which the series f (z) converges uniformly.

• g(z) is bounded on C by some M > 0.

Let C have length L and let ϵ > 0 be given. Since f (z) converges uniformly when

z − z

≤ R

∃N such that n > N =⇒

( z)

Now take integrals and moduli; if n > N, then



g(z) f (z) dz −

∑

k=0

g(z)(z − z

)



g(z)

f (z) −

∑

k=0

( z −z

)



g(z)ρ

( z) dz



< M ·

· L = ϵ

Everything we want now follows by choosing speciﬁc functions g(z) in Theorem 5.14!

Corollary 5.15. Suppose f (z) =

∑

( z −z

)

has positive radius of convergence R

1. (Term-by-term integration) Let g(z) = 1 to see that

f (z) dz =

∞

∑

n=0

( z −z

)

dz =

∞

∑

n=0

n + 1

( z −z

)

n+1



C(end)

C(start)

2. (Holomorphicity) By part 1,

f is path-independent for any contour C in the open disk of

convergence. We conclude (Summary, page 57) that f is holomorphic on said disk. In particular,

every analytic function is holomorphic.

3. (Term-by-term differentiation) Given

w − z

< R

, let g(z) =

2πi(z−w)

and apply Cauchy’s integral formula on a small circle around w:

′

( w) =

2πi

f (z)

( z −w)

dz =

∑

2πi

( z −z

)

( z −w)

∑



z=w

( z −z

)

∑

n(w −z

)

n−1

4. (Unique representation) The power series is the Taylor series of f (z): that is, a

(n)

)

In Exercise 7 proves unique representation. Since analytic and holomorphic are now equivalent, we’ll

retire the latter term for the rest of these notes.

Examples 5.16. We may now compute Taylor & Maclaurin series using algebra, integration and

differentiation: if a function equals a series, that’s the one we want regardless of how we found it!

1. f (z) = z

= z

∞

∑

n=0

)

∞

∑

n=0

2n+3

is the Maclaurin series of f (z). Since the radius of con-

vergence is inﬁnite, the function equals its Maclaurin series everywhere on C.

2. The function f ( z) =

(

sin z

if z = 0

1 if z = 0

is entire since it equals the power series

∞

∑

n=0

(−1)

(2n+1)!

3. We ﬁnd the Maclaurin series of f (z) =

+16i

algebraically:

f (z) =

16i



1 −

−16i



16i

∞

∑

n=0



−16i



∞

∑

n=0

n−1

n+1

This converges whenever



−16i



< 1 ⇐⇒

< 2, equalling the

distance from the center to the nearest point(s) where f (z) is unde-

ﬁned. If C is the straight line from z = 0 to z = 1 + i, then

−2 −1 1 2

−2i

−i

f (z) dz =

∞

∑

n=0

n−1

n+1

dz =

∞

∑

n=0

n−1

(1 + i)

4n+1

n+1

(4n + 1)

∞

∑

n=0

1 −i

16( 4n + 1)



−i



Zeros of Analytic Functions & the Analytic Continuation

The theory we’ve developed has huge implications. As an example, we apply our discussion to the

zeros of an analytic function. If a function is analytic at z

, then (Deﬁnition 5.3, Theorem 5.7) there is

some disk on which it equals its Taylor series:

z − z

< R =⇒ f (z) =

∞

∑

n=0

(n)

( z

)

( z −z

)

(∗)

Suppose f (z

) = 0. We ask if/when the Taylor series starts to get interesting. . .

Deﬁnition 5.17. Suppose z

is a zero of an analytic function f (z).

1. We say that z

is a zero of order m ≥ 1 if f

(m)

( z

) is the ﬁrst non-zero derivative. A zero of order

1 is also called a simple zero.

2. If all derivatives are zero, z

is a non-isolated zero: plainly f (z) ≡ 0 on some disk

z − z

< R.

We are used to the idea of a polynomial having a zero z

if and only if we can factorize out z − z

The tight link-up with Taylor series makes essentially this observation hold for any analytic function!

Theorem 5.18. An analytic function f (z) has a zero z

of order m if and only if f (z) = (z −z

)

ψ(z)

where ψ(z) is analytic at z

and ψ(z

) = 0. Indeed, on some disk

z − z

< R,

f (z) =

∞

∑

n=m

(n)

( z

)

( z −z

)

= (z − z

)

ψ(z)

Proof. (⇒) Suppose z

is a zero of order m. Deﬁne

ψ(z) =











f (z)

( z −z

)

whenever z = z

(m)

( z

)

when z = z

On the disk

z − z

< R, (∗) says that this function equals the power series

ψ(z) =

∞

∑

n=m

(n)

( z

)

( z −z

)

n−m

and is therefore analytic. Moreover, f (z) = (z −z

)

ψ(z) everywhere.

(⇐) If ψ(z) is analytic at z

, then it equals its Taylor series on some disk

z − z

< R. But then

f (z) := (z −z

)

ψ(z) =

∞

∑

n=0m

(n)

( z

)

( z −z

)

n+m

is a power series on this disk. By uniqueness of representation (Corollary 5.15), this is the Taylor

series of f (z). If ψ(z

) = 0, then f

(m)

( z

) = m!ψ(z

) is plainly the ﬁrst non-zero derivative.

Examples 5.19. 1. f ( z) = z

( z −2i)

= z

( z) = (z − 2i)

( z) has two zeros:

Order four at z

= 0: ψ

( z) = (z − 2i)

is analytic with ψ

(0) = −1024 = 0.

Order ten at z

= 2i: ψ

( z) = z

is analytic with ψ

(2i) = 16 = 0.

2. g(z) = 17(z −4i)

sin z has a zero of order three at 4i, and simple zeros at each integer multiple

of π. For instance, the Taylor series for sine centered at z

= π yields

g(z) = 17(z −4i)

sin(z − π + π) = −17(z −4i)

sin(z − π)

= −17(z −4i)

∞

∑

n=0

( −1)

(2n + 1)!

( z −π)

2n+1

= (z − π)ψ(z)

where ψ(z) = −17(z −4i)

∞

∑

n=0

(−1)

(2n+1)!

( z −π)

is analytic and ψ(π) = −17(π −4i)

= 0.

The examples are typical in that the zeros of an analytic function tend to be isolated. Indeed, analytic

functions with non-isolated zeros are very boring. . .

Theorem 5.20. Suppose z

is a zero of an analytic function f (z).

1. If z

has order m, then there exists a punctured disk 0 <

z − z

< R on which f (z) = 0.

Otherwise said: z

is isolated.

2. If z

is non-isolated and the domain D of f (z) is open and connected, then f (z) ≡ 0 on D.

Example 5.21. Analyticity is crucial! For instance, the non-analytic function f (z) = z + z = 2x is

zero precisely along the real axis: the zero is not isolated but neither is the function identically zero

on any open domain. There really is something to prove here!

Proof. 1. Write f (z) = (z −z

)

ψ(z), where ψ is analytic and non-zero at z

. Since ψ is continuous,

it is non-zero on some disk

z − z

< R.

2. As observed in Deﬁnition 5.17, f ( z) ≡ 0 on some disk

z − z

< R.

To ﬁnish, we extend by an argument similar to that used to prove the Maximum Modulus

Principle (Theorem 4.32). Given any w = z

, join z

to w by a path in D which we cover by

ﬁnitely many disks; since f (z) ≡ 0 on all these disks, we conclude that f (w) = 0.

By applying part 2 to a function f − g, we obtain the proof of a powerful and previously heralded

result.

Corollary 5.22 (Theorem 2.36—Analytic Continuation). Suppose f (z), g( z) are analytic on an open

connected domain D, and that f (z) = g(z) on some contour C in D. Then f (z) = g(z) on D.

As the proof of Theorem 5.20 shows, it is enough that the set of points where f (z) = g(z) have a limit

point z

∈ D (i.e., that f − g has a non-isolated zero).

Example 5.23. The function g(z) =

1−z

is analytic on the punctured plane D = C \ {1} and equals

f (z) =

∑

on the disk

< 1. Even though the power series f (z) does not converge outside the

disk, we conclude that g(z) is the unique analytic continuation of f ( z) to the larger domain D.

Exercises 5.3. 1. Find a power series representation and the radius of convergence:

(a) f (z) =

4−z

about z

= 0 (b) f (z) = z sin z

about z

= 0

iπ

2. (a) Without computing derivatives, ﬁnd the Taylor series for f (z) =

about z

= 0.

(b) Differentiate your answer term-by-term to ﬁnd the Taylor series of

about z

3. By expressing f (z) as a Maclaurin series, show that it is entire:

f (z) =

(

(1 − cos z) if z = 0

if z = 0

4. (a) By integrating the Taylor series for z

−1

about z

= 1, prove that

Log z =

∞

∑

n=1

( −1)

n+1

( z −1)

whenever

z − 1

< 1

(b) Prove that the following function is analytic on the domain 0 <

, Arg z ∈ (−π, π):

f (z) =

(

Log z

z−1

if z = 1

1 if z = 1

5. Suppose f ( z) is analytic and non-constant at z

( f

′

( z

) = 0). Prove that

∃R > 0 such that 0 <

z − z

< ϵ =⇒ f (z) = f (z

)

To what extent can you weaken the hypothesis f

′

( z

) = 0?

6. Consider the Maclaurin series f (z) =

∑

∞

n=0

( −1)

on the disk

< 1. Show that h(z) =

is the analytic continuation of f (z) to C \ {i, −i}.

7. (a) Prove part 4 of Corollary 5.15: if f (z) =

∑

( z −z

)

, prove that f

(m)

( z

) = m!a

so that

the series really is the Taylor series of f (z).

(Hint: let g(z) =

2πi(z−z

)

m+1

in Theorem 5.14)

(b) Explain carefully why every power series deﬁnes an analytic function.

(Think carefully about the deﬁnitions and what we’ve proved in the last two sections!)

8. Suppose that the series

∑

( z −z

)

has radius of convergence R

and that f (z) =

∑

( z −z

)

whenever

z − z

< R

. Prove that

= inf



z − z

: f (z) non-analytic or undeﬁned at



is essentially the distance from z

to the nearest point at which f (z) is non-analytic)

9. (Hard) Consider f (z) =

1−z

∞

∑

n=0

< 1.

(a) Verify uniform convergence when

≤ R

< 1: given ϵ > 0, ﬁnd an explicit N such that

n > N =⇒

( z)



f (z) −

∑

k=0



< ϵ whenever

≤ R

(b) Prove that f (z) is not uniformly convergent on

< 1 (Let ϵ = 1, then. . . ).

6 Laurent Series, Residues and Poles

While Taylor series are undeniably useful, they also have weaknesses, not least because their domains

are disks. We motivate a more general construction with an example.

Example 6.1. f (z) =

z(2−z)

can be written as a Taylor series centered at z = 1:

f (z) =

1 − (z −1)

∞

∑

n=0

( z −1)

whenever

z − 1

< 1

However, we are also interested in the behavior of f (z) near the points

z = 0, 2. Due to their disk-domains, we can’t use Taylor series to loop

around these points.

As an alternative, start with the Maclaurin series for

2−z

(valid when

< 2) and divide through by z to obtain a new expression:

−2 −1 1 2

−i

−2i

f (z) =

2z(1 −

)

∞

∑

n=0





∞

∑

n=−1

n+2

+ ···

By construction, this ‘series with negative terms’ is valid on the punctured disk 0 <

< 2.

6.1 Laurent Series

Series with negative terms are very useful in complex analysis. In Example 6.1, the larger domain

encircling the origin provides an obvious advantage over the Taylor series. That such a series con-

verges on an annulus rather than a disk is typical. We omit the full details, but by splitting a general

series into positive and negative powers, substituting w = (z − z

)

−1

, and applying Theorems 5.5,

5.12 and 5.13 to the resulting power series in w

∞

∑

n=−∞

( z −z

)

∞

∑

n=0

( z −z

)

−1

∑

n=−∞

( z −z

)

∞

∑

n=0

( z −z

)

∞

∑

n=1

−n

one obtains the natural extensions of these results.

Theorem 6.2 (Annulus of convergence). Given a series

∞

∑

n=−∞

( z −z

)

, deﬁne

= inf



z − z

: f (z) converges



= sup



z − z

: f (z) converges



1. The series converges absolutely to a continuous function on the

(open) annulus of convergence R

z − z

< R

. The conver-

gence is moreover uniform on any closed sub-annulus.

2. If

z − z

< R

z − z

> R

, the series diverges.

As with power series, convergence on the boundary circles must be checked separately.

As in Example 6.1, the inner radius can be R

= 0 and the domain a punctured disk. Moreover, the

outer radius can be R

= ∞.

As with Taylor series, our goal is often to ﬁnd a series representation of a given function.

Deﬁnition 6.3. Suppose f (z) is analytic on an annulus R

z − z

< R

Its Laurent series centered at z

is the expression

∞

∑

n=−∞

( z −z

)

where a

2πi

f (z)

( z −z

)

n+1

and C is any simple closed contour encircling z

within the annulus.

• If you prefer, write

∞

∑

n=0

( z −z

)

∞

∑

n=1

(z−z

)

where b

2πi

( z −z

)

n−1

f (z) dz.

• The coefﬁcients a

are independent of the choice of contour C.

To see this, suppose D is another simple closed curve encircling z

and choose a circle E outside both C and D. Since

f (z)

(z−z

)

n+1

is analytic

on the annulus, two applications of Cauchy–Goursat yield

f (z)

( z −z

)

n+1

dz =

f (z)

( z −z

)

n+1

dz =

f (z)

( z −z

)

n+1

• The Laurent coefﬁcients are inspired by our discussion of Taylor series. If f (z) happens to be

analytic on the disk

z − z

< R

, then its Laurent series equals the Taylor series:

(

n ≥ 0 =⇒ a

(n)

)

(Cauchy’s integral formula)

n < 0 =⇒ a

= 0 (Cauchy–Goursat)

Typically, however, f (z) is not deﬁned (or analytic) at z

. Laurent series therefore generalize

and replace Taylor series in such situations.

Example (6.1, cont.). f (z) =

z(2−z)

is certainly analytic on the annulus 0 <

< 2.

By ﬁrst writing f (z) =



2−z



using partial fractions, we compute

the Laurent series for f on the annulus using the unit circle centered at

the origin:

2πi

f (z)

n+1

dz =

4πi

n+2

(2 − z)z

n+1

The ﬁrst integral evaluates to

when n = −1 and to zero otherwise.

The second integral may be found using Cauchy’s integral formula:

−2 −1 1 2

−i

−2i

4πi

(2 − z)z

n+1

dz =







0 if n ≤ −1

2(n!)



z=0

(2 − z)

−1

n+2

if n ≥ 0

We conclude that a

n+2

: the Laurent series of f (z) is precisely the series computed previously!

The Laurent series centered at z

= 2 can be computed similarly, both using the integral method and

our original approach.

The example is typical. Computing a Laurent series directly from the deﬁnition is usually ugly given

how many contour integrals must be evaluated! Thankfully, as we’ll see shortly, all the standard facts

regarding Taylor series translate to this new situation. In particular, if f (z) =

∑

( z − z

)

equals a

series with negative terms, then the series will turn out to be the Laurent series of f (z). We’ll deal

with the theory shortly, but ﬁrst a few more examples following from this observation to get used to

the idea.

Examples 6.4. 1. On the disk

< 1, we have the Maclaurin series

z − i

−i(1 −

)

= i

∞

∑

n=0

( −iz)

= i + z −iz

−z

+ iz

+ ···

On the annulus

> 1, we have the Laurent series

z − i

(1 −

)

∞

∑

n=0

−n−1

−

+ ···

2. On the 1 <

< 2, we have the Laurent series

(2 − z)(1 + z)

2 − z

1 + z

2( 1 −

)

z(1 +

)

∞

∑

n=0





∞

∑

m=0

( −z)

−m

= ··· + z

−3

−z

−2

+ z

−1

z +

+ ···

−2 −1 1 2

−i

−2i

−2 −1 1 2

−i

−2i

3. Since e

∑

is valid on the entire complex plane, we obtain the Laurent series expansion

∞

∑

n=0

−n

= 1 +

+ ···

valid on the punctured plane z = 0.

4. Again by substituting in a known Maclaurin series, we obtain another Laurent series valid on

the punctured plane z = 0:

sin z

∞

∑

n=0

( −1)

(2n + 1)!

4n−5

= z

−5

−

−1

120

−

5040

+ ···

5. Multiplying term-by-term, and since we need both Maclaurin series to be valid, we obtain (the

ﬁrst few terms of) a Laurent series valid on the punctured disk 0 <

< 1:

z(z −1)(z −2i)

∞

∑

n=0

( −1)

∞

∑

m=0







1 − z + z

−z

+ ···





1 +

z −

−

+ ···





−1 +





−



z +



−



+ ···

We now state and prove the main properties of Laurent series. These are very similar to the cor-

responding statements & arguments for Taylor series; the additional challenge mostly comes from

keeping track of two series at once.

Theorem 6.5 (Laurent’s Theorem). An analytic function on an open annulus equals its Laurent

series.

Proof. It is enough to prove when the annulus is centered at z

= 0. Let w in the annulus be given.

Since the annulus is open, we may choose three non-overlapping circles

α, β, γ with radii R

, R

as in the picture:

• γ a small circle centered at w inside the annulus;

• α, β centered at 0, α inside and β outside w (thus R

< R

Since

f (z)

z−w

is analytic on/inside the closed region with boundaries α, β, γ,

Cauchy–Goursat and the integral formula tell us that



−



f (z)

z − w

dz = 0 =⇒ f (w) =

2πi

f (z)

z − w

dz =

2πi



−



f (z)

z − w

dz (∗)

As in the proof of Taylor’s theorem, we expand

z−w

, this time in two ways:

z − w

n−1

∑

k=0





z − w





= −

∑

k=1





k−1

z − w





These expansions allow us to attack the two integrals in (∗):

2πi

f (z)

z − w

dz =

n−1

∑

k=0

2πi

f (z)

k+1

| {z }

2πi

f (z)

( z −w)

−1

2πi

f (z)

z − w

dz =

∑

k=1

2πiw

k−1

f (z) dz

| {z }

−k

−

2πiw

f (z)

z − w

We ﬁnish by summing and estimating, using the facts that z ∈ α ∪ β =⇒

z − w

> R

, and that f (z)

is bounded (by some M > 0) on the closed bounded annulus between α, β:



f (w) −

n−1

∑

k=−n



2πi

f (z)

( z −w)

dz −

2πiw

f (z)

z − w



△

≤

2π



f (z)

( z −w)



2π



f (z)

z − w



≤

2π

·2πR

2π

·2πR

(Theorem 4.16)







+ R







−−−→

n→∞

Our ﬁnal corollary summarizes the remaining core properties of Laurent series.

Corollary 6.6. Let f (z) =

∑

( z −z

)

be a series. Then:

1. (Term-by-term Integration) If g(z) is continuous on a contour C lying inside the annulus, then

g(z) f (z) dz =

∞

∑

n=−∞

g(z)(z − z

)

In particular, f (z) may be integrated term-by-term along C.

2. (Analyticity/Derivatives) f (z) is analytic on the annulus and f

′

( z) =

∞

∑

n=−∞

n(z −z

)

n−1

3. (Uniqueness)

∑

( z −z

)

is the Laurent series of f (z) (as in Deﬁnition 6.3).

The remainder are the analogues of Theorems 5.14 and Corollary 5.15: some details are in Exercise 7.

Examples 6.7. The corollary formally justiﬁes Examples 6.4. Here are two extensions.

1. In accordance with part 2 of the corollary,

1/z

= −

1/z

∞

∑

n=0

−n

∞

∑

n=1

−1−n

( n −1)!

∞

∑

n=1

−(n−1)

( n −1)!

∞

∑

n=0

−n

24z

+ ···

We need not have differentiated: by part 3 we could instead simply multiply the known series

for e

1/z

term-by-term to obtain the result!

2. We use part 1 to compute the integral around a contour C encircling the origin:

sin z

dz =

∞

∑

n=0

( −1)

(2n + 1)!

4n−5

dz =

( −1)

(2 + 1)!

4−5

= −

πi

since all but one of the integrals evaluates to zero.

Exercises 6.1. 1. (Example 6.1, cont.) The function f (z) =

z(2−z)

is analytic everywhere except at

z = 0 and 2.

(a) Use Deﬁnition 6.3 to directly compute the Laurent series of f (z) on the punctured disk

0 <

z − 2

< 2: use the circle radius 1 centered at z

= 2.

(b) Use the Taylor series of

centered at z

= 2 to more rapidly compute the Laurent series in

part (a).

2. Find a Laurent series representation for each function. Also ﬁnd

f (z) dz where C is a simple

closed curve in the given domain encircling the origin.

(a) f (z) =

when

> 0 (b) f (z) = cos

when

> 0

1+z

when 1 <

(Hint: let w = z

−1

)

3. On each domain, ﬁnd a Laurent series centered at z

= 0 for the function

f (z) =

z(z −2i)



−

z − 2i



(a) D

= {z : 0 <

< 2} (b) D

= {z :

> 2} (again let w = z

−1

)

4. Repeat the previous question for

f (z) =

1 −2i

( z −1)(z − 2i)

z − 1

−

z − 2i

Also ﬁnd

f (z) dz where C is a simple closed curve in the given domain encircling the origin.

(a) D

= {z : 0 <

< 1} (b) D

= {z : 1 <

< 2} (c) D

= {z :

> 2}

5. Show that when 0 <

z − 1

< 2, we have

( z −1)(z − 3)

= −

2(z −1)

−3

∞

∑

n=0

( z −1)

n+2

6. Let a be complex number. Show that

z − a

∞

∑

n=1

whenever

7. Suppose f ( z) =

∑

( z −z

)

is a series as in Corollary 6.6.

(a) Suppose part 1 has been proved. Explain why the function f (z) −a

−1

( z −z

)

−1

is analytic

on the annulus. Hence conclude that f ( z) is analytic on the annulus.

(This is different to Corollary 5.15 since a

−1

( z −z

)

−1

has no anti-derivative on the annulus!)

(b) In order to mimic the proof of Corollary 5.15 to show that f (z) is differentiable term-by-

term, what properties must the curve C have?

(Hint: Recall Exercise 5.3.7 - the same hint works!).

6.2 Singularities and Cauchy’s Residue Theorem

One practical purpose of this chapter is the development of an efﬁcient method for computing con-

tour integrals of analytic functions. Essentially everything depends on a few crucial facts:

1. If C encircles z

, then

( z −z

)

dz = 0 when n = −1, and

( z −z

)

−1

dz = 2πi.

2. Cauchy–Goursat (Theorem 4.19) ( f analytic on/inside C =⇒

f = 0) and its extension

(Corollary 4.21) to regions with ﬁnitely many interior boundary curves.

3. The existence of Taylor/Laurent series expansions.

Make sure you are familiar with these ideas before proceeding!

Example 6.8. The rational function

f (z) =

z − 2

z − 1 −2i

is analytic except at the points z

= 0, z

= 2, z

= 1 + 2i.

Several curves are drawn. Integrating round the small circle C

easy using the ﬁrst fact (above):

f (z) dz = 3

5i dz

z − 2

z − 1 −2i

= 3 ·2πi + 0 + 0 + 0 = 6πi

−1 1 2 3

−i

since the latter three integrands are analytic on/inside C

. Similarly,

f (z) dz = 5i

z − 2

= −10π,

f (z) dz =

z − 1 −2i

= 2πi

The curves C

and C

are more interesting. Since f (z) is analytic on and between C

and C

Cauchy–Goursat tells us that

f (z) dz =

f (z) dz +

f (z) dz = −10π + 2πi = 2π(i − 5)

appears trickier, though it becomes easy once you visualize it as two contours: the ﬁrst encircles

= 2 counter-clockwise while the second passes clockwise around z

= 0. We conclude that

f (z) dz =

f (z) dz −

f (z) dz = −10π − 6πi = −2π(5 + 3i)

The example suggests that the value of any integral round a simple closed contour can be evaluated

as a linear combination

f (z) dz = λ

f + λ

f = 6πiλ

−10πλ

+ 2πiλ

where λ

denotes the number of times C orbits z

in a counter-clockwise direction.

To properly develop the idea in the example, some new language is helpful.

Deﬁnition 6.9 (Isolated Singularities). We say that z

is an isolated singularity of f (z) if the function

is analytic on a punctured disk 0 <

z − z

< R, but not at z

itself.

By Laurent’s Theorem (6.5), f (z) equals its Laurent series on this domain:

f (z) =

∞

∑

n=−∞

( z −z

)

where a

2πi

f (z)

( z −z

)

n+1

and C encircles z

. The residue of f (z) at z

is the −1-coefﬁcient

Res

z=z

f (z) = a

−1

2πi

f (z) dz

The type of isolated singularity is determined by the rest of the Laurent series:

Removable Singularity The Laurent series is a Taylor series (the residue is necessarily zero). There are

no negative powers; the series f (z) extends analytically to z

Pole of order m The highest negative power in the Laurent series is (z − z

)

−m

. A pole of order 1 is

typically called a simple pole, order 2 a double pole, etc.

Essential Singularity The Laurent series has inﬁnitely many negative terms.

Examples 6.10. 1. The series f (z) =

∞

∑

n=0

−n

( z −2i)

deﬁned on the punctured disk 0 <

z − 2i

< 3

has a removable singularity at z

= 2i (residue Res

z=2i

f (z) = 0). Indeed the function is a geometric

series and thus equals

f (z) =

1 −

z−2i

3 + 2i −z

on the punctured disk. Certainly this extends analytically to f (2i) = 1.

2. e

1/z

∞

∑

n=0

−n

= 1 +

+ ··· has an essential singularity at zero with Res

z=0

1/z

= 1.

3. (Example 6.8) The function f (z) =

z−2

z−1−2i

is analytic on the punctured disk

0 <

< 0.3 (inside the circle C

). Since

z−2

z−1−2i

is also analytic at zero, the Laurent series

of f (z) about z

= 0 has the form

f (z) =

∞

∑

n=0

We conclude that f (z) has a pole of order 2 at z

= 0 and residue Res

z=0

f (z) = 3. Similarly, f (z)

has simple poles (order 1) at z

= 2 and z

= 1 + 2i with

Res

z=2

f (z) = 5i, Res

z=1+2i

f (z) = 1

Theorem 6.11 (Cauchy’s Residue Theorem). Let f (z) be analytic on

and inside a simple closed contour C, except at ﬁnitely many singular-

ities z

, . . . , z

. Then

f (z) dz = 2πi

∑

k=1

Res

z=z

f (z)

More generally, if C is closed and orbits the point z

counter-clockwise

times (the winding number of C about z

), then

f (z) dz = 2πi

∑

k=1

Res

z=z

f (z)

···

Proof. First, center a circle C

at each z

such that no other singularities lie on/inside C

and apply

Cauchy–Goursat. The general argument follows by considering paths like C

in Example 6.8.

Examples 6.12. Let C be the circle with radius 4 centered at the origin and E the green curve drawn.

1. The rational function f (z) =

3(1+iz)

z(z−3i)

has simple poles at z

= 0 and z

= 3i. There are several

ways to compute the residues and thus the integrals

f (z)dz and

f (z)dz.

Partial Fractions It’s just high-school algebra!

f (z) =

z − 3i

=⇒ Res

z=0

f (z) = i, Res

z=3i

f (z) = 2i

=⇒

f (z) dz = 2πi(i + 2i) = −6π

The curve E orbits twice clockwise around z

and once counter-

clockwise around z

. Thus

f (z) dz = 2πi



−2 Res

z=0

f (z) + Res

z=3i

f (z)



= 0

−4 −2 2 4

−2i

−4i

Laurent series Remember that we only need to ﬁnd the z

−1

terms for the residues!

3( 1 + iz)

z(z −3i)

i −z

z(1 −

)



−1



∞

∑

n=0





+ power series

3( 1 + iz)

z(z −3i)

z − 3i + 2i

(1 +

z−3i

)(z −3i)



z − 3i

+ 1



∞

∑

n=0



3i −z



z − 3i

+ power series

Cauchy’s integral formula Let C

be a small circle around z

, then

Res

z=0

f (z) =

2πi

f (z) dz =

2πi

3( 1 + iz)

z(z −3i)

dz =

3( 1 + iz)

z − 3i



z=0

= i

Res

z=3i

f (z) =

2πi

f (z) dz =

2πi

3( 1 + iz)

z(z −3i)

dz =

3( 1 + iz)



z=3i

= 2i

2. f (z) = z

sin

has one isolated singularity at the origin. Using the Maclaurin series for sin z,

we see that this is an essential singularity. Moreover,

sin

∞

∑

n=0

( −1)

(2n + 1)!

1−2n

=⇒

sin

dz = 2πi Res

z=0



sin



= −

πi

Since E loops twice clockwise around the origin, we obtain

sin

dz = 2πi · (−2) Res

z=0



sin



2πi

3. f (z) = 3e

1/z

z−7i

z+1

has an essential singularity at the origin and two simple poles at −1

and 7i. Since the last of these lies outside the curves C, E, it does not contribute to either integral.

Moreover, since E loops twice clockwise around the origin and once clockwise around z

= −1.

We therefore have

f (z) dz = 2πi



Res

z=0

1/z

+ Res

z=−1

z + 1



= 2πi(3 + 2i)

f (z) dz = 2πi



−2 Res

z=0

1/z

+ (−1) Res

z=−1

z + 1



= 2πi(−6 −2i) = 4π(1 −3i)

Non-isolated Singularities

Revisit Deﬁnition 6.9. A point z

is said to be a non-isolated singularity of f (z) if the function is non-

analytic (including undeﬁned) at z

and at some point on every punctured disk 0 <

z − z

< R.

Such a function is necessarily non-analytic at inﬁnitely many points and does not have a Laurent

series centered at z

. Non-isolated singularities typically appear in two ﬂavors.

Examples 6.13. 1. f (z) =



2πi

− 1



−1

has singularities at z

= 0 and at z

for each non-zero

integer n. Each non-zero singularity is isolated (choose e.g., R

(

+1)

in Deﬁnition 6.9). The

limit point z

= 0 is non-isolated: for any R > 0, the punctured disk 0 <

< R contains other

singularities.

−1 0 1

−

−1

2. The square-root function f (z) = z

1/2

has a branch point z

= 0. For

f to be analytic, we need to make a branch cut: for instance the non-

positive real axis for the principal branch.

Since any punctured disk 0 <

< R contains other points of

the branch cut (where f is non-analytic), the branch point is a non-

isolated singularity for any branch of f .

Exercises 6.2. 1. For each type of singularity, what, if anything, can be said about the value of the

residue Res

z=z

f (z)? Choose from ‘Equals zero,’ ‘Non-zero,’ or ‘No restriction’.

(a) Removable singularity. (b) Simple pole.

2. Find the residue at z = 0 of each function:

(a) f (z) =

z + 3z

(b) g(z) = z cos

z − sin z

3. Let C be the circle

= 3. Evaluate the integrals using Cauchy’s residue theorem:

(a)

−z

dz (b)

−z

( z −1)

dz (c)

1/z

dz (d)

z + 1

−2z

4. Suppose a closed contour C loops twice counter-clockwise around z = i and three times clock-

wise around z = 2. Use residues to compute the integral

z + 3

( z −2)

( z −i)

5. Identify the type of singular point of each function and determine the residue:

(a)

1 −cosh z

(b)

1 − e

(c)

( z −1)

6. Suppose f ( z) is analytic at z

and deﬁne g(z) = (z − z

)

−1

f (z). Prove:

(a) If f (z

) = 0, then z

is a simple pole of g(z) with Res

z=z

g(z) = f (z

);

(b) If f (z

) = 0, then z

is a removable singularity of g(z).

7. A function f (z) is said to have an isolated singularity at ∞ if it is analytic on some inﬁnite annulus

> R. On this domain, we have a Laurent series f (z) =

∑

∞

n=−∞

(a) What properties do you think this series should have if we were to describe the singularity

at ∞ to be:

i. Removable? ii. A pole of order m? iii. Essential?

(b) The residue at ∞ of the above function is deﬁned to be

Res

z=∞

f (z) := −Res

z=0



−2

f (z

−1

)



(∗)

i. Prove that Res

z=∞

f (z) = −a

−1

, where a

−1

is the coefﬁcient in the above Laurent series.

ii. Suppose f ( z) is analytic on C except at ﬁnitely many singularities. By substituting

w =

in (∗), prove that the sum of all residues, including at inﬁnity, is zero.

8. Let P(z) and Q(z) be polynomials and assume C is a simple closed contour such that all zeros

of Q(z) lie interior to C.

(a) If deg Q ≥ 2 + deg P, prove that

P(z)

Q(z)

dz = 0

(b) What can you conclude if deg Q = 1 + deg P?

(Hint: Combine Exercises 6 & 7, or try the substitution w =

directly)

6.3 Poles & Zeros

If the order of a pole is known, residues may often be computed quite efﬁciently. For instance, if f (z)

has a pole of order m, Laurent’s Theorem tells us that

f (z) =

∞

∑

n=−m

( z −z

)

= (z − z

)

−m

ϕ(z), where ϕ(z) =

∞

∑

n=0

n−m

( z −z

)

(∗)

Plainly ϕ(z) is analytic at z

and ϕ(z

) = a

−m

= 0. In fact this property categorizes poles of order m

(analogous to Theorem 5.18 for zeros), providing a simple formula for the computation of residues.

Theorem 6.14. A function f ( z) has a pole of order m at z

if and only if f (z) = (z −z

)

−m

ϕ(z) where

ϕ(z) is analytic at z

and ϕ(z

) = 0. In such situations, the residue is a Taylor coefﬁcient of ϕ(z):

f (z) = (z − z

)

−m

∞

∑

n=0

(n)

( z

)

( z −z

)

=⇒ Res

z=z

f (z) =

( m −1)!

(m−1)

( z

)

This specializes to Res

z=z

f (z) = ϕ(z

) for a simple pole.

Proof. The (⇒) direction is above. For the converse: ϕ(z) equals its Taylor series and so (∗) describes

the Laurent series of f (z) = (z − z

)

−m

ϕ(z) (Corollary 6.6, uniqueness), which moreover has a pole

of order m since a

−m

= ϕ(z

) = 0.

Examples 6.15. 1. (Example 6.12 revisited) f (z) =

3(1+iz)

z(z−3i)

has two poles:

Simple pole at z

= 0: Write f (z) = z

−1

( z), where ϕ

( z) =

3(1+iz)

z−3i

. This is analytic and non-

zero at z

= 0, whence

Res

z=0

f (z) = ϕ

(0) =

−3i

= i

Simple pole at z

= 3i: This time, f (z) = (z −3i)

−1

( z), where ϕ

( z) =

3(1+iz)

. Thus

Res

z=3i

f (z) = ϕ

(3i) =

3( 1 −3)

= 2i

2. f (z) =

1−2iz

(z−1)(z−2i)

also has two poles.

Simple pole at z

= 1: Write f (z) = (z −1)

−1

( z), where ϕ

( z) =

1−2iz

(z−2i)

. Thus

Res

z=1

f (z) = ϕ

(1) =

1 −2i

(1 − 2i)

4i −3

Pole of order three at z

= 2i: f (z) = (z −2i)

−1

( z), where ϕ

( z) =

1−2iz

z−1

= −2i +

1−2i

z−1

. Thus

Res

z=2i

f (z) =

(3 − 1)!

′′

(2i) =

1 −2i

( z −1)



z=2i

−1

(2i −1)

3 −4i

The formula works whenever f (z) = (z −z

)

−m

ϕ(z) with ϕ analytic, even if ϕ(z

) = 0. However, a na

ıve application

means you won’t know the order of the pole and you’ll have to differentiate more times than necessary!

As the previous examples show, the method is very effective for rational with low-order poles; as a

bonus, it saves us from partial fractions! Its utility is more variable for other functions. . .

Examples 6.16. 1. The function f (z) =

(z−1)

(z+1)

= (z −1)

−2

ϕ(z) has a double pole at z

= 1, with

Res

z=1

f (z) =

(2 − 1)!

′

(1) =

( z + 1)



z=1

2. Don’t let the denominator fool you! At ﬁrst glance we appear to have a pole of order six:

f (z) =

6 sin z −6z + z

ϕ(z)

=⇒ Res

z=0

f (z) =

(5)

(0) =

120

However, if we apply the Maclaurin series for sine, we instead ﬁnd a simple pole:

f (z) =

∞

∑

n=0

( −1)

(2n + 1)!

2n+1

−6z + z

= z

−6

∞

∑

n=2

6(−1)

(2n + 1)!

2n+1

= z

−1

∞

∑

m=0

6(−1)

(2m + 5)!

20z

840

+ ··· =⇒ Res

z=0

f (z) =

Even though the ﬁrst calculation produced the correct residue (see footnote 16), the function

was incorrect (

ϕ(0) = 0). The correct function is the series ϕ( z) = 6

∞

∑

m=0

(−1)

(2m+5)!

For certain functions with simple poles, the method is even easier.

Corollary 6.17. Suppose p(z), q(z) are analytic at z

, that p(z

) = 0 and that q(z) has a simple zero.

Then f (z) =

p(z)

q(z)

has a simple pole at z

, and

Res

z=z

p(z)

q(z)

p(z

)

′

( z

)

Proof. By Theorem 5.18, q(z) = (z − z

) ψ(z) where ψ is analytic and ψ(z

) = 0. Take ϕ(z) =

p(z)

ψ(z)

Theorem 6.14 (ϕ is analytic and non-zero at z

), to see that

f (z) =

p(z)

q(z)

= (z − z

)

−1

p(z)

ψ(z)

and Res

z=z

p(z)

q(z)

= ϕ(z

) =

p(z

)

ψ(z

)

p(z

)

′

( z

)

Examples 6.18. 1. The function f (z) =

p(z)

q(z)

sin z

(z−2i)(z+2i)

has simple poles at ±2i with

Res

z=2i

f (z) = Res

z=−2i

f (z) =

p(±2i)

′

( ±2i)

sin 2i

( e

−e

−2

) =

sinh 2

The reciprocal g(z) =

q(z)

p(z)

sin z

has simple poles at z = nπ for every n ∈ Z, with

Res

z=nπ

+ 4

sin z

q(nπ)

′

( nπ)

+ 4

cos nπ

= (−1)

( n

+ 4)

2. Since q(z) = e

−1 =

∞

∑

n=1

= z

∞

∑

n=0

(n+1)!

has a simple zero at z = 0, we see that

f (z) =

√

z + 4i

( z + i)

Log(z + 2)(e

−1)

p(z)

q(z)



p(z) =

√

z+4i

(z+i)

Log(z+2)



has a simple pole at z = 0 and may easily compute

Res

z=0

f (z) =

p(0)

′

(0)

iπ

ln 2

1 + i

√

2 ln 2

We could instead have chosen q(z) = (z + i)

Log(z + 2)(e

−1), but computing q

′

( z) would

have been disgusting!

Counting the Number of Poles and Zeros

A sneaky application of our pole/zero categorization permits us to count them.

Suppose f has a zero of order m at z

. Write f (z) = (z −z

)

ψ(z) where ψ(z) is analytic and non-zero

on some closed disk

z − z

≤ R. Let C

be the boundary of the disk and compute:

′

( z)

f (z)

dz =

m(z −z

)

m−1

ψ(z) + (z −z

)

′

( z)

( z −z

)

ψ(z)

z − z

′

( z)

ψ(z)

dz = 2πim

where we used the fact that

′

(z)

ψ(z)

is analytic on and inside C

. The integral

2πi

′

(z)

f (z)

dz therefore

counts the multiplicity of the zero. By performing a similar calculation for poles and combining the

results for several poles and zeros, we recover a famous result.

Theorem 6.19 (Cauchy’s Argument Principle). Suppose f is analytic except at poles,

on and inside

a simple closed curve C, and that f (z) has no poles or zeros on C. Then

2πi

′

( z)

f (z)

dz = Z − P

where Z and P are the number of zeros and poles of f inside C, counted up to multiplicity.

Example 6.20. Consider f (z) =

(z−i)

sin z

(z−5)

where C is a large circle surrounding the points 0, i and 5.

Plainly Z = 2 + 1 = 3 and P = 4. We may compute the integral explicitly,

2πi

′

( z)

f (z)

dz =

2πi

z − i

cos z

sin z

−

z − 5

dz = 2 + cos 0 − 4 = −1 = Z −P

(Use logarithmic differentiation unless you’re feeling masochistic!)

Such a function is termed meromorphic. The result is called the argument principle because

′

(z)

f (z)

log f ( z) means

that the integral is calculating the net change in the logarithm, and thus argument, of f (z) round the curve C.

Properties of Isolated Singularities

Recall (Deﬁnition 6.9) that a function f (z) analytic on a punctured disk 0 <

z − z

< R can have one

of three types of isolated singularity at z

: removable, pole, and essential. We ﬁnish by considering

some conditions/properties for each type.

Theorem 6.21 (Removable Singularities). Suppose f (z) has an isolated singularity at z

. The fol-

lowing are equivalent:

1. The singularity is removable.

2. lim

z→z

f (z) exists and is ﬁnite.

3. There exists a punctured disk 0 <

z − z

< δ on which f (z) is bounded.

Proof. For simplicity, suppose that z

= 0 and that f (z) is analytic on the punctured disk 0 <

< R.

(1 ⇒ 2) Since z

is removable, the Laurent series f (z) =

∞

∑

n=0

on 0 <

< R has no negative

terms. It follows that lim

z→0

f (z) = a

(2 ⇒ 3) This is almost tautological but bears repeating: If lim

z→0

f (z) = a

is ﬁnite then choose ϵ =

in the deﬁnition;

∃δ > 0 such that 0 <

< δ =⇒

f (z) − a

=⇒

f (z)

f (z) − a

+ a

△

< 2

(3 ⇒ 1) Suppose f (z) is bounded on 0 <

< δ, and consider

g(z) =

(

f (z) if 0 < z < δ

0 if z = 0

Since f is bounded, we may compute the limit

lim

z→0

g(z) − g(0)

= lim

z→0

z f (z) = 0

whence g(z) is differentiable at zero! Since g(z) is already differentiable on the punctured disk

0 <

< min(δ, R), it is therefore analytic on the disk

< min{δ, R} and equals its Maclaurin

series

g(z) =

∞

∑

n=0

∞

∑

n=2

= z

∞

∑

m=0

m−2

= g(0) = 0 and b

= g

′

(0) = 0)

We conclude that

f (z) =

∞

∑

m=0

m−2

whenever 0 <

< min(δ, R)

whence f has a removable singularity at zero.

We leave the next result as an exercise.

Theorem 6.22. Suppose f has an isolated singularity at z = z

1. z

is a pole if and only if lim

z→z

f (z) = ∞.

2. Suppose z

is essential and that w ∈ C ∪ {∞} is given. Then there exists a sequence (z

)

converging to z

for which lim

n→∞

f (z

) = w.

Combined with the previous result, we see that z

is an essential singularity if and only if lim

z→z

f (z)

does not exist.

The second result is the Casorati–Weierstrass Theorem; the range of f (z) is dense in any neighborhood

of an essential singularity. A stronger result is available, through its proof is beyond us.

Theorem 6.23 (Picard). If z

is an essential singularity of f (z), then, on any neighborhood of z

, f (z)

takes every complex value except at most one.

Example 6.24. We verify Picard’s Theorem for the essential singularity of f (z) = e

1/z

at z

= 0. Let

w ∈ C \ {0} be given, and write w = re

iθ

where 0 ≤ θ < 2π. Then

f (z) = w ⇐⇒ e

1/z

= e

ln r+iθ

⇐⇒

= ln r + iθ + 2πin

for some integer n. If n > 0, observe that



(ln r)

+ (θ + 2πn)

≥ 2πn =⇒

2πn

whence

can be chosen arbitrarily small. A suitable value z therefore exists in any punctured disk

0 <

< δ. The function f (z) = e

1/z

thus takes on every complex value except one (namely w = 0).

Exercises 6.3. 1. Determine the order of each pole and its residue.

(a) f (z) =

z + 1

+ 9

(b) f (z) =



2z + 1



2. Verify the value of each residue:

(a) Res

z=−1

1/4

z + 1

1 + i

√

when

> 0 and arg z ∈ (0, 2π) (b) Res

z=i

Log z

( z

+ 1)

π + 2i

z=z



z sec z



= (−1)

n+1

, where z

+ nπ and n ∈ Z

3. Find the value of the integral round each circle:

+ 2

( z −1)(z

+ 9)

(a)

z − 2

= 2 (b)

= 4

4. Let C be the circle

= 2. Evaluate

tan z dz.

5. Suppose functions f , g satisfy g(z) =

f (z)

. Prove:

At z

, f (z) has a zero of order m ⇐⇒ g(z) has a pole of order m

6. (a) Evaluate Res



z sin



−1

for each n ∈ Z.

(b) Why doesn’t Res

z=0



z sin



−1

make sense?

7. Suppose that C is the rectangle whose sides are the lines x = ±2, y = 0 and y = 1. Prove that

( z

−1)

+ 3

√

(Hint: the integrand has four simple poles, only two of which lie inside C)

8. (Hard) Suppose f (z) and the contour C satisfy the hypotheses of Cauchy’s Argument Principle

(Theorem 6.19).

(a) Explain why we can be sure that there are only ﬁnitely many poles/zeros inside C and that

all zeros are isolated.

(b) Complete the proof of the argument principle by performing a suitable integral on a small

circle round each pole and applying Cauchy–Goursat and/or the Residue Theorem.

2πi

′

(z)

f (z)

dz counts the number times γ orbits the origin counter-clockwise.

(d) i. (Rouch

e’s Theorem) Suppose f is analytic on/inside C and that

g(z)

f (z)

for all

z ∈ C. Prove that the number of zeros of f + g inside C equals that of f .

(Hints: Apply the argument principle to f + g = f · (1 +

) and consider why 1 +

has

winding number zero—why is Re(1 +

) > 0?)

ii. How many solutions (up to multiplicity) are there to the equation z

+ 4z

+ 2i = 0

on the domain

< 1?

9. (Hard) Prove both parts of Theorem 6.22; for simplicity, assume z

= 0.

(Hint 1: f (z) has a pole if and only if

f (z)

has a zero.

Hint 2: If no such sequence exists, show that g(z) :=

f (z)−w

is analytic and bounded.)

6.4 Improper Integrals

The theory of residues may be applied to help evaluate certain real improper integrals. We start with

an alternative deﬁnition of improper integral suited to our purposes.

Deﬁnition 6.25. Suppose f : R → R is integrable. Provided the limit exists, the Cauchy principal

value of the improper integral

∞

−∞

f (x) dx is the limit

P.V.

∞

−∞

f (x) dx = lim

R→∞

−R

f (x) dx

This deﬁnition is potentially misleading. In standard calculus such an improper integral requires two

limits, both of which must exist for the integral to converge:

∞

−∞

f (x) dx := lim

→∞

−R

f (x) dx + lim

→∞

f (x) dx (∗)

If the standard integral (∗) converges, then it equals its Cauchy principal value. However, the con-

verse doesn’t necessarily hold.

Example 6.26. If f (x) is any odd function



f (−x) = −f (x)



, then

P.V.

∞

−∞

f (x) dx = lim

R→∞

−R

f (x) dx = lim

R→∞

0 = 0

However, if one (necessarily both) of the 1-sided improper integrals diverges, then the full integral

also diverges: for instance,

P.V.

∞

−∞

dx = 0 but

∞

dx diverges =⇒

∞

−∞

dx diverges

Residue theory supplies a neat trick for computing Cauchy principal values:

1. Suppose f (x) is the restriction to the real line of a complex func-

tion f (z) which is analytic on the upper half-plane (Im z ≥ 0)

except at ﬁnitely many poles z

, . . . , z

, none of which lie on the

real axis.

2. Choose R > 0 so that all poles z

lie inside the curve formed by

the real axis and the semi-circle C

with radius R. By Cauchy’s

Residue Theorem,

−R

f (x) dx +

f (z) dz = 2πi

∑

k=1

Res

z=z

f (z)

R−R

3. If lim

R→∞

f (z) dz = 0, then P.V.

∞

−∞

f (x) dx = 2πi

∑

k=1

Res

z=z

f (z)

The method is easiest when applied to rational functions f (z) =

p(z)

q(z)

, though it works more generally.

Beyond the relative ease of residue calculation, one reason to stick to rational functions is that deg q ≥

deg p + 2 is enough to guarantee convergence in step 3 (Exercise 5).

Examples 6.27. 1. f ( z) =

(z−i)(z+i)

has simple poles at ±i. Provided

= R > 1,



+ 1



≥



−1



= R

−1 =⇒

+ 1

≤

−1

=⇒



f (z) dz



≤

πR

−1

−−−→

R→∞

=⇒ P.V.

∞

−∞

+ 1

dx = 2πi Res

z=i

f (z) =

2πi

= π

Compare with the usual calculus method:

∞

−∞

+ 1

dx = tan

−1



→∞

−R

→−∞

= π

−1 0 1

R−R

2. f (z) =

4(z

−1)

+16

has simple poles at ±2ζ, ±2ζ

where ζ = e

πi

Let p(z) = 16(z

−1) and q(z) = z

+ 16, so that

Res

z=z

f (z) =

p(z

)

′

( z

)

−1

When

= R > 2, we see that

−2 −1 0 1 2

R−R

2ζ



+ 16



≥ R

−16 =⇒



f (z) dz



≤

4πR(R

+ 1)

−16

→ 0

P.V.

∞

−∞

f (x) dx = 2πi



Res

z=2ζ

f (z) + Res

z=2ζ

f (z)



= 2πi



4ζ

−1

8ζ

4ζ

−1

8ζ



3π

√

3. Variations are possible, for instance by taking only part of a semi-circular arc. The function

f (z) =

has ﬁve simple poles: the ﬁfth-roots of −1.

Since the pole ζω

= −1 lies on the negative real axis, the

integral

∞

−∞

f (x) dx diverges. Instead consider the arcs in the

picture when R > 1. Parametrizing C

via z(t) = tω,

+ 1

dz =

+ 1

dt = −ω

+ 1

= −ω

+ 1

−1 0 1

ζω

Rω

=⇒ (1 − ω)

+ 1

dx +

+ 1

dz = 2πi Res

z=ζ

+ 1

2πi

5ζ

2πi

5ω

When

= R > 1, we see that



+ 1



≥ R

−1 =⇒



≤

2πR

5(R

−1)

−−−→

R→∞

0, and we

conclude

∞

+ 1

dx =

2πi

5(ω

−ω

)

2πi

5ζω

( ζ

−1

−ζ)

2πi

5( 2i sin

)

csc

Jordan’s Lemma

It is often desired, particularly when computing Fourier transforms,

to evaluate integrals of the

form

∞

−∞

f (x)e

iax

dx =

∞

−∞

f (x) cos ax dx + i

∞

−∞

f (x) sin ax dx

where a > 0 is a real constant and f : R → C is a given function. If f (x) is real-valued, then the

above breaks the integral into real and imaginary parts. Given reasonable conditions on f (x), the

principal-value method can often be employed.

Example 6.28. The function f (z) =

is analytic on the upper half-plane except at the simple pole

z = 2i. With R > 2 and C

the usual semi-circle, we see that



3iz



= e

−3y

≤ 1 =⇒



f (z)e

3iz



≤

πR

−4

−−−→

R→∞

=⇒ P.V.

∞

−∞

3ix

+ 4

dx = lim

R→∞

−R

3ix

+ 4

dx = 2πi Res

z=2i

3iz

+ 4

= 2πi

−6

πe

−6

Since this is real, we see that the result is in fact the integral

∞

−∞

cos 3x

dx. We don’t need the Cauchy

principal value here since the full improper integral converges. The corresponding imaginary inte-

gral is trivially zero since

sin 3x

is an odd function.

The following result, which we state without proof, assists with such computations; as usual, for any

R > 0, C

= {z :

= R, Im z ≥ 0} is the counter-clockwise semi-circle in the upper half-plane.

Theorem 6.29 (Jordan’s Lemma). Let a, R

> 0 be constants and f (z) a function. Suppose:

1. f (z) is analytic on the upper half-plane outside the semi-circle C

2. Whenever

= R > R

, f (z) is bounded by some function M(R) with lim

R→∞

M(R) = 0.

Then lim

R→∞

f (z)e

iaz

dz = 0. If f (z) moreover satisﬁes the hypotheses of our method, then

P.V.

∞

−∞

f (x)e

iax

dx = 2πi

∑

k=1

Res

z=z

f (z)e

iaz

Example 6.30. If f (x) =

x+2

and R > 3, then

f (z)

z + 2

+ 9

≤

R + 2

−9

= M

−−−→

R→∞

=⇒ P.V.

∞

−∞

(x + 2)e

iax

+ 9

dx = 2πi Res

z=3i

( z + 2)e

iaz

+ 9

2πi(2 + 3i) e

−3a

π(2 + 3i)

−3a

By considering even and odd functions, etc., we can rewrite this as

∞

cos ax

+ 9

dx =

−3a

∞

x sin ax

+ 9

dx =

−3a

The Fourier transform of f (x) is the function

f (ξ) :=

∞

−∞

f (x)e

−2πixξ

dx.

Indented paths Another modiﬁcation allows f (z) to have a simple pole on the real axis.

Lemma 6.31. Let D be the disk

z − z

≤ ϵ, let δ < ϵ, and let C

be the

clockwise semi-circle pictured.

1. If ϕ(z) is analytic on D, then lim

δ→0

ϕ(z) dz = 0.

2. If f (z) is analytic on D \{z

} with a simple pole at z

, then

lim

δ→0

f (z) dz = −πi Res

z=z

f (z)

+ δz

− δ

More generally, if C

spans θ radians clockwise round z

, then lim

δ→0

f (z) dz = −iθ Res

z=z

f (z).

Proof. 1. ϕ is continuous on D and is thus bounded by some M; but then



ϕ(z) dz



≤ Mπδ.

2. The Laurent series expansion of f (z) on D \ {z

} is

f (z) =

−1

z − z

+ ϕ(z)

where a

−1

= Res

z=z

f (z) and ϕ( z) is analytic on D. Now observe that

−1

z − z

dz = a

−1

δe

iθ

iδe

iθ

dθ = −ia

−1

dθ = −πia

−1

is independent of δ. By part (a), we conclude that lim

δ→0

f (z) dz = −πi Res

z=z

f (z).

Example 6.32. Consider f (z) =

, which has a simple pole at z

= 0. If 0 < δ < R, then



−δ

−R



f (x) dx =



−δ

−R



cos x + i sin x

dx = 2i

sin x

by even/oddness. Moreover, by Lemma 6.31,

lim

δ→0

dz = −iπ Res

z=0

f (z) = −iπ

Since

f (z)

−y

≤

on C

, Jordan’s lemma tells us that

0 = 2i

∞

sin x

dx − iπ =⇒

∞

sin x

dx =

δ−δ R−R

The example relied on the evenness of

sin x

and on the fact that the region of the half-plane between

and C

contains no poles of f (z). We essentially evaluated

sin x

dx =

−R

sin x

dx using an

indented path lying on the x-axis but dodging round the simple pole at zero. Many other versions of

this trick are possible!

Exercises 6.4. Many of these problems require extensive calculations using residues. Take your time

and use them as an excuse to practice.

1. Use residues to verify each improper integral:

(a)

∞

+ 1)

(b)

∞

+ 1

(c)

∞

+ 1)(x

+ 4)

(d)

∞

+ 9)(x

+ 4)

200

2. Find the Cauchy principal value of each integral:

(a)

∞

−∞

+ 2x + 2

(b)

∞

−∞

x dx

+ 1)(x

+ 2x + 2)

3. Let m, n be integers where 0 ≤ m ≤ n −2. By mimicking Example 6.27.3, prove that

∞

+ 1

dx =

csc

( m + 1)π

4. (a) If

∞

−∞

f (x) dx converges, prove that it equals its Cauchy principal value.

(b) Suppose f (x) is an even function and that P.V.

∞

−∞

f (x) dx exists. Prove that

∞

−∞

f (x) dx

exists and has the same value.

5. Suppose f (x) =

p(x)

q(x)

is a rational function where q(x) has no (real) zeros and 2 + deg p ≤ deg q.

Prove that

∞

f (x) dx converges.

(Hint: let p, q be monic and recall the comparison test for improper integrals)

6. Prove the identities:

(a)

∞

−∞

cos x dx

+ a

)( x

+ b

)

−b



−b

−

−a



if a > b > 0

(b)

∞

cos ax dx

+ b

)

(1 + ab)e

−ab

if a, b > 0

7. Evaluate the integrals:

(a)

∞

−∞

x sin x dx

+ 1)(x

+ 4)

(b)

∞

sin x dx

+ 1)(x

+ 9)

8. If a is any real number and b > 0, ﬁnd the Cauchy principal value of

∞

−∞

cos x dx

(x + a)

+ b

9. Use f ( z) = z

−2

( e

iaz

−e

ibz

) and an indented contour around z

= 0 to prove that

∞

cos ax −cos bx

( b − a) whenever a, b ≥ 0

10. By integrating f (z) =

−1/2

exp(−

log z)

where arg z ∈ (−

3π

) along an indented contour,

prove that

∞

√

x(x

+ 1)

√

11. What happens to part 2 of Lemma 6.31 if f (z) is analytic on D \ {z

} but has a pole of order

m ≥ 2 at z

12. (Hard) A similar trick can be applied with sequences of boundary curves C

. For instance,

for each N ∈ N, let C

denote the positively oriented boundary of the square whose edges lie

along the lines x, y = ±



N +



π. Prove that

sin z

= 2πi

+ 2

∑

n=1

( −1)

Hence conclude that

∞

∑

n=1

(−1)

n−1

, and

∞

∑

n=1