contentsquigg/teach/courses/572/572 2018...mat 572 lecture notes john quigg it’s an...

MAT 572 Lecture Notes John Quigg

Contents

1 Introduction 2

2 Complex numbers 2

3 Analysis in C — first steps 7

4 Topology of C 9

5 Basic analysis in C — functions 15

6 Complex differentiation 17

7 Special functions 21

8 Fractional linear transformations 23

9 Paths 25

10 Local Cauchy Theorem 31

11 First Flurry of Consequences 35

12 Liouville’s Theorem 42

13 Isolated singularities 44

14 Global Cauchy Theorem 47

15 Laurent series 52

16 Calculus of residues 55

c©2018 Arizona State University School of Mathematical & Statistical Sciences 1


1 Introduction

Most of these notes closely follow the books by Rudin, Ahlfors, Lang, and Conway [Rud87,Ahl78, Lan85, Con78], with some help from the notes by Spielberg [Spi].

Whenever a formal result (e.g., theorem, lemma, corollary, proposition) is presentedwithout proof, the implication is that you should be able to prove it yourself (and youshould do it!), or that the result is from prerequisite material (e.g., elementary analysis inR or Rn). Some of the exercises ask you to prove facts that will be used later in the formaldevelopment of the theory. Although I will probably not assign every exercise as homework,you are expected to read all the exercises carefully, and attempt as many as you have timefor. You know by now that you really only learn something by using it in exercises. Youalso know that you must practice writing the definitions and theorems, many times, so thatyou can give polished statements of them.

2 Complex numbers

I presume that you have had some exposure to complex numbers. In high school (maybebefore that?) you learned the quadratic formula, and saw that sometimes the solutions ofa quadratic equation would be complex numbers — the most elementary example beingx2 + 1 = 0. If you took a course in differential equations you learned a bit more aboutcomplex numbers when you studied linear ODE’s with constant coefficients. Also in linearalgebra when you studied eigenvalues. You have surely seen Euler’s equation

eiπ = −1

and more generallyeiθ = cos θ + i sin θ.

Here I’ll very briefly review the relevant definitions, which I’m sure you’ve seen before.

The set of complex numbers, written C, is R2 with the usual addition, and multiplicationgiven by

(x, y)(u, v) = (xu− yv, xv + yu).

A real number x is identified with the complex number (x, 0), and this preserves all arithmeticoperations. The particular complex number (0, 1) is written as i, and is a square root of −1.The other square root of −1 is −i = (0,−1). We can write

(x, y) = (x, 0) + (0, y) = x(1, 0) + y(0, 1) = x+ iy,

and then addition and multiplication in C are exactly what is needed for the usual rules ofalgebra to hold, together with the relation i2 = −1.



It’s an under-appreciated fact that the map

x+ iy 7→(x −yy x

)is an isomorphism of C into (not onto!) the ring M2(R) of 2× 2 matrices over R.

The complex conjugate of z = x+ iy is

z = x− iy,

the absolute value, also called the modulus, of z is

|z| =√zz =

√x2 + y2,

and the distance between z = x+ iy and w = u+ iv is

d(z, w) = |z − w| =√

(x− u)2 + (y − v)2.

The real part of z = x+ iy is

Re z = x =z + z

2,

and the imaginary part is

Im z = y =z − z

2i,

so that z = Re z + i Im z.

C can be represented geometrically as the complex plane, with real axis

R = {(x, 0) : x ∈ R}

and imaginary axisiR = {(0, y) : y ∈ R},

and then the real and imaginary parts are the coordinates of z = x + iy. The positive realaxis is [0,∞), and the negative real axis is (−∞, 0].

C is a field, just like R:

• addition and multiplication are commutative and associative,

• the real number 0 is the additive identity, i.e., 0 + z = z for all z ∈ C,

• if z = x + iy with x, y ∈ R then −z = −x − iy is the additive inverse of z, i.e.,−z + z = 0,

• the real number 1 is the multiplicative identity, i.e., 1z = z for all z ∈ C,



• and if z = x+ iy 6= 0 then

1

z=

z

zz=

x− iyx2 + y2

=x

x2 + y2+ i

−yx2 + y2

is the multiplicative inverse of z, i.e., z(1/z) = 1.

Conjugation obeys the laws

z + w = z + w

zw = (z)(w),

and this is why complex roots of polynomial equations with real coefficients come in conjugatepairs.

The absolute value satisfies

|z| ≥ 0 (and |z| = 0 if and only if z = 0)

|z + w| ≤ |z|+ |w| (the triangle inequality)

|zw| = |z||w||z| = |z|

|z + w|2 = |z|2 + |w|2 + 2 Re zw.

From linear algebra you know that C is a 1-dimensional vector space over itself, and isin fact a complex inner product space:

〈z, w〉 = zw,

and the norm is the absolute value:

‖z‖ = |z| =√〈z, z〉.

Consequently, we of course have the Cauchy-Schwarz inequality |〈z, w〉| ≤ |z||w|, but in factwe have equality, by the properties of absolute value. We will not have much need of it, butoccasionally it is useful to have n-dimensional coordinate space:

Cn = {(z1, . . . , zn) : zj ∈ C for j = 1, . . . , n}

is also a complex inner product space:

〈z, w〉 =n∑j=1

zjwj,

and the Cauchy-Schwarz inequality then be written in the form∣∣∣∣∣n∑j=1

zjwj

∣∣∣∣∣2

≤n∑j=1

|zj|2n∑j=1

|wj|2.



Occasionally we need to refer to C2 as a normed space, with norm coming from the innerproduct:

‖(z, w)‖ =√〈(z, w), (z, w)〉 =

√|z|2 + |w|2.

The argument of a nonzero complex number z, denoted by arg z, is the angle measuredin the positive (counterclockwise) direction from the positive real axis to the ray starting at0 and going through z. Of course, this is only determined up to addition by an arbitraryinteger multiple of 2π.

If θ ∈ R we writeeiθ = cos θ + i sin θ,

which initially is merely a suggestive notation whose rigorous definition will come later. Theusual laws of exponents hold, at least in the following limited form:

eiθeiφ = ei(θ+φ)

(eiθ)n = einθ for n ∈ Z1

eiθ= eiθ = e−iθ.

The second of the above laws can be expressed as de Moivre’s Formula:

(cos θ + i sin θ)n = cosnθ + i sinnθ.

The unit circle is

T = {z ∈ C : |z| = 1} = {cos θ + i sin θ : θ ∈ R} = {eiθ : θ ∈ R}.

The unit disk isD = {z ∈ C : |z| < 1}.

The polar form of z isz = reiθ,

where r = |z| and θ = arg z. Then complex multiplication can be interpreted geometrically:if z = reiθ and w = seiφ then

zw =(reiθ)(seiφ

)= rsei(θ+φ),

so that the absolute values multiply and the arguments add. In particular, multiplicationby i preserves the absolute value, and rotates by π/2 (counterclockwise). Also, conjugationcan be interpreted geometrically:

z = reiθ = re−iθ

is the reflection of z across the real axis. If z ∈ C, then

• z is real if and only if z = z,

• z is imaginary if and only if z = −z, and



• z ∈ T if and only if z = 1/z.

For every n ∈ N, the number e2πi/n ∈ T is an n-th root of 1, and in fact T containsexactly n of solutions of the equation zn = 1:

e2πki/n for k = 0, 1, . . . , n− 1,

called the n-th roots of unity. These roots of unity can also be expressed as 1, ω, ω2, . . . , ωn−1,where ω = e2πi/n is a primitive n-th root of unity1. More generally, every nonzero z = reiθ

has exactly n distinct n-th roots,

{r1/nei(θ+2πk)/n : k = 0, . . . , n− 1},

where adopt the convention that for r ≥ 0 we write r1/n for the nonnegative n-th root of r.The n-th roots are uniformly spaced on a circle centered at 0 with radius r1/n.

Occasionally it is convenient to identify C with the horizontal coordinate plane in R3,and use the unit sphere

S2 = {(x, y, z) ∈ R3 : x2 + y2 + z2 = 1}

to get a different picture of C, in the following way: consider any complex number

z = x+ iy = (x, y, 0) ∈ R3,

and now consider the line through z and the north pole (0, 0, 1). This line intersects thesphere S2 in two points: the north pole and another point. We regard z as the stereographicprojection of this other point in S2. This identifies C with the Riemann sphere S2 minus thenorth pole; we regard the north pole as the point at infinity, written ∞, and the Riemannsphere as a compactification of C. C is homeomorphic to its image in the sphere, but thisimage, as a subset of the metric space R2, is not complete, whereas C is a complete metricspace. Nevertheless, the Riemann sphere, also called the extended complex plane, gives usefulgeometric intuition for many properties of C. To be clear: S2 is itself just a sphere; we onlycall it the Riemann sphere if we’re identifying S2 \ {north pole} with C by stereographicprojection.

Exercises

1. For which complex numbers z is−z = eπi/2z?

2. Let n ∈ {2, 3, 4, . . . }, and assume that zn = 1 and z 6= 1. Prove that

n−1∑k=0

zk = 0.

1more completely, the primitive n-th roots of unity are ωj for j relatively prime to n, equivalently thosez for which zn = 0 but zk 6= 0 for k = 0, 1, . . . , n− 1



3 Analysis in C — first steps

We record various definitions and theorems from elementary analysis in R2, applied to thecomplex plane C. Much of the following will look very similar to the corresponding theoryfor analysis in R.

Definition 3.1. Let (zn) be a sequence in C and a ∈ C. Then (zn) converges to a, writtenzn → a, if for all ε > 0 there exists k ∈ N such that

|zn − a| < ε for all n ≥ k.

Lemma 3.2. If zn = xn + iyn and a = b + ic are the Cartesian forms, then zn → a if andonly if xn → b and yn → c.

We say a sequence is convergent if it converges, and otherwise it diverges, or is divergent.

Definition 3.3. A sequence (zn) is Cauchy if for all ε > 0 there exists k ∈ N such that

|zn − zj| < ε for all n, j ≥ k.

Theorem 3.4. A sequence in C is convergent if and only if it is Cauchy.

The above theorem expresses the fact that C is complete.

A convenient way to express the Cauchy condition for a series∑∞

n=1 zn is that for allε > 0 there exists k ∈ N such that |

∑qn=p zn| < ε for all q ≥ p ≥ k. A series

∑∞n=1 zn

converges absolutely if∑∞

n=1 |zn| <∞, and the Cauchy criterion makes it obvious that everyabsolutely convergent series converges.

The basic rules of sequences and series extend (except in cases that are obviously inap-propriate) from real to complex numbers.

Lemma 3.5 (Arithmetic of Convergence). Let zn → a and wn → b in C. Then:

(1) zn + wn → a+ b;

(2) znwn → ab;

(3) czn → cz if c ∈ C;

(4) znwn→ a

bif b 6= 0 and wn 6= 0 for all n.

However, since there is no order relation in C like there is in R, there are no “monotonicityof convergence” or “squeeze theorem”.



Definition 3.6. Let (zn) and (wk) be sequences. Then (wk) is a subsequence of (zn) if thereexist n1 < n2 < · · · in N such that

wk = znkfor all k ∈ N.

Definition 3.7. E is bounded if supz∈E |z| <∞.

Definition 3.8. The diameter of E is

diamE = supz,w∈E

|z − w|.

Lemma 3.9. E is bounded if and only if diamE <∞.

Theorem 3.10 (Bolzano-Weierstrass Theorem). Every bounded sequence in C has a con-vergent subsequence.

Remark 3.11. Many basic facts about analysis in C extend routinely to Cn. Occasionallywe will apply this without comment in C2.

In single-variable advanced calculus, a lot of attention is focused upon infinite limits andlimits at (plus or minus) infinity. In multivariable analysis, not so much. The complex planeis essentially the same as R2 as far as analysis is concerned. But we will still need versionsof limits involving infinity. The exercises explore the basics.

Definition 3.12. Let (zn) be a sequence in C. Then |zn| goes to ∞, written |zn| → ∞, iffor all R > 0 there exists k ∈ N such that

|zn| > R for all n ≥ k.

Note that we do not say “zn → ∞”, because there is no “∞”, unlike the ±∞ that canbe added to R.

Exercises

1. (Arithmetic of Infinite Limits) Let (zn) and (wn) be sequences in C. Suppose that |zn| → ∞.Prove:

(a) |zn + wn| → ∞ if (wn) is bounded;

(b) |znwn| → ∞ if there exists r > 0 and k ∈ N such that |wn| > r for all n ≥ k;

(c) wn/zn → 0 if (wn) is bounded and zn 6= 0 for all n.

2. Let (zn) be a sequence in C, and let p(z) be a nonconstant polynomial. Prove that if |zn| → ∞then |p(zn)| → ∞.



4 Topology of C

Throughout this section, we are interested in subsets of C. Any set will tacitly be assumedto be a subset of C, unless otherwise specified. If A ⊆ C, then the complement will be takenrelative to C:

Ac = C \ A.

Definition 4.1. If r > 0 and a ∈ C,

• the open disk of radius r centered at a is Dr(a) = {z ∈ C : |z − a| < r},

• the closed disk of radius r centered at a is Dr(a) = {z ∈ C : |z − a| ≤ r}, and

• the punctured disk of radius r centered at a is Dr(a) \ {a} = {z ∈ C : 0 < |z− a| < r}.

When a = 0 we use a further abbreviation:

Dr = Dr(0) and Dr = Dr(0).

Recall that we write D for the unit disk D1.

Definition 4.2. E ⊆ C is open if for all a ∈ E there exists r > 0 such that Dr(a) ⊆ E. Thetopology of C is the family of open subsets of C.

Proposition 4.3. (1) ∅ and C are open.

(2) Every union of open sets is open.

(3) Every finite intersection of open sets is open.

(4) Every open disk is open.

Definition 4.4. E ⊆ C is closed if Ec is open.

Proposition 4.5. (1) ∅ and C are closed.

(2) Every intersection of closed sets is closed.

(3) Every finite union of closed sets is closed.

(4) Every closed disk is closed.

(5) Every point is closed.

Lemma 4.6. E ⊆ C is closed if and only if for every sequence (zn) in E and all a ∈ C,zn → a implies a ∈ E.

Definition 4.7. The closure of E is

E =⋂{A : A is closed and E ⊆ A}.



Remark 4.8. E is the smallest closed superset of E.

Example 4.9. Dr(a) is the closure of Dr(a).

Lemma 4.10. a ∈ E if and only if Dr(a) ∩ E 6= ∅ for all r > 0.

Definition 4.11. K ⊆ C is compact if whenever K ⊆⋃α∈I Gα and Gα is open for all α,

there exists a finite subset F ⊆ I such that K ⊆⋃α∈F Gα.

We say that {Gα}α∈I is a cover of K if K ⊆⋃α∈I Gα, and the cover is open if Gα is open

for all α, and the above definition is commonly expressed as “every open cover of K has afinite subcover”.

Definition 4.12. If a ∈ E ⊆ C then a is an interior point of E if there exists r > 0 suchthat Dr(a) ⊆ E, in which case we call E a neighborhood of a. The interior of E, denotedby E◦, is the set of interior points of E.

Remark 4.13. E◦ is the union of all open subsets of E, and is the largest open subset ofE.

Definition 4.14. The boundary of E is

∂E = E ∩ Ec.

A boundary point of E is an element of ∂E.

Lemma 4.15. a ∈ ∂E if and only if for all r > 0 we have Dr(a)∩E 6= ∅ and Dr(a)\E 6= ∅.

Theorem 4.16 (Heine-Borel Theorem). A subset of C is compact if and only if it is closedand bounded.

Definition 4.17. E ⊆ C is totally bounded if for all r > 0 there exists a finite set F ⊆ Csuch that

E ⊆⋃z∈F

Dr(z).

Theorem 4.18. E has compact closure if and only if it is totally bounded.

Theorem 4.19. A subset K of C is compact if and only if every sequence in K has asubsequence that converges to a point of K.

The above theorem is a form of the Bolzano-Weierstrass Theorem.

Proposition 4.20. Every closed subset of a compact set is compact.

Proposition 4.21. If K is compact and a /∈ K, then there exist disjoint open sets U and Vsuch that K ⊆ U and a ∈ V .

Corollary 4.22. Every compact set is closed.



Theorem 4.23. If {Kα}α∈I is a family of compact sets such that⋂α∈F Kα 6= ∅ for every

finite F ⊆ I, then⋂α∈I Kα 6= ∅.

The condition “⋂α∈F Kα 6= ∅ for every finite F ⊆ I” is called the finite intersection

property, and the above theorem is commonly expressed as “every family of compact setswith the finite intersection property has nonempty intersection”.

Definition 4.24. A sequence (An) of sets is decreasing if An ⊇ An+1 for all n.

Theorem 4.25. Let (Kn) be a decreasing sequence of nonempty compact sets. If diamKn →0 then

⋂Kn consists of a single point.

Theorem 4.26. Let K ⊆ U with K compact and U open. Then there exists an open set Vwith compact closure such that

K ⊆ V ⊆ V ⊆ U.

Definition 4.27. If E ⊆ C, then

• E is disconnected if there exist nonempty sets A,B such that E = A ∪B and

A ∩B = A ∩B = ∅,

and

• E is connected if it is not disconnected.

Remark 4.28. If E is closed, then E is disconnected if and only if it is a union of twodisjoint nonempty closed sets, and similarly if E is open, then E is disconnected if and onlyif it is a union of two disjoint nonempty open sets.

Lemma 4.29. (1) If A ⊆ B ⊆ A and A is connected, then B is connected.

(2) If {Aα}α∈I is family of connected sets and⋂α∈I 6= ∅, then

⋃α∈I Aα is connected.

Definition 4.30. Let A ⊆ E ⊆ C. Then A is a connected component of E if

(1) A is connected, and

(2) whenever A ⊆ B ⊆ E, if B is connected then A = B.

Thus, a connected component of E is a maximal connected subset of E.

Remark 4.31. Any subset E ⊆ C is the disjoint union of its connected components.

Definition 4.32. A region is a nonempty connected open subset of C.

Remark 4.33. The components of an open set are open, and hence are regions.



Definition 4.34. Let E, a ∈ E, and f : E → C. Then f is continuous at a if for all ε > 0there exists δ > 0 such that for all z ∈ E, |z − a| < δ implies |f(z)− f(a)| < ε. We say thatf is continuous if it is continuous at every point of E, and we write C(E) for the set of allcontinuous functions on E.

Remark 4.35. Since R ⊆ C, the above definition applies to complex-valued functionsdefined on a subset of R, and also to real-valued functions defined on a subset of C.

Example 4.36. Re, Im, conjugation, and | · | are continuous.

Proposition 4.37. If E ⊆ C, a ∈ E, and f : E → C, then f is continuous at a if and onlyif f(zn)→ f(a) whenever zn → a.

Proposition 4.38. For f : E → C, the following are equivalent:

(1) f is continuous.

(2) f−1(U) is open for all open U .

(3) f−1(C) is closed for all closed C.

(4) Re f and Im f are continuous.

Theorem 4.39. If f : E → C is continuous and K ⊆ E is compact, then f(K) is compact.

Corollary 4.40. If f : K → R is continuous and K is compact, then f has both a maximumand a minimum.

Theorem 4.41. If f : E → C is continuous and S ⊆ E is connected, then f(S) is connected.

Definition 4.42. If a ∈ C and E ⊆ C then a is an accumulation point of E if

Dr(a) ∩ E \ {a} 6= ∅ for all r > 0.

Lemma 4.43. a is an accumulation point of E if and only if Dr(a)∩E is infinite for everyr > 0.

Lemma 4.44. E is closed if and only if it contains all its accumulation points.

Definition 4.45. If f : E → C, a is an accumulation point of E, and b ∈ C, then b is thelimit of f at a, written b = limz→a f(z), if for all ε > 0 there exists δ > 0 such that for allz ∈ E, 0 < |z − a| < δ implies |f(x)− b| < ε.

Many of the usual facts about limits carry over from advanced calculus in R:

Lemma 4.46 (Sequential Characterization of Limits). Let E ⊆ C, f : E → C, and a bean accumulation point of E. Then limz→a f(z) exists if and only if for every sequence (zn)in E \ {a} converging to a, the sequence (f(zn)) converges, in which case limz→a f(z) =limn→∞ f(zn).



Lemma 4.47 (Arithmetic of Limits). Let A ⊆ C, f, g : A → C, and a be an accumulationpoint of A. If limz→a f(z) = u and limz→a g(z) = v, then

(1) limz→a(f(z) + g(z)) = u+ v;

(2) limz→a cf(z) = cu if c ∈ C;

(3) limz→a f(z)g(z) = uv;

(4) limz→af(z)g(z)

= uv

if v 6= 0 and 0 /∈ g(A).

Lemma 4.48 (Composition of Limits). Let A,B ⊆ C, a be an accumulation point of A, bbe an accumulation point of B, f : A → B \ {b}, and g : B → C. If limz→a f(z) = b andlimw→b g(w) = u, then

limz→a

g ◦ f(z) = u.

As with sequences, there are no results concerning inequalities with complex limits.

There are infinite limits and limits at infinity:

Definition 4.49. Let A ⊆ C, f : A → C, and a be an accumulation point of A. Thenlimz→a |f(z)| = ∞ if for all R > 0 there exists δ > 0 such that for all z ∈ A \ {a}, if|z − a| < δ then |f(z)| > R.

Definition 4.50. Let A ⊆ C be unbounded, f : A→ C, and u ∈ C. Then lim|z|→∞ f(z) = uif for all ε > 0 there exists R > 0 such that for all z ∈ A, if |z| > R then |f(z − u)| < ε.

These variations on limits have obvious sequential characterizations. Also, both |z| and|f | could go to to ∞, with the obvious definition.

Example 4.51. If p(z) is a nonconstant polynomial, then lim|z|→∞ |p(z)| = ∞. The sametechnique as in the solution of Problem 2 in Section 3 can be used, alternatively the result ofthat problem and the sequential characterization of infinite limits at infinity could be used.

Lemma 4.52. If a ∈ E ⊆ C, f : E → C, and a is an accumulation point of E, then f iscontinuous at a if and only if limz→a f(z) = f(a).

Theorem 4.53. Every bounded infinite subset of C has an accumulation point.

The above theorem is equivalent to the Bolzano-Weierstrass Theorem, and in fact is oftenregarded as an alternative form of this theorem.

Definition 4.54. If a ∈ E ⊆ C, then a is an isolated point of E if it is not an accumulationpoint of E.

Remark 4.55. a is an isolated point of E if and only if there exists r > 0 such thatDr(a) ∩ E = {a}.



Remark 4.56. Every function is automatically continuous at every isolated point of itsdomain.

Definition 4.57. The distance between subsets A,B ⊆ C is

d(A,B) = inf{|z − w| : z ∈ A,w ∈ B}

If a ∈ C we write d(a,B) = d({a}, B).

Proposition 4.58. If A and B are nonempty, A is compact, B is closed, and A ∩ B = ∅,then there exist z ∈ A and w ∈ B such that

d(A,B) = |z − w|.

Exercises

1. Give a detailed proof of the assertion in Example 4.51 using the same technique as in thesolution of Problem 2 in Section 3.

2. Let p(z) and q(z) be polynomials of degrees n and k and leading coefficients a and b, respec-tively. Prove:

(a) If n < k then

lim|z|→∞

p(z)

q(z)= 0.

(b) If n = k then

lim|z|→∞

p(z)

q(z)=a

b.

(c) If n > k then

lim|z|→∞

∣∣∣∣p(z)

q(z)

∣∣∣∣ =∞.

3. Let zn → a in C. Prove that the set {zn : n ∈ N} ∪ {a} is compact.

4. Let A ⊆ C be closed and t /∈ A. Prove that d(t, A) > 0.

5. Prove that every subset A ⊆ C consisting only of isolated points must be countable.

6. Prove that an open subset of C has only countably many components.

7. Let A ⊆ C. Prove that the set of accumulation points of A is closed.

8. Let D ⊆ C and f : D → C. We say that f is bounded away from 0 on D if

inf{|f(z)| : z ∈ D} > 0.

Prove that this is equivalent to 1/f being bounded on D.



5 Basic analysis in C — functions

Definition 5.1. Let E ⊆ C and f : E → C. Then f is uniformly continuous if for all ε > 0there exists δ > 0 such that for all z, w ∈ E, |z − w| < δ implies |f(z)− f(w)| < ε.

Theorem 5.2. Every continuous function on a compact set is uniformly continuous.

Definition 5.3. Let E ⊆ C, let (fn) be a sequence of functions from E to C, and letf : E → C. Then (fn) converges uniformly to f , written fn → f uniformly, if for all ε > 0there exists k ∈ N such that |fn(z)− f(z)| < ε for all n ≥ k and all z ∈ E.

A series of functions converges uniformly if the sequence of partial sums does so.

Definition 5.4. A sequence (fn) of functions on a set E is uniformly Cauchy if for all ε > 0there exists k ∈ N such that |fn(z)− fj(z)| < ε for all n, j ≥ k and all z ∈ E.

A series of functions is uniformly Cauchy if the sequence of partial sums is.

Remark 5.5. A convenient way to express the uniform Cauchy condition for a series of

functions∑∞

n=1 fn on a set E is that for all ε > 0 there exists k ∈ N such that∣∣∣∑q

n=p fn(z)∣∣∣ <

ε for all q ≥ p ≥ k and all z ∈ E.

Lemma 5.6. A sequence (fn) converges uniformly if and only if it is uniformly Cauchy.Similarly for series.

Theorem 5.7. A uniform limit of continuous functions is continuous.

There is a companion result for limits, which is not quite equivalent to the above theorem,and which is significantly fussier to state2:

Theorem 5.8. Let fn → f uniformly on E, and let a be an accumulation point of E. Iflimz→a fn(z) exists for every n, then

limz→a

f(z) = limn→∞

limz→a

fn(z).

Since sequences and series are two ways of looking at the same thing, every theoremabout sequences of functions has a companion theorem for series. The following is a specialcase that does not have a convenient version for sequences:

Theorem 5.9 (Weierstrass M -Test). Let∑fn be a series of functions on E, and suppose

that there exist constants Mn such

• |fn(z)| ≤Mn for all n ∈ N and z ∈ E, and

•∑Mn <∞.

Then∑fn converges uniformly.

2but which is nevertheless occasionally necessary



Exercises

1. Let E ⊆ C, let a be an accumulation point of E, and let f : E → C. Prove that limz→a f(z)exists if and only if for every sequence (zn) in E \ {a} converging to a the sequence (f(zn))converges, in which case limz→a f(z) = limn→∞ f(zn).

2. Let E,K ⊆ C, with K compact, and let f : E × K → C be continuous. Prove that ifzn → z in E then f(zn, w)→ f(z, w) uniformly for w ∈ K. Hint: use contradiction and theBolzano-Weierstrass Theorem.

3. If A,B ⊆ C, f : A × B → C, g : B → C, and t is an accumulation point of A, we say thatf(z, w) → g(w) uniformly for w ∈ B as z → t if for all ε > 0 there exists δ > 0 such thatfor all (z, w) ∈ A×B with 0 < |z − t| < δ we have

|f(z, w)− g(w)| < ε.

Prove that this is equivalent to the following: for every sequence (zn) in A \ {t} convergingto t, the functions gn : B → C defined by

gn(w) = f(zn, w)

converge uniformly to g.

4. Similarly to the preceding problem, if A,B ⊆ C, f : A × B → C, g : B → C, and A isunbounded, we say that f(z, w) → g(w) uniformly for w ∈ B as |z| → ∞ if for all ε > 0there exists R > 0 such that for all (z, w) ∈ A×B with |z| > R we have

|f(z, w)− g(w)| < ε.

Prove that this is equivalent to the following: for every sequence (zn) in A with |zn| → ∞,the functions gn : B → C defined by

gn(w) = f(zn, w)

converge uniformly to g.

5. Let f(z) = p(z)/q(z) be a rational function, where the polynomials p(z) and q(z) havedegrees n and k, respectively. Assume that n < k. Prove that there exist M,K > 0 suchthat

|f(z)| ≤ M

|z|k−nif |z| > K.

6. Let K ⊆ C be compact. Prove that

lim|z|→∞

1

w − z= 0

uniformly for w ∈ K.



6 Complex differentiation

Definition 6.1. Let U ⊆ C be open, f : U → C, and a ∈ U . Then f is differentiable at a ifthe derivative

f ′(a) = limz→a

f(z)− f(a)

z − aof f at a exists. We say that f is holomorphic if it is differentiable at every point of U . Wewrite H(U) for the set of all holomorphic functions on U . If f ∈ H(C) it is called entire.

We also sometimes write df/dz for f ′(z). We also say that f is holomorphic at a if it isholomorphic on some open neighborhood of a.

Proposition 6.2. If f is differentiable at a then it is continuous at a.

Proposition 6.3. If f and g are differentiable at a, then:

(1) f + g is differentiable at a, and (f + g)′(a) = f ′(a) + g′(a).

(2) fg is differentiable at a, and (fg)′(a) = f ′(a)g(a) + f(a)g′(a).

(3) If c ∈ C, then cf is differentiable at a, and (cf)′(a) = cf ′(a).

(4) Provided g is never 0, f/g is differentiable at a, and(f

g

)′(a) =

f ′(a)g(a)− f(a)g′(a)

g(a)2.

Proposition 6.4. If f is differentiable at a and g is differentiable at f(a), then g ◦ f isdifferentiable at a, and

(g ◦ f)′(a) = g′(f(a))f ′(a).

Example 6.5. If n ∈ Z \ {0}, then

d

dzzn = nzn−1.

Thus, polynomials and rational functions are holomorphic.

Definition 6.6. If a ∈ C and (cn) is a sequence in C, the power series with center a andcoefficients cn is

∑∞n=0 cn(z − a)n.

Theorem 6.7. For every power series∑∞

n=0 cn(z − a)n there exists a unique R ∈ [0,∞]such that the series converges absolutely when |z − a| < R and diverges when |z − a| > R.Moreover,

R =1

lim sup |cn|1/n,

interpreted as 0 if lim sup |cn|1/n = ∞ and as ∞ if lim sup |cn|1/n = 0. Finally, the powerseries converges uniformly on Dr(a) for all r < R.



Definition 6.8. In the above theorem, R is the radius of convergence of the power series.

Definition 6.9. If U ⊆ C is open, f : U → C, and a ∈ U , then f is analytic at a if thereexists r > 0 such that

f(z) =∞∑n=0

cn(z − a)n for all z ∈ Dr(a).

f is analytic if it is analytic at every point of U .

Of course every linear combination of analytic functions is analytic.

Theorem 6.10. Let the power series

∞∑n=0

cn(z − a)n (1)

have radius of convergence R, and assume that R > 0. Then the power series

∞∑n=1

ncn(z − a)n−1 (2)

also has radius of convergence R, and moreover

d

dz

∞∑n=0

cn(z − a)n =∞∑n=1

ncn(z − a)n−1 for all z ∈ DR(a).

Proof. First, the power series in (2) has the same radius of convergence as (1), because

lim sup |ncn|1/n = lim sup |cn|1/n

since n1/n → 1.

Define f : DR(a)→ C by f(z) =∑∞

n=0 cn(z−a)n. Since differentiation is a local process,it suffices to restrict z to a disk Dr(a) with 0 < r < R. For w ∈ Dr(a) \ {z}, we have

f(w)− f(z)

w − z=∞∑n=0

cn(w − a)n − (z − a)n

w − z. (3)

For each n ≥ 1,∣∣∣∣(w − a)n − (z − a)n

w − z

∣∣∣∣ ≤ n−1∑k=0

∣∣(w − a)k(z − a)n−1−k∣∣ ≤ nrn−1,

and the series∞∑n=0

|cn|nrn−1



converges because the power series

∞∑n=0

ncn(z − a)n−1

converges absolutely at z = a + r. Thus by the Weierstrass M -Test the series on the right-hand side of (3) converges uniformly for w ∈ Dr(a) \ {z}, and hence

limw→z

f(w)− f(z)

w − z= lim

w→z

∞∑n=0

cn(w − a)n − (z − a)n

w − z

=∞∑n=0

limw→z

cn(w − a)n − (z − a)n

w − z

=∞∑n=0

d

dz[cn(z − a)n]

=∞∑n=0

ncn(z − a)n−1,

and we are done.

The above proof might appear overly complicated compared to the corresponding proofin real variables. This is essentially due to the lack of a Mean Value theorem (see Exercise 7in Section 7).

Example 6.11. The geometric series∞∑n=0

zn

has radius of convergence 1, because

lim sup 11/n = 1.

Moreover, on the unit disk D we have

d

dz

∞∑n=0

zn =∞∑n=1

nzn−1 =∞∑n=0

(n+ 1)zn.

Corollary 6.12. Let f : U → C be analytic. Then

(1) f is holomorphic.

(2) f is infinitely differentiable, and f (n) is analytic for all n ∈ N.

(3) If f(z) =∑∞

n=0 cn(z − a)n near a, then

cn =f (n)(a)

n!for all n.



Exercises

1. Prove that the function f(z) = z is not differentiable anywhere.

2. Let f ∈ H(U). The real part of f is the function u = Re f : E → R defined by

u(z) = Re(f(z)) =f(z) + f(z)

2,

and the imaginary part is the function v = Im f : E → R defined by

v(z) = Im(f(z)) =f(z)− f(z)

2i.

Write z = x + iy with x, y ∈ R, and regard x, y as independent real variables. In this waywe can regard u and v as functions of two real variables:

u(x, y) = u(x+ iy) and v(x, y) = v(x+ iy).

Show that for any a ∈ U ,

f ′(a) =∂u

∂x(a) + i

∂v

∂x(a)

= −i∂u∂y

(a) +∂v

∂x(a).

Deduce that u and v satisfy the Cauchy-Riemann equations :

∂u

∂x=∂v

∂y∂u

∂y= −∂v

∂x.

3. Prove that if |cn+1/cn| → L then the radius of convergence of∑∞

n=0 cn(z − a)n is 1/L.

4. Expand the following in power series about the given a, and give the radius of convergence.Be sure to justify your steps.

(a) 11+z

; a = 0

(b) 1z; a = 1

(c) 1π−z ; a = 3

(d) 1(1−z)2 ; a = 0

5. If cn is 2−n for n odd and 3−n for n even, what is the radius of convergence of the powerseries

∑∞n=0 cnz

n?



7 Special functions

Definition 7.1. The functions exp, cos, and sin are defined on C by

exp(z) = ez =∞∑n=0

zn

n!

cos z =∞∑n=0

(−1)nz2n

(2n)!

sin z =∞∑n=0

(−1)nz2n+1

(2n+ 1)!

Note that this definition makes sense because the three power series have radius of conver-gence ∞.

Thus the functions exp, sin, and cos are entire.

The following properties follow from routine computations with the power series.

eiz = cos z + i sin z for all z ∈ C. (Euler’s Formula)

d

dzez = ez,

d

dzsin z = cos z, and

d

dzcos z = − sin z.

In fact, all properties of exp, cos, and sin can be derived from the above.

The propertyezew = ez+w

can be verified using the differential equation f ′(z) = f(z).

Thus, e−z = 1/ez, and ez 6= 0 for all z ∈ C.

For all θ ∈ R, eiθ = e−iθ, so|eiθ|2 = eiθeiθ = 1,

and hence eiθ is in the unit circle T. In fact, since eiθ = cos θ + i sin θ,

T = {eiθ : 0 ≤ θ < 2π}.

For z = x+ iy, we have |ez| = ex, so the polar form of ez is

ez = exeiy.



Consequently, ez = 1 if and only if z ∈ 2πiZ.

The range of the exponential function is the punctured plane C \ {0}. For any z 6= 0, alogarithm of z is any w for which ew = z, equivalently, writing w = u+ iv with u, v ∈ R,

u = log |z| and v = arg z,

where log |z| here denotes the unique real number u with eu = |z|. Recall that arg z is onlydetermined up to addition by an integer multiple of 2π.

Definition 7.2. A branch of log on a region U is a continuous function f : U → C such thatef(z) = z for all z ∈ U . The principal branch of log is defined on the slit (or cut) plane

U = C \ (−∞, 0]

bylog z = log |z|+ i arg z,

where arg z is chosen in (−π, π).

Another common choice of domain for a branch of log is C \ [0,∞). Note that there isno branch of log on the punctured disk D \ {0}.

Definition 7.3. For z 6= 0 and a ∈ C, we define

za = ea log z.

If f is a branch of log on a region U , then the function g : U → C defined by

g(z) = eaf(z)

is a branch of za.

Example 7.4. If n ∈ {2, 3, 4, . . . }, the principal branch of z1/n is

z 7→ |z|1/nei arg z/n,

with | arg z| < π.

It will follow later from the Inverse Function Theorem that every branch of log or of za

is holomorphic, withd

dzlog z =

1

zand

d

dzza = aza−1,

where in the latter derivative formula we take the branches of za and za−1 from the samebranch of log.



Exercises

1. Compute cos i and sin i.

2. Prove that ez is periodic with period 2πi.

3. Prove that ez maps the strip 0 < | Im z| < 2π bijectively onto C \ [0,∞).

4. Prove that there is a branch of z1/3 on the slit plane C \ (−∞, 0], and find its range.

5. Compute all the values of log i.

6. Compute all the values of ii.

7. Prove that there is no simple-minded adaptation of the Mean Value Theorem to complexfunctions of a complex variable, by considering f(z) = ez and the points z = 0, 2πi. For thisreason, in complex-variable theory no attempt is typically made to use techniques involvingany version of the Mean Value Theorem.

8 Fractional linear transformations

In this section I’ll list without proof the important properties of a special class of functions,which we will only need once, in Problem 29 in Section 16.

Definition 8.1. A fractional linear transformation is a function of the form

T (z) =az + b

cz + d,

where a, b, c, d ∈ C with ad− bc 6= 0.

If we want to remember the coefficients we write Ta,b,c,d. If c = 0 then T is entire (andmay be regarded as of the form az+b), while if c 6= 0 then T is holomorphic on the puncturedplane C \ {−d/c}.

Two fractional linear transformations (az+ b)/(cx+ d) and (a′z+ b′)/(c′z+ d′) are equalif and only if there exists λ ∈ C such that

a′ = λa, b′ = λb, c′ = λc, d′ = λd.

The map (a bc d

)7→ Ta,b,c,d



from the set GL2(C) of invertible 2-by-2 complex matrices takes matrix multiplication tofunction composition. Thus the set of fractional linear transformations is a group undercomposition, and is a quotient of the group GL2(C).

T (z) = z + a is a translation, T (z) = az is a dilation, and is also a rotation if |a| = 1,and T (z) = 1/z is the inversion.

It is convenient to regard fractional linear transformations as maps from the Riemannsphere S2 to itself, with

T

(−dc

)=∞ and T (∞) =

a

c.

Moreover, we regard a straight line in C as a circle in S2 containing the point at infinity.With this convention, fractional linear transformations map circles to circles. Moreover, ifC and C ′ are two circles, the first containing distinct points z2, z3, z4 (in S2) and the secondcontaining distinct points w2, w3, w4, then there is a unique fractional linear transformationtaking C to C ′ that maps zj to wj for j = 1, 2, 3. Letting w2, w3, w4 = 1, 0,∞ (in that order),we have

S(z) =

(z − z2

z − z4

) / ( z2 − z3

x2 − z4

)if z2, z3, z4 ∈ C

z − z3

z − z4

if z2 =∞

z2 − z4

z − z4

if z3 =∞

z − z3

z2 − z3

if z4 =∞.

Definition 8.2. With the above notation, the cross ratio of z1, z2, z3, z4 is

(z1, z2, z3, z4) = S(z1).

The following fact is fundamental: if T is a fractional linear transformation then

(T (z1), T (z2), T (z3), T (z4)) = (z1, z2, z3, z4).

The ordering of the points zj gives an orientation to the line C, and the region to theright of C is

{z ∈ S2 : Im(z, z2, z3, z4) > 0}If T is a fractional linear transformation taking C to C ′, then T maps the region to the rightof C onto the region to the right of C ′.

Definition 8.3. If C is a circle then two points z, z∗ /∈ C are symmetric with respect to Cif for some (and hence any) distinct points z2, z3, z4 ∈ C we have

(z∗, z2, z3, z4) = (z, z2, z3, z4).

If T is a fractional linear transformation and z, z∗ are symmetric with respect to a circleC, then T (z), T (z∗) are symmetric with respect to the image circle T (C).



Exercises

1. Find a fractional linear transformation that maps the upper half-plane Im z > 0 onto thedisk |z| < 1.

2. Given any two points a, b in the half-plane Im z > 0, find a fractional linear transformationthat maps the half-plane onto itself and takes a to b. (For example, you could use a horizontalshift and a radial dilation.)

3. Given any two points a, b in an open disk D, prove that there is a fractional linear transfor-mation that maps the disk onto itself and takes a to b. You can either do it directly or useProblems 1 and 2.

9 Paths

Definition 9.1. A curve is a continuous function γ : [a, b]→ C (where we assume that a < bin R). We write γ∗ for the range of γ, and we say that γ is a parametrization of γ∗, withparameter interval [a, b]. We say γ is a curve in U if γ∗ ⊆ U . The initial and final points ofthe curve are γ(a) and γ(b), respectively. and we sometimes say that γ is a curve from γ(a)to γ(b). A curve is closed if its initial and final points coincide.

A curve γ is differentiable at t if the derivative

γ′(t) = limh→0

γ(t+ h)− γ(t)

h

exists, and is continuously differentiable, or C1, if γ′ is continuous on the parameter intervalof γ.

A path is a piecewise C1 curve, i.e., a curve γ : [a, b] → C for which there is a partitiona = t0 < t1 < · · · < tn = b such that γ is C1 on [tj−1, tj] for each j = 1, . . . , n. Note that foreach j = 1, . . . , n the function γ is left-differentiable at tj, and similarly is right-differentiableat tj for j = 0, . . . , n − 1, but for 0 < j < n the left and right derivatives of γ might notagree.

Two paths γ : [a, b]→ C and φ : [c, d]→ C are equivalent if there is a strictly increasingpiecewise C1 bijection ψ : [c, d] → [a, b] such that γ ◦ ψ = φ, in which case we call φ areparametrization of γ.

Remark 9.2. Write γ = u+ iv with u, v real-valued. Then γ is differentiable if and only ifu and v are differentiable, in which case γ′ = u′ + iv′, and is C1, if and only if u and v are.

Remark 9.3. The usual rules for derivatives involving sums, products, and quotients applyto paths. We also have the following forms of the chain rule:



(1) If γ is a path in U and f ∈ H(U), then f ◦ γ is a path, and

(f ◦ γ)′(t) = f ′(γ(t))γ′(t)

wherever γ is differentiable.

(2) If γ is a path in U and ψ : [c, d]→ [a, b] is a strictly increasing piecewise C1 bijection,then γ ◦ ψ is a path, and

(γ ◦ ψ)′(t) = γ′(ψ(t))ψ′(t)

wherever γ and ψ are differentiable.

Definition 9.4. Let γ : [a, b]→ C be a path and let f ∈ C(γ∗). The integral of f over γ is∫γ

f(z) dz =

∫ b

a

f(γ(t))γ′(t) dt,

and we frequently abbreviate this as∫γf .

In the above integral, we tacitly understand that we may have to break the interval [a, b]using a partition, because γ′ may fail to exist at finitely many points. This also applies tothe verifications of many of the elementary properties of integrals over paths.

The above integral only depends upon the equivalence class of the path:

Lemma 9.5 (Invariance under Reparametrization). If γ and φ are equivalent paths andf ∈ C(γ∗), then ∫

γ◦φf =

∫γ

f.

Clearly path integration is linear:∫γ

(f + g) =

∫γ

f +

∫γ

g∫γ

cf = c

∫γ

f for c ∈ C.

Lemma 9.6 (Fubini’s Theorem). If γ and ζ are paths and f ∈ C(γ∗ × ζ∗) then∫γ

∫ζ

f(z, w) dw dz =

∫ζ

∫γ

f(z, w) dz dw.

Lemma 9.7 (Fundamental Theorem of Calculus). (1) If f is C1 on U and γ : [a, b]→ Uis a path then ∫

γ

f ′ = f(γ(b))− f(γ(a)).



(2) If γ : [a, b]→ C is a path and f ∈ C(γ∗), then

d

dt

∫ t

a

f(γ(s))γ′(s) ds = f(γ(t))γ′(t)

at every t where γ′ is continuous.

We will need a form of “arithmetic” of paths:

Definition 9.8. Let γ : [a, b]→ C be a path

(1) Let a < c < b, and let φ = γ|[a,c] and ψ = γ|[c,b]. We write γ = φ + ψ and call this asubdivision of γ.

(2) The path −γ : [−b,−a]→ C is defined by

(−γ)(t) = γ(−t).

(3) If ζ is a path with the same final point as γ, then we write γ − ζ = γ + (−ζ).

Remark 9.9. We can reverse the subdivision procedure of (1), joining suitably compatiblepaths φ, ψ to make the path φ+ ψ.

Lemma 9.10. With the above arithmetic of paths and any appropriate function f ,

(1)∫γ+ζ

f =∫γf +

∫ζf

(2)∫−γ f = −

∫γf

(3)∫γ−ζ f =

∫γf −

∫ζf .

Definition 9.11. The length of a path γ : [a, b]→ C is

L(γ) =

∫ b

a

|γ′(t)| dt.

Lemma 9.12. If γ be a path and f ∈ C(γ∗), then∣∣∣∣∫γ

f

∣∣∣∣ ≤ ‖f‖L(γ),

where ‖f‖ is the sup norm of f .



Lemma 9.12 is interesting here, because, unlike many of the basic results I list withoutproof, it does not follow easily from the techniques that you learned in (single-variable)advanced calculus. In that course you did learn that if u : [a, b]→ R is continuous then∣∣∣∣∫ b

a

u(t) dt

∣∣∣∣ ≤ ∫ b

a

|u(t)| dt,

which of course immediately gives the inequality |∫ bau(t) dt| ≤ ‖u‖(b − a) for real-valued

continuous functions. While it’s true that in Lemma 9.12 we can decompose f = u+ iv intoreal and imaginary parts, this does not help in any obvious way: a simple-minded estimateshows ∣∣∣∣∫

γ

f

∣∣∣∣ =

∣∣∣∣∫γ

u+ i

∫γ

v

∣∣∣∣≤∣∣∣∣∫γ

u

∣∣∣∣+

∣∣∣∣∫γ

v

∣∣∣∣ ,but even if we knew the desired inequality for real-valued functions:∣∣∣∣∫

γ

u

∣∣∣∣ ≤ ‖u‖L(γ),

this would not give us what we want; instead we would get∣∣∣∣∫γ

f

∣∣∣∣ ≤ (‖u‖+ ‖v‖)L(γ),

and it is false that‖u‖+ ‖v‖ ≤ ‖u+ iv‖

in general. So, to prove Lemma 9.12 we need a new approach. First, note that if we knewthat for every continuous function g : [a, b]→ C we had∣∣∣∣∫ b

a

g(t) dt

∣∣∣∣ ≤ ∫ b

a

|g(t)| dt,

then this would be enough, since in the context of Lemma 9.12, if the parameter interval ofthe path γ is [a, b], then we would have∣∣∣∣∫

γ

f

∣∣∣∣ =

∣∣∣∣∫ b

a

f(γ(t))γ′(t) dt

∣∣∣∣≤∫ b

a

|f(γ(t))γ′(t)| dt

=

∫ b

a

|f(γ(t))||γ′(t)| dt

≤∫ b

a

‖f‖|γ′(t)| dt



= ‖f‖L(γ).

So now we work on g : [a, b] → C. It is easy to see that, letting u = Re g and v = Im g,since the integrals of u and v are limits of Riemann sums, so it the integral of g. For everypartition {tj}nj=0 of [a, b], we have∣∣∣∣∣

n∑j=1

g(tj)∆tj

∣∣∣∣∣ ≤n∑j=1

|g(tj)|∆tj,

and taking limits gives ∣∣∣∣∫ b

a

g(t) dt

∣∣∣∣ ≤ ∫ b

a

|g(t)| dt.

Example 9.13. If a ∈ C and r > 0, the positively oriented circle with center a and radiusr is the closed curve

γ(t) = a+ reit for 0 ≤ t ≤ 2π.

We have ∫γ

f = ir

∫ 2π

0

f(a+ reiθ)eiθ dθ,

and we sometimes denote this integral by∫|z−a|=r f .

Example 9.14. If a, b ∈ C, the oriented interval [a, b] is the curve

γ(t) = a+ (b− a)t for 0 ≤ t ≤ 1.

We have ∫γ

f = (b− a)

∫ 1

0

f(a+ (b− a)t

)dt.

Note that −[a, b] = [b, a] and L(γ) = |b− a|.

Example 9.15. If (a, b, c) ∈ C3, the closed triangle

∆ = ∆(a, b, c)

is the convex hull of {a, b, c}, with boundary path

∂∆ = [a, b] + [b, c] + [c, a].

Theorem 9.16. Let γ be a closed path, and let U = C \ γ∗. Define

Wγ(z) =1

2πi

∫γ

dw

w − zfor z ∈ U.

Then the function Wγ is Z-valued and continuous, and is 0 on the unbounded component ofU .



Proof. If γ : [a, b]→ C, fix z ∈ U and define h : [a, b]→ C by

h(t) =

∫ t

a

γ′(s)

γ(s)− zds.

By the Fundamental Theorem of Calculus, differentiating gives

h′(t) =γ′(t)

γ(t)− z

at every t where γ′ is continuous.

Now define g : [a, b]→ C by

g(t) = e−h(t)(γ(t)− z).

Then g′ = 0 except at finitely many points, and g is continuous. Therefore g is constant.Since

g(a) = e−h(a)(γ(a)− z) = γ(a)− zg(b) = e−h(b)(γ(b)− z)

and γ(a) = γ(b), we must have e−h(b) = 1, and so

Wγ(z) = h(b) ∈ 2πiZ.

By Problem 5, the integer-valued function Wγ is continuous, and hence is constant oncomponents of U . Since

lim|z|→∞

1

w − z= 0

uniformly for w ∈ γ∗, we see that Wγ must be 0 on the unbounded component of U .

Definition 9.17. With the above notation, Wγ(z) is the winding number of γ with respectto z.

Example 9.18. If γ is the positively oriented circle with center a and radius r, then

Wγ(z) =

{1 if |z − a| < r

0 if |z − a| > r.

Exercises

1. Prove that an open set U ⊆ C is connected if and only if it is path connected, meaning thatfor all a, b ∈ U there is a path in U from a to b. Hint: fix a ∈ U , and let

B = {z ∈ U : there is a path in U form a to z}.

Prove that B and U \B are open.



2. Find a closed path γ with Wγ(2) = 1 and Wγ(−3) = −1.

3. Let p(z) =∏n

j=1(z − aj), and let γ be a closed path missing the aj’s. Prove that

1

2πi

∫γ

p′

p=

n∑j=1

Wγ(aj)

Hint:(fg)′

fg=f ′

f+g′

g.

4. Let γ be a path, and let fn be a sequence of continuous functions on γ∗ converging uniformlyto f . Prove that ∫

γ

fn →∫γ

f.

5. (Continuous Integral Lemma) Let E ⊆ C, let γ be a curve, and let f : E × γ∗ → C becontinuous. Define g : E → C by

g(z) =

∫γ

f(z, w) dw.

Prove that g is continuous. Hint: you could use Problem 4

6. Let U be open and f : U → C be continuous. Let γ be a path, and for each y ∈ R let γy bethe path given by

γy(t) = γ(t) + iy.

Let δ > 0, and suppose that γ∗y ⊆ U for all 0 ≤ y ≤ δ. Prove that

limy↓0

∫γy

f =

∫γ

f.

Note that γ0 = γ. Don’t forget that γ is only piecewise C1.

10 Local Cauchy Theorem

Definition 10.1. A function F ∈ H(U) is a primitive for a function f : U → C if F ′ = fon U .

Theorem 10.2. If f ∈ C(U) has a primitive, then∫γ

f = 0

for every closed path γ in U .



Proof. Immediate from the Fundamental Theorem of Calculus.

Example 10.3. If n ∈ Z \ {−1}, then∫γzn dz = 0 for every closed path, provided that

0 /∈ γ∗ when n < 0. However, 1/z is holomorphic on the punctured plane C \ {0}, but∫|z|=r 1/z, dz = 2πi. We conclude that 1/z has no primitive on the punctured plane.

We are ready for our first form of Cauchy’s Theorem (see Theorems 10.5 and 14.6 forthe others). In the following statement we include the exceptional point a because it doesnot cause much trouble and it will be convenient later. The proof, due to Goursat, was asignificant improvement over Cauchy’s, because it showed that the hypothesis that f ′ be con-tinuous is redundant — in the Higher Order Cauchy’s Formula we’ll see that holomorphicityimplies not only continuity of the derivative, but in fact infinite differentiability.

Theorem 10.4 (Cauchy’s Theorem for a Triangle). If U is open, a ∈ U , f is continuouson U and holomorphic on U \ {a}, and ∆ is a closed triangle in U , then∫

∂∆

f = 0.

Proof. Suppose first that a /∈ ∆ = ∆(b, c, d). Let b′, c′, d′ be the midpoints of the sidesopposite b, c, d, respectively, and subdivide ∆ into 4 subtriangles ∆1, . . . ,∆4 with verticesgiven by the ordered triples

(b, d′, c′), (c, b′, d′), (b′, d, c′), (b′, c′, d′).

Then by cancellation ∫∂∆

f =4∑j=1

∫∂∆j

f,

so we can let ∆1 be one of ∆1, . . . ,∆4 such that∣∣∣∣∫∂∆1

f

∣∣∣∣ ≥ 1

4

∣∣∣∣∫∂∆

f

∣∣∣∣ .Continuing inductively, we get a decreasing sequence

∆ ⊇ ∆1 ⊇ ∆2 ⊇ · · ·

such that for every nL(∂∆n) = 2−nL(∂∆)

and ∣∣∣∣∫∂∆

f

∣∣∣∣ ≤ 4n∣∣∣∣∫∂∆n

f

∣∣∣∣Since diam ∆n → 0 and the ∆n are compact, we can let

{t} =⋂n

∆n.



Then t ∈ ∆, so f is differentiable at t.

Let ε > 0. We can choose r > 0 such that

|f(z)− f(t)− f ′(t)(z − t)| < ε|z − t| for all z ∈ Dr(t),

and then we can choose n ∈ N such that ∆n ⊆ Dr(t). Then for all z ∈ ∆n we have

|z − t| ≤ 2−nL(∂∆).

Thus ∣∣∣∣∫∂∆n

f

∣∣∣∣ =

∣∣∣∣∫∂∆n

(f(z)− f(t)− f ′(t)(z − t)

)dt

∣∣∣∣(since f(t)− f ′(t)(z − t) is a polynomial in z)

≤ ε(2−nL(∂∆)

)2,

and hence ∣∣∣∣∫∆

f

∣∣∣∣ ≤ εL(∂∆)2.

Since ε > 0 was arbitrary we must have∫

∆f = 0.

Next, if a is a vertex of ∆, say a = b, choose x ∈ [b, c] and y ∈ [d, b], both close to butnot equal to b, and note that∫

∂∆

f =

∫∂∆(b,x,y)

f +

∫∂∆(x,c,y)

f +

∫∂∆(c,d,y)

f =

∫∂∆(b,x,y)

f,

by the first part of the proof. Letting x, y → b, the integral∫∂∆(b,x,y)

f goes to 0,

so∫

∆f = 0. Finally, if a ∈ ∆ is arbitrary, apply the preceding to the triangles

∆(b, c, a),∆(c, d, a),∆(d, b, a).

Theorem 10.5 (Cauchy’s Theorem in a Disk). If U is an open disk, a ∈ U , f ∈ C(U), andf ∈ H(U \ {a}), then f has a primitive on U , so∫

γ

f = 0.

for every closed path γ in U .

Proof. Fix a ∈ U . Since U is convex, it contains every interval [a, z] for z ∈ U , so we candefine F : U → C by

F (z) =

∫[a,z]

f.

Now fix z ∈ U . Then for every w ∈ U the triangle ∆(a, z, w) is in U , and so it follows fromCauchy’s Theorem for a Triangle that∣∣∣∣F (w)− F (z)

w − z− f(z)

∣∣∣∣ =

∣∣∣∣ 1

w − z

∫[z,w]

(f(u)− f(z)

)du

∣∣∣∣c©2018 Arizona State University School of Mathematical & Statistical Sciences 33


≤ maxu∈[z,w]

|f(u)− f(z)|w→z−−−→ 0.

Thus F ′ = f on U , and so the other part follows from Theorem 10.2.

The following is our first form of Cauchy’s Formula (see Theorems 11.2 and 14.4 forothers), which should be regarded as an alternative form of Cauchy’s Theorem.

Theorem 10.6 (Cauchy’s Formula in a Disk). If γ is a closed path in an open disk U , andf ∈ H(U), then

f(z)Wγ(z) =1

2πi

∫γ

f(w)

w − zdw for all z ∈ U \ γ∗. (4)

Proof. Fix z ∈ U \ γ∗, and define g : U → C by

g(w) =

f(w)− f(z)

w − zif w 6= z

f ′(z) if w = z.

Then the hypotheses of Cauchy’s Theorem in a Disk are satisfied with f and a replaced byg and z, respectively, and (4) follows.

Exercises

1. Find ∫|z|=2

z3ez dz

2. Find ∫|z|=1

ez

zdz

3. Find ∫|z|=2

dz

z2 − 1

4. Let f ∈ H(U) and a ∈ U . Prove that if f(a) = 0 and f ′(a) 6= 0 then∫|z−a|=r

1

f=

2πi

f ′(a)for small r > 0.

5. Is there a sequence of polynomials converging uniformly to 1/z on the unit circle |z| = 1?



11 First Flurry of Consequences

Lemma 11.1 (Analytic Integral Lemma). If γ is a path and f ∈ C(γ∗), then the functiong : C \ γ∗ → C defined by

g(z) =

∫γ

f(w)

w − zdw

is analytic. In fact, if a ∈ C \ γ∗ and r = d(a, γ∗), then g has a power series expansion at awith radius of convergence at least r. Moreover, the derivatives of g are given by

g(n)(z) = n!

∫γ

f(w)

(w − z)n+1dw. (5)

Proof. Let a and r be as in the statement of the lemma. Let z ∈ Dr(a). Then for all w ∈ γ∗,∣∣∣∣ z − aw − a

∣∣∣∣ ≤ |z − a|r< 1,

so by the Weierstrass M -Test the geometric series

1

w − z=∞∑n=0

(z − a)n

(w − a)n+1

converges uniformly for w ∈ γ∗. Thus∫γ

f(w)

w − zdw =

∫γ

∞∑n=0

f(w)(z − a)n

(w − a)n+1dw

=∞∑n=0

∫γ

f(w)

(w − a)n+1dw(z − a)n

=∞∑n=0

cn(z − a)n,

where

cn =

∫γ

f(w)

(w − a)n+1dw.

Since

cn =g(n)(a)

n!,

the last part now follows when a is replaced by z.

Theorem 11.2 (Higher Order Cauchy’s Formula). Every holomorphic function is analytic.In fact, if f ∈ H(U) and a ∈ U , then f has a power series expansion at a with radius ofconvergence at least d(a, ∂U). Moreover, f is infinitely differentiable, and if Dr(a) ⊆ U then

f (n)(z) =n!

2πi

∫|z−a|=r

f(w)

(w − z)n+1dw for all z ∈ Dr(a), n = 0, 1, 2, . . . . (6)



Proof. Let f , a, and r be as in the statement of the theorem. By Cauchy’s Formula in aDisk

f(z) =1

2πi

∫|z−a|=r

f(w)

w − zdw for all z ∈ Dr(a),

and the theorem now follows from Lemma 11.1.

Remark 11.3. I separated Lemma 11.1 from the proof of the Higher Order Cauchy’s The-orem because occasionally we will encounter a function defined by an integral, and then theAnalytic Integral Lemma will immediately tell us that the function is analytic, and henceholomorphic. Note that Theorem 6.12, Lemma 11.1, Cauchy’s Formula in a Disk, and theHigher Order Cauchy’s Theorem taken together tell us that a function is holomorphic if andonly if it is analytic, if and only if it can be expressed as an integral of the form given in thelemma. In particular, this is why the words holomorphic and analytic are often regarded assynonyms.

Corollary 11.4. If f is entire then for all a ∈ C, f has a power series expansion at a thatconverges for all z ∈ C.

Example 11.5. For any n ∈ N, ∫|z|=1

ezz−n dz =2πi

(n− 1)!,

because ez is entire and coincides with its (n− 1)st derivative.

Theorem 11.6 (Morera’s Theorem). Let U be open and f ∈ C(U). If∫∂∆

f = 0

for every closed triangle ∆ in U , then f is holomorphic.

Proof. Since holomorphicity is a local property, we can restrict to a disk D ⊆ U . Assumethe condition regarding triangles. As in the proof of Cauchy’s Theorem in a Disk, f has aprimitive F on D. Since F is holomorphic on D, so is f = F ′ by the Higher Order Cauchy’sFormula.

Theorem 11.7 (Isolated Zeros Theorem). Let U be a region, and let f ∈ H(U) be notidentically 0. Then the set of zeros of f has no accumulation point in U , and for every zeroa of f there exists m ∈ N such that

f(z) = (z − a)mg(z) for all z ∈ U, (7)

where g ∈ H(U) and g(a) 6= 0.

Proof. First we show that for all a ∈ U there exists n ≥ 0 such that f (n)(a) 6= 0. To see this,let

A = {a ∈ U : f (n)(a) = 0 for all n ≥ 0}.



Since f is holomorphic, it is analytic by the Higher Order Cauchy’s Formula. Thus for alla ∈ A there exists r > 0 such that

f(z) =∞∑n=0

f (n)(a)

n!(z − a)n = 0 for all z ∈ Dr(a).

Thus A is open. On the other hand,

U \ A =⋃n≥0

(f (n)

)−1(C \ {0})

is open by continuity of the functions f (n). Since f is not identically 0, we have A 6= U .Therefore A = ∅ since U is a region.

Let Z be the set of zeros of f . By continuity, any accumulation point of Z in U will bean element of Z. On the other hand, if (7) holds then g is nonzero in a neighborhood ofa, and it follows that a is not an accumulation point of Z. Thus it suffices to show that ifa ∈ Z then (7) holds for suitable m, g.

Choose r > 0 such that Dr(a) ⊆ U . Then by the Higher Order Cauchy’s Formula we canexpand f as

f(z) =∞∑n=0

cn(z − a)n for z ∈ Dr(a).

By the first part of the proof, we can choose the smallest m such that cm 6= 0. Then m ≥ 1since c0 = f(a) = 0. Thus

f(z) =∞∑n=m

cn(z − a)n = (z − a)m∞∑n=m

cn(z − a)n−m.

Define g : U → C by

g(z) =

{(z − a)−mf(z) if z 6= a

cm if z = a.

Then (7) holds, g(a) 6= 0, and g ∈ H(U \ {a}). Since

g(z) =∞∑n=m

cn(z − a)n−m for all z ∈ Dr(a) \ {a},

and the functions on both sides are continuous on Dr(a), g equals the power series on thewhole disk Dr(a), and hence is holomorphic on U .

Definition 11.8. In the notation of the Isolated Zeros Theorem, m is the order of the zeroa of f , written m = Ord(f, a).

Remark 11.9. In the proof of the Isolated Zeros Theorem, we could define g in the diskDr(a) by the power series

∑∞n=m cn(z−a)n−m; however, this does not give us g on the whole

region U .



Remark 11.10. In the Isolated Zeros Theorem, for any k ≤ Ord(f, a) we could also factorf(x) = (z − a)kg(z) for some g ∈ H(U), in which case a might also be a zero of g, and wewould say that f has a zero of order at least k at a. When we apply this fact, we will justsay “by the Isolated Zeros Theorem”.

How can we check whether f has a zero of order at least k at a? One sufficient conditionis that there exists c > 0 such that |f(z)| ≤ c|z − a|k for z near a (exercise). Anothersufficient condition, which is obvious by considering the power series of f about a, is thatf (j)(a) = 0 for all j = 0, 1, . . . , k.

Corollary 11.11. If U is a region, f, g ∈ H(U), and the set where f = g has an accumula-tion point in U , then f = g.

Proof. By the Isolated Zeros Theorem, f − g is identically 0.

The following theorem (together with its corollary) is a preliminary version of the Argu-ment Principle (see Theorem 16.10 and the comment following its proof).

Theorem 11.12. Let f be nonconstant and holomorphic on a disk D, and let {aj} be thezeros of f , counted according to their multiplicity. Then for every closed path γ in D thatmisses every aj, ∑

j

Wγ(aj) =1

2πi

∫γ

f ′

f.

Proof. By Problem 8, Wγ(aj) is only finitely nonzero, so the sum is finite. Applying theIsolated Zeros Theorem finitely many times, we find a holomorphic function g with no zerosin D \ {a : Wγ(a) 6= 0} such that

f(z) =∏j

(z − aj)g(z),

and hencef ′(z)

f(z)=∑j

1

z − aj+g′(z)

g(z)

for all z ∈ γ∗. Since g′/g ∈ H(D), by Cauchy’s Theorem in a Disk we have∫γg′/g = 0, so∫

γ

f ′

f=∑j

∫γ

dz

z − aj=∑j

2πiWγ(aj).

Theorem 11.12 can be alternatively formulated as following: letting Z be the set of zerosof f in D,

1

2πi

∫γ

f ′

f=∑a∈Z

Ord(f, a)Wγ(a).



Corollary 11.13. With the hypotheses of Theorem 11.12, if ζ = f ◦ γ, then

Wζ(0) =∑j

Wγ(aj).

Proof. If the parameter interval of γ is [a, b], we have∫ζ

dw

w=

∫ b

a

ζ ′(t)

ζ(t)dt

=

∫ b

a

f ′(γ(t))

f(γ(t))γ′(t) dt

=

∫γ

f ′

f,

so the corollary follows from Theorem 11.12.

Definition 11.14. If f is a function and b ∈ C, a simple root of the equation f(z) = b is azero of the function f − b of order 1.

Corollary 11.15. Let f ∈ H(U), a ∈ U , and f(a) = b, and let the order of the zero of f − bat a be m. Then there exist ε, δ > 0 such that for all w ∈ Dδ(b) \ {b} the equation f(z) = whas exactly m roots in Dε(a) \ {a}, and moreover all these roots are simple.

Proof. By the Isolated Zeros Theorem, we can choose ε > 0 such that Dε(a) ⊆ U and forall z ∈ Dε(a) \ {a} we have both f(z) 6= b and f ′(z) 6= 0. Let γ be the positively orientedboundary circle |z − a| = ε, and let ζ = f ◦ γ. Since b /∈ ζ∗, we can choose δ > 0 such thatDδ(b) ∩ ζ∗ = ∅. Then the disk Dδ(b) is contained in a component of C \ ζ∗, so the windingnumber Wζ(w) is constant for w in this disk. Applying Corollary 11.13 to the function f− b,but with the same image path ζ = f ◦ γ, we have

Wζ(b) = m,

because Wγ(a) = 1 and the root a of f(z) = b has multiplicity m. Thus for all w ∈ Dδ(b) wehave Wζ(w) = m, so again by Corollary 11.13 the equation f(z) = w has exactly m roots inDε(a) counting multiplicity. In particular, for w ∈ Dδ(b)\{b}, every root z is in Dε(a)\{a},and hence is simple since f ′(z) 6= 0.

Corollary 11.16 (Open Mapping Theorem). If f ∈ H(U) is nonconstant, then f(U) isopen.

Proof. In Corollary 11.15 we have f(Dε(a)) ⊇ Dδ(b).

Corollary 11.17 (Inverse Function Theorem). If f ∈ H(U), a ∈ U , and f ′(a) 6= 0, thenthere exist open sets V,W such that

(1) a ∈ V ⊆ U ,



(2) f maps V 1-1 onto W , and

(3) the inverse function g : W → V is holomorphic.

Proof. In Corollary 11.15 we have m = 1, and we take

W = Dδ(b) and V = f−1(W ) ∩Dε(a).

By Corollary 11.15 and the Open Mapping Theorem, f : V → W is an open bijection, sog : W → V is a continuous bijection. Fix d ∈ W . Then f(c) = d for a unique c ∈ V . Ifw ∈ W and g(w) = z ∈ V , we have

g(w)− g(d)

w − d=

z − cf(z)− f(c)

.

Since g is continuous, z → c when w → d. If c = a, then f ′(c) 6= 0 by hypothesis, while ifc 6= a then f ′(c) 6= 0 by the construction in the proof of Corollary 11.15. Thus g′(d) = 1/f ′(c).Therefore g ∈ H(W ).

Corollary 11.18. If f ∈ H(U) is 1-1, then f ′ is never 0, and the inverse of f is holomorphic.

Proof. If f(a) = b and f ′(a) = 0, then the power series for f − b looks like

f(z)− b =∞∑n=2

cn(z − a)n = (z − a)2

∞∑n=0

cn+2(z − a)n,

so a is a zero of f − b with order at least 2. But then by Corollary 11.15 there exists m ≥ 2such that f is m to 1 in some punctured disk Dr(a) \ {a}, which is a contradiction. Now theholomorphicity of f−1 follows from the Inverse Function Theorem.

Corollary 11.19. (1) ez is holomorphic and 1-1 on the strip | Im z| < π, and the imageis the slit plane C \ (−∞, 0]. Therefore there is a holomorphic branch of log z on thisslit plane, with range the strip.

(2) Similarly, for all n ∈ N, n√z has a holomorphic branch on the slit plane C \ (−∞, 0],

with range the wedge | arg z| < π/n.

(3) Moreover, every holomorphic function whose range is contained in the slit plane has aholomorphic log and a holomorphic nth root.

Remark 11.20. In Corollary 11.19 the plane could be slit along any ray starting at 0.

Corollary 11.21 (Local Map Theorem). Let f ∈ H(U) be nonconstant, a ∈ U , and b =f(a). If m is the order of the zero of f − b at a, then there exist an open neighborhood V ofa and g ∈ H(V ) such that f(z) = b+ g(z)m for all z ∈ V .



Proof. By the Isolated Zeros Theorem we can find a holomorphic function h such thath(a) 6= 0 and

f(z)− b = (z − a)mh(z) for all z near a.

Restrict to an open neighborhood V of a such that

|h(z)− h(a)| < |h(a)| for all z ∈ V.

Then by Corollary 11.19 we can choose a holomorphic branch of h1/m on V , and we can take

g(z) = (z − a)h1/m(z).

Corollary 11.22 (Maximum Modulus Theorem). If U is a region and f ∈ H(U) is non-constant, then |f | has no local maximum on U .

Proof. This is obvious, since by the Open Mapping Theorem every value f(a) has a neigh-borhood V ⊆ f(U).

Corollary 11.23 (Alternative Maximum Modulus Theorem). If U is a bounded region andf is nonconstant and holomorphic on U and continuous on U , then the maximum of |f |occurs only on ∂U .

Proof. By way of contradiction, suppose that a ∈ U and

|f(a)| = max{|f(z)| : z ∈ U}.

Then in particular |f | has a local maximum at a, which contradicts the Maximum ModulusTheorem.

Corollary 11.24 (Minimum Modulus Theorem). Let U be a bounded region and f be non-constant and holomorphic on U and continuous on U . If f has no zeros in U then theminimum of |f | occurs only on ∂U .

Proof. Assume that f has no zeros in U , and suppose that |f | has a minimum at a ∈ U .Then we can restrict to a closed disk Dr(a) ⊆ U , and then |1/f | has a local maximum at a,which is a contradiction.

Corollary 11.25 (Schwarz Lemma). If f ∈ H(D), f(0) = 0, and |f | ≤ 1, then |f(z)| ≤ |z|for all z ∈ D, and |f ′(0)| ≤ 1. Moreover, if |f(z)| = |z| for some z 6= 0, or if |f ′(0)| = 1,then there is a constant c ∈ T such that f(z) = cz.

Proof. By the Isolated Zeros Theorem, we can factor f(z) = zg(z) for some g ∈ H(D).Moreover, the hypotheses imply that |g| ≤ 1 on D\{0}, and hence on D by continuity. Thus|f(z)| ≤ |z| on D. Also, since g(0) = f ′(0), we get |f ′(0)| ≤ 1.

Now suppose that |f(z)| = |z| for some z 6= 0. Then |g(z)| = 1, so by the MaximumModulus Theorem g is constant, with constant value c ∈ T, and hence f(z) = cz on D. Onthe other hand, if we instead assume that |f ′(0)| = 1, then |g(0)| = 1, and again we getg = c and hence f(z) = cz on D.



Exercises

1. Find ∫|z−1|=1

cos(πz/4)

(z − 1)5dz

2. Let f be holomorphic on |z| < 2, and assume that |f(z) − z| < |z| on T. Prove that|f ′(1/2)| ≤ 8.

3. Let f be entire, and suppose that for every a ∈ C the power series expansion of f about ahas at least one zero coefficient. Prove that f is a polynomial.

4. Let f ∈ H(U), a ∈ U , and k ∈ N. Suppose that there is c > 0 such that |f(z)| ≤ c|z − a|kfor z near a. Prove that f has a zero of order at least k at a.

5. Let f be holomorphic on the unit disk D, and n ∈ N, and assume that |f(z)| ≤ |z|n on D,and that f(a) = an for some a 6= 0. Prove that f(z) = zn on D.

6. Let f ∈ H(U), 0 ∈ U , f(0) = 0, and f ′(0) 6= 0. Let n ∈ N. Prove that there exist an openneighborhood V of 0 and g ∈ H(V ) such that

f(zn) = g(z)n for all z ∈ V.

7. Let U be a bounded region, and let (fn) be a sequence of holomorphic functions on U thatare continuous on U . Prove that if the sequence converges uniformly on ∂U then it convergesuniformly on U .

8. Let f be nonconstant and holomorphic on a disk D = Dr(a), let Z be the set of zeros of fin D, and let γ be a closed path in D with γ∗ ∩ Z = ∅. Let B = {a ∈ Z : Wγ(a) 6= 0}.Show that B is finite.

12 Liouville’s Theorem

Theorem 12.1 (Cauchy’s Estimates). If f ∈ H(Dr(a)) and |f | ≤M on Dr(a), then

|f (n)(a)| ≤ n!M

rnfor all n ∈ N.

Proof. For 0 < s < r, by the Higher Order Cauchy’s Formula

f (n)(a) =n!

2πi

∫|z−a|=s

f(w)

(w − a)n+1dw,

so

|f (n)(a)| ≤ n!Ms

sn+1=n!M

sn.

Letting s ↑ r gives the theorem.



Example 12.2. If f is entire and there exists n ∈ N such that |f(z)| ≤ |z|n for all large z,then f is a polynomial, because for all k > n and all large r, by Cauchy’s Estimates

|f (k)(0)| ≤ k!rn

rk=

k!

rk−nr→∞−−−→ 0,

so the power series expansion

f(z) =∞∑k=0

f (k)(0)

k!zk

is finite. In fact, the above reasoning shows that f is a polynomial of degree at most n.

Corollary 12.3 (Liouville’s Theorem). Every bounded entire function is constant.

Proof. Letting n = 1 and r →∞ in Cauchy’s Estimates, we see that f ′ is identically 0, andhence f is constant.

Corollary 12.4 (Fundamental Theorem of Algebra). Every nonconstant polynomial has azero.

Proof. Suppose that p is a nonconstant polynomial with no zero. Then 1/p is entire, and isbounded since lim|z|→∞ |p(z)| = ∞. But then by Liouville’s Theorem 1/p, and hence p, isconstant.

Definition 12.5. A sequence of functions on a region U converges uniformly on compactsets if it converges uniformly on every compact subset of U .

Theorem 12.6 (Weierstrass’ Convergence Theorem). If (fn) is a sequence in H(U), and if

fn → f uniformly on compact sets, then f ∈ H(U), and f(k)n → f (k) uniformly on compact

sets for every k ∈ N.

Proof. On each closed disk in U , f is continuous because it is a uniform limit of continuousfunctions. If ∆ is a triangle in U , then∫

∆

f = lim

∫∆

fn = 0,

so f is holomorphic by Morera’s Theorem.

For the uniform convergence of the derivatives, let K ⊆ U be compact, and let a ∈ K.Choose r > 0 such that Dr(a) ⊆ U . Then for 0 < s < r the formula

f ′n(z) =1

2πi

∫|z−a|=r

fn(w)

(w − z)2dw,

and similarly for f ′, implies that f ′n → f ′ uniformly on Ds(a). Since K can be covered byfinitely many such open disks, f ′n → f ′ uniformly on K.

Since the derivatives are holomorphic, an induction argument now shows that f(k)n → f (k)

uniformly on compact sets for every k ∈ N.

Remark 12.7. In Weierstrass’ Convergence Theorem the hypotheses can be slightly weak-ened by not requiring the fn to be holomorphic on all of U . For example, it is enough tohave fn ∈ H(Un) with Un increasing to U .



Exercises

1. Let f ∈ C(C) be analytic off an interval [a, b] with a, b ∈ C. Prove that f is entire.

2. Can every continuous function on D be approximated uniformly by polynomials?

3. Let f be entire, and suppose that |f(z)|/|z|7 has a finite limit as z →∞. Prove that f is apolynomial of degree at most 7.

4. Prove that∞∑n=1

1

nz

is a holomorphic function of z on the half-plane Re z > 1. Also, find the derivative.

5. (Hurwitz’ Theorem) Let U be a region and fn ∈ H(U) for all n ∈ N. Suppose that fn → funiformly on compact sets, and that none of the functions fn has a zero in U . Prove that iff has a zero in U then f is constant.

6. Show how Weierstrass’ Convergence Theorem could be proved using Cauchy’s Formula,the Higher Order Cauchy’s Formula, and Lemma 11.1, together with uniform convergencearguments.

13 Isolated singularities

Definition 13.1. If U is open, a ∈ U , and f ∈ H(U \{a}), then f has an isolated singularityat a. The singularity is called removable if f is bounded near a, a pole if limz→a |f(z)| =∞,and essential if it is neither removable nor a pole.

Theorem 13.2. If f ∈ H(U\{a}) has a removable singularity at a, then f has a holomorphicextension to U .

Proof. Choose a punctured disk Dr(a) \ {a} on which f is bounded. Define g : U → C by

g(z) =

{(z − a)2f(z) if z 6= a

0 if z = a.

Our boundedness assumption implies that g′(a) = 0. Then g ∈ H(U) has a zero at a oforder at least 2, so we can write

g(z) =∞∑n=2

cn(z − a)n for all z ∈ Dr(a).



Extend f to U by definingf(a) = c2.

Then

f(z) =∞∑n=0

cn+2(z − a)n for all z ∈ Dr(a),

so f ∈ H(U).

Theorem 13.3. If f has a pole at a, then there exist n ∈ N and a holomorphic function gsuch that g(a) 6= 0 and

f(z) =g(z)

(z − a)nfor z near a.

Proof. By the Isolated Zeros Theorem we can choose r > 0 such that f(z) 6= 0 for 0 <|z − a| < r. Then in this punctured disk the function h = 1/f has a removable singularityat a, because h→ 0 since |f | → ∞. So we can extend h holomorphically to Dr(a), and thenh has a zero at a; let n be its order. Then h(z) = (z − a)nk(z) with k holomorphic andk(a) 6= 0, and we can take g = 1/k.

Definition 13.4. In Theorem 13.3, n is called order of the pole, written n = Ord(f, a).Dividing the power series of g at a by (z − a)n, we get c1, . . . , cn ∈ C such that cn 6= 0 and

f(z)−n∑k=1

ck(z − a)k

has a removable singularity at a. The function

n∑k=1

ck(z − a)k

is the principal part (also called the singular part) of f at a. A pole is simple if its orderis 1. An isolated singularity that is either removable or a pole is called at most a pole. Afunction f is meromorphic on a region U if is it holomorphic on U except for poles.

Theorem 13.5 (Casorati-Weierstrass Theorem). If f has an essential singularity at a, thenfor every punctured disk D = Dr(a) \ {a} on which f is holomorphic, the image f(D) isdense in C.

Proof. Suppose not, and choose c ∈ C such that |f − c| is bounded away from 0 on D. Theng = 1/(f − c) is bounded on D, so g has a removable singularity at a. But then it followsfrom Problem 2 that f = c+ 1/g has at most a pole at a, which is a contradiction.

Remark 13.6. If f is holomorphic on |z| > R, then we say that f has an isolated singularityat ∞. Then the function g(z) = f(1/z) has an isolated singularity at 0. If f is boundedas |z| → ∞, then g has a removable singularity at 0, and we say that f has a removablesingularity at ∞, and moreover we have lim|z|→∞ f(z) = limz→0 g(z). If |f(z)| → ∞ as



|z| → ∞, then g has a pole at 0, and we say that f has a pole at ∞, and moreover if theprincipal part of g at 0 is a polynomial p in 1/z, we say that the principal part of f at infinityis the polynomial p itself. If the singularity of f at ∞ is neither removable nor a pole, wesay it is essential, and in that case the Casorati-Weierstrass Theorem implies (via g) that ftakes values arbitrarily close to every complex number as |z| → ∞. If f is meromorphic inC and has a pole at ∞, then it must be rational by Problem 5, and if we further assumethat in fact f is entire then it is a polynomial by Problem 4.

Example 13.7. ez, cos z, and sin z have essential singularities at ∞.

Exercises

1. In each part, the function f has an isolated singularity at z = 0. Classify this singularity asremovable, pole, or essential. If 0 is removable, find the value that must be assigned to f at0 to get a holomorphic extension. If 0 is a pole, find its order and principal part.

(a) f(z) = sin zz

(b) f(z) = cos zz

(c) f(z) = log(1+2z)z

(d) f(z) = e1/z

(e) f(z) = z2+1z2(z+1)

(f) f(z) = zsin z

(g) f(z) = 1sin z

2. Let f ∈ H(U \ {a}). Prove that if f has at most a pole at a, at a, then 1/f also has at mosta pole at a.

3. Prove that if f ∈ H(U) is nonconstant and has a zero at a, then 1/f has a pole at a, withorder equal to the order of the zero of f .

4. Prove that an entire function with a pole at ∞ is a polynomial.

5. Prove that a meromorphic function with a pole at ∞ is rational.

6. Prove that a nonconstant entire function has range dense in C.



14 Global Cauchy Theorem

Definition 14.1. A chain is a formal Z-linear combination

γ =n∑j=1

mjγj (8)

of paths γj, where mj ∈ Z. We write γ∗ =⋃nj=1 γ

∗j , and we define∫

γ

f =n∑j=1

mj

∫γj

f for f ∈ C(γ∗).

Z-linear combinations of chains are defined in the obvious way, and we define −γ = (−1)γ.A chain γ is a cycle if it can be expressed in the form (8) where each path γj is closed, andγ is in U if each path γj is in U . Note that a cycle can be a sum of nonclosed paths.

Definition 14.2. If γ is a cycle and a /∈ γ∗, the winding number of γ with respect to a is

Wγ(a) =1

2πi

∫γ

dz

z − a=

n∑1

mjWγj(a).

Definition 14.3. Two cycles γ and η in an open set U are homologous in U if

Wγ(a) = Wη(a) for all a /∈ U,

and γ is homologous to 0 in U if

Wγ(a) = 0 for all a /∈ U.

Theorem 14.4 (Cauchy’s Formula). If f ∈ H(U) and γ is homologous to 0 in U , then

f(z)Wγ(z) =1

2πi

∫γ

f(w)

w − zdw for all z ∈ U \ γ∗.

Proof. Define g : U × U → C by

g(z, w) =

f(w)− f(z)

w − zif w 6= z

f ′(z) if w = z.

We need to know that g is continuous. Only the continuity at points (a, a) is nonobvious.Fix a ∈ U , and let ε > 0. Choose r > 0 such that Dr(a) ⊆ U and |f ′(u) − f ′(a)| < ε forall u ∈ Dr(a). If z 6= w in Dr(a), and γ = [z, w], then γ∗ ⊆ Dr(a), so by the FundamentalTheorem of Calculus

|g(z, w)− g(a, a)| =∣∣∣∣∫ 1

0

(f ′(γ(t))− f ′(a)

)dt

∣∣∣∣ < ε.



On the other hand, if z ∈ Dr(a) then

|g(z, z)− g(a, a)| = |f ′(z)− f ′(a)| < ε.

Thus g is continuous at (a, a).

Let V = {z ∈ C : Wγ(z) = 0}, and note that U ∪ V = C, since γ is homologous to 0 inU . Define h : C→ C by

h(z) =

{∫γg(z, w) dw if z ∈ U∫

γf(w)w−z dw if z ∈ V,

which is well-defined since the two integrals coincide for z ∈ U ∩ V . Then h is continuous,by two applications of Problem 5 in Section 9. We need to know that h is holomorphic. Itis holomorphic on V by Lemma 11.1, but holomorphicity on U takes a bit more work. If ∆is a closed triangle in U , then by Fubini’s Theorem∫

∂∆

h =1

2πi

∫∂∆

∫γ

g(z, w) dw dz =1

2πi

∫γ

(∫∂∆

g(z, w) dz

)dw. (9)

For each w ∈ U , the function z 7→ g(z, w) is holomorphic, since the singularity at z = w isremovable. Thus by Cauchy’s Theorem for a Triangle, the inner integral on the right-handside of (9) is 0 for all w ∈ γ∗. Thus by Morera’s Theorem h is holomorphic on U .

Thus h is entire. Since V contains the unbounded component of C \ γ∗,

lim|z|→∞

h(z) = 0,

since

lim|z|→∞

f(w)

w − z= 0 uniformly for w ∈ γ∗.

Thus by Liouville’s Theorem h is identically 0, and hence∫γg(z, w) dw = 0 for all z ∈ U .

The theorem follows, since for all z ∈ U \ γ∗,

h(z) =

∫γ

f(w)

w − zdw − f(z)2πiWγ(z).

Example 14.5. By Cauchy’s Formula,∫|z|=1

cos z

zdz = 2πi cos 0 = 2πi,

because cos z is entire, the circle is homologous to 0 in C, and the winding number of thecircle with respect to 0 is 1.

Theorem 14.6 (Cauchy’s Theorem). If f ∈ H(U) and γ is homologous to 0 in U , then∫γ

f = 0.



Proof. Fix a ∈ U \ γ∗, and define g ∈ H(U) by g(z) = (z − a)f(z). Then by Cauchy’sFormula

1

2πi

∫γ

f =1

2πi

∫γ

g(z)

z − adz = g(a)Wγ(a) = 0

because g(a) = 0.

Here is a common use of Cauchy’s Theorem: if f ∈ H(U) and γ and ζ are homologouscycles in U , then

∫γf =

∫ζf , by Cauchy’s Theorem applied to γ − ζ.

Example 14.7. Let γ be homologous to 0 in an open set U , and let f be holomorphic onU except at z1, . . . , zn not in γ∗. For each j = 1, . . . , n let γi be the positively orientedboundary of a closed disk Dj centered at zj and contained in U . Choose the radii so thatthe disks are pairwise disjoint. Let V = U \ {z1, . . . , zn}, and for each j let mj = Wγ(zj).Then γ is homologous to

∑n1 mjγj in V , and∫

γ

f =n∑1

mj

∫γj

f.

To see this, let η = γ −∑n

1 mjγj. If a /∈ U , then

Wη(a) = Wγ(a)−n∑1

mjWγj(a) = 0

because Wγj(a) = 0 for every j. On the other hand, if a = zk for some k, then

Wγj(a) =

{1 if j = k

0 if j 6= k,

and soWη(a) = Wγ(a)−mk = 0.

Therefore η is homologous to 0 in V . Since f ∈ H(V ), the result now follows from Cauchy’sTheorem.

Remark 14.8. How do we easily see that two cycles are homologous? Although we won’tuse it, homotopy is effective: two closed curves γ, ζ in a region U , both with parameterinterval I = [0, 1], are U-homotopic if there exists a continuous H : I2 → C such that

H(s, 0) = γ(s), H(s, 1) = ζ(s), and H(0, t) = H(1, t)

for all s, t ∈ I. It is not hard to prove that if γ, ζ are closed paths with parameter interval[0, 1], a ∈ C, and

|ζ(s)− γ(s)| < |a− γ(s)| for 0 ≤ s ≤ 1,

then Wζ(a) = Wγ(a). Applying this with cleverly chosen auxiliary paths, one can prove thatU -homotopic paths are homologous in U . The subtlety here is that the curves in a homotopyare not necessarily all paths. A closed curve γ is null-homotopic in U if it is U -homotopic to



a point — i.e. to a constant curve. A region U is simply connected if every closed curve in Uis U -homotopic to a point. Cauchy’s Theorem and Formula take simpler forms in a simplyconnected region. We will not prove it, but it is a fact that a region U is simply connectedif and only if every cycle in U is homologous to 0 in U . In view of this, for our purposes itis more convenient to adopt the following definition:

Definition 14.9. A region U is simply connected if every cycle in U is homologous to 0 inU .

Example 14.10. Every convex open set is simply connected.

Example 14.11. The slit plane C \ [0,∞) is simply connected.

Example 14.12. The punctured plane C \ {0} is not simply connected.

The following theorem extends Cauchy’s Theorem in a Disk to arbitrary simply connectedregions.

Theorem 14.13. If U is simply connected and f ∈ H(U), then f has a primitive on U .

Proof. Fix a ∈ U , and define g : U → C by

g(z) =

∫γ

f,

where γ is any path in U from a to z. By Cauchy’s Theorem and simple connectedness, thisis well-defined, since if γ and ζ are both paths from a to z in U then γ − ζ is a cycle, and so∫γf =

∫ζf . Let z ∈ U , and choose r > 0 such that Dr(z) ⊆ U . Then 0 < |h| < r implies

g(z + h)− g(z)

h=

1

h

∫[z,z+h]

f.

Thus ∣∣∣∣g(z + h)− g(z)

h− f(z)

∣∣∣∣ =

∣∣∣∣∫ 1

0

(f(z + th)− f(z)

)dt

∣∣∣∣≤ max{|f(z + th)− f(z)| : 0 ≤ t ≤ 1}h→0−−→ 0,

so g′(z) = f(z).

Corollary 14.14. If U is simply connected and f ∈ H(U) is never 0, then f there areholomorphic branches on U of log f and f 1/n for any n ∈ N.

Proof. Since f ′/f ∈ H(U), by Theorem 14.13 it has a primitive g. The derivative of fe−g is0, so (modulo adding a suitable constant to g) f = eg. Thus g is a holomorphic branch oflog f , and then eg/n is a holomorphic branch of f 1/n.



Exercises

1. Find the winding number of γ with respect to a in the image below:

2. (a) Find the winding numbers of γ with respect to a and b in the image below:

(b) Find the winding numbers of ζ with respect to a and b in the image below:

(c) Find the winding numbers of γ with respect to a and b in the image below:



3. Find the largest region in which the function

f(z) =

∫ 1

0

dt

1 + tz

is holomorphic. Use Morera’s Theorem and Fubini’s Theorem.

4. Let γ be a closed path, and let U = C \ γ∗. Then U is an open set. Prove that U can haveinfinitely many connected components. (But of course there will only be countably many.)

15 Laurent series

Definition 15.1. Given a doubly infinite sequence (zn)∞n=−∞ in C, the series∑∞

n=−∞ znconverges absolutely if both series

∑n≥0 zn and

∑n<0 zn converge absolutely, in which case

the sum is∞∑

n=−∞

zn =∑n≥0

zn +∑n<0

zn.

Given functions (fn)∞−∞ defined on E ⊆ C, if the series∑∞−∞ fn(z) converges absolutely for

each z ∈ E, then the series∑∞

n=−∞ fn converges uniformly if both∑

n≥0 fn and∑

n<0 fnconverge uniformly. A Laurent series is a series of the form

∞∑n=−∞

cnzn,

with coefficients cn.

Definition 15.2. Given a ∈ C and 0 ≤ r < R, the open annulus with center a, inner radiusr, and outer radius R is the set

A(a, r, R) = {z ∈ C : r < |a− z| < R}.

Theorem 15.3. Let A = A(a,R1, R2), and let f ∈ H(A). Then f has a Laurent seriesexpansion

f(z) =∞∑

n=−∞

cn(z − a)n



that converges absolutely on A and uniformly on compact subsets of A. Moreover, for any rwith R1 < r < R2,

cn =1

2πi

∫|z−a|=r

f(w)

(w − z)n+1dw for all n ∈ Z. (10)

Note that the last part of Theorem 15.3 says in particular that on any annulus the Laurentexpansion of f is unique.

Proof. Note first that by Cauchy’s Theorem the integrals in (10) are independent of thechoice of r, since any two of the circles are homologous in A.

LetR1 < r1 < r2 < R2,

and for j = 1, 2 let γj = {z : |z − a| = rj} and define fj on C \ γ∗j by

fj(z) =1

2πi

∫γj

f(w)

w − zdw.

Then fj ∈ H(C \ γ∗j ), and since the cycle γ2 − γ1 is homologous to 0 in A, by Cauchy’sFormula

f = f2 − f1

on A(a, r1, r2).

We already know that

f2(z) =∞∑n=0

1

2πi

∫γ2

f(w)


on {z : |z − a| < r2}. The expansion

f1(z) = −∑n<0

1

2πi

∫γ1

f(w)


on {z : |z−a| > r1} is similar, and can be justified using a geometric series in (w−a)/(z−a).This gives the Laurent expansion of f on A1 ∩ A2, converging absolutely, and uniformly oncompact subsets. Since r1 < r2 were arbitrary, the theorem follows.

Remark 15.4. If R1 = 0, then a is an isolated singularity of f , that is

• removable if cn = 0 for all n < 0,

• a pole of order m if c−m 6= 0 and cn = 0 for all n < −m,

• and essential if cn 6= 0 for infinitely many n < 0.



Example 15.5. The Laurent series of e1/z about 0 is computed by the change of variablez 7→ 1/z in the power series for ez:

e1/z =0∑

n=−∞

zn

n!for 0 < |z|,

By Theorem 15.3 the series converges uniformly on compact subsets of C \ {0}. In fact, inthis case it converges uniformly on the complement of any disk Dr.

Exercises

1. Find ∫|z|=1

e1/z dz

2. Find ∫|z|=1

e1/z2 dz

3. Find the Laurent series of sin(1/z), about 0, on the punctured plane C \ {0}.

4. Find ∫|z|=1

sin 1/z dz

5. Find ∫|z|=1

cos 1/z dz

6. Prove that if f is holomorphic on an annulus, then the Laurent series of f ′ is gotten bydifferentiating the Laurent series of f term-by-term.

7. Find the Laurent series for

f(z) =1

(z − 1)(z − 2)

in the:

(a) disk |z| < 1

(b) annulus 1 < |z| < 2

(c) region 2 < |z|

8. Toward the end of the proof of Theorem 15.3, I said that the expansion of f1 can be justifiedusing a geometric series in (w − a)/(z − a). Give the precise details of how this goes.



16 Calculus of residues

Definition 16.1. Let f have an isolated singularity at a, and let

f(z) =∞∑

n=−∞

cn(z − a)n

be its Laurent series at a. The residue of f at a is

Res(f, a) = c−1.

If the singularity of f at a is removable, then of course Res(f, a) = 0. If the singularityis a pole of order m, then we let g(z) = (z − a)mf(z) and compute the residue as

Res(f, a) =g(m−1)(a)

(m− 1)!.

In particular, at a simple pole the residue is limz→a(z − a)f(z). At an essential singularity,the residue is not so easy to compute in general, unless we can find the Laurent series.(However, see Problems 2, 3, and 4.)

Theorem 16.2 (Residue Formula). Let f be holomorphic in a region U except for a set Aof isolated singularities, and let γ be a cycle in U \ A that is homologous to 0 in U . Then

1

2πi

∫γ

f =∑a∈A

Res(f, a)Wγ(a).

Proof. Let B = {a ∈ A : Wγ(a) 6= 0}. By Problem 32, B is finite, so we can writeB = {a1, . . . , ak}. Let D1, . . . , Dk be disjoint closed disks centered at the aj, with boundary

circles γj. Then γ is homologous to∑k

1 Wγ(aj)γj in U \ A, so by Cauchy’s Theorem∫γ

f =k∑1

Wγ(aj)

∫γj

f.

For each j = 1, . . . , k, the Laurent series f(z) =∑∞

n=−∞ cn(z − aj)n converges uniformly onγ∗j , so ∫

γj

f =∞∑

n=−∞

cn

∫γj

(z − aj)n dz.

Since ∫γj

(z − aj)n dz =

{2πi if n = −1

0 if n 6= −1,

the theorem follows.



The Residue Formula is useful in evaluating many integrals of functions of one realvariable. In the following examples we illustrate a few of the most important techniques.There will be a lot of exposition in the examples. In practice, most of the explanation canbe omitted, but the steps still must be included, at least in abbreviated form.

Definition 16.3. Let γ be a cycle such that Wγ(z) = 0 or 1 for every z /∈ γ∗. A set A isenclosed by γ if Wγ(z) = 1 for all z ∈ A.

Example 16.4. We can use the calculus of residues to handle the integral from 0 to 2π ofany rational function of cos and sin, as indicated in the following example:∫ 2π

0

1

2 + cos θdθ.

We want to substitute z = eiθ. We have not recorded any result involving such a change ofvariables, so we pause to justify it in general: Suppose that R(x, y) is a rational function oftwo real variables x, y that is continuous on the unit circle x2 + y2 = 1. Define

f(z) =R(z+1/z

2, z−1/z

2i

)iz

.

Then f is a rational function with no poles on T, and for z = eiθ in T we have

cos θ = Re z =z + 1/z

2and sin θ = Im z =

z − 1/z

2i,

and hence ∫Tf(z) dz =

∫ 2π

0

f(eit)ieit dt =

∫ 2π

0

R (cos θ, sin θ) dt.

Given the integral∫ 2π

0R(cos θ, sin θ) dθ, we want to replace it with

∫T f . In practice, rather

than remember the above form of f(z), it is easier so just substitute cos θ = (z+ 1/z)/2 andsin θ = (z − 1/z)/2i and use the rule that if z = eiθ then dz = ieiθdθ = izdθ, so that

dθ =dz

iz.

In the present example we have∫ 2π

0

1

2 + cos θdθ =

∫T

1

2 + z+1/z2

· dziz

=

∫T

−2i

z2 + 4z + 1dz

=

∫Tf,



where

f(z) =−2i

z2 + 4z + 1.

By the Residue Formula we have∫ 2π

0

1

2 + cos θdθ = 2πi(sum of the residues of f enclosed by T).

To find the poles we factor the denominator:

z2 + 4z + 1 = (z − a)(z − b),

wherea = −2 +

√3 and b = −2−

√3.

Thus f has simple poles a, b. Since |a| < 1 and |b| > 1, only the pole a is enclosed by T, andwe have

Res(f, a) = limz→a

(z − a)f(z) =−2i

a− b=−2i

2√

3=−i√

3.

Therefore ∫ 2π

0

1

2 + cos θdθ =

2πi(−i)√3

=2π√

3.

Note that if we wanted the integral from 0 to π instead of 0 to 2π, we could use the propertythat cos θ is symmetric with respect to θ = π, so that we would only have to divide by 2:∫ π

0

1

2 + cos θdθ =

π√3.

Example 16.5. We can use the calculus of residues to integrate over the real line rationalfunctions with a zero at ∞ of order at least 2. For example, consider∫ ∞

−∞

1

(x2 + 1)2dx.

The function f(z) = 1/(z2 + 1)2 has a zero of order 4 at ∞, so the integral converges, with∫ ∞−∞

1

(x2 + 1)2dx = lim

r→∞

∫ r

−rf(x) dx.

Note that f has poles at ±i, both of order 2. For r > 1 consider the cycle

γr = [−r, r] + Cr,

where Cr is the upper semicircle z = reit for 0 ≤ t ≤ π, oriented counterclockwise. Forz ∈ C∗r ,

|f(z)| = 1

|z2 + 1|2≤ 1

(|z|2 − 1)2=

1

(r2 − 1)2,



keeping in mind that we are only considering r > 1. Thus∣∣∣∣∫Cr

f

∣∣∣∣ ≤ L(Cr)

(r2 − 1)2=

πr

(r2 − 1)2

r→∞−−−→ 0.

We must calculate the residue of f at i (the only pole in the upper half plane). Since theorder of the pole is 2, we let

g(z) = (z − i)2f(z) =1

(z + i)2.

Then

g′(z) =−2

(z + i)3,

so

Res(f, i) = g′(i) =−2

(2i)3=−1

4i3=−i4.

Since r > 1, the cycle γr encloses the pole i, so by the Residue Formula∫γr

f = 2πiRes(f, i) =π

2.

Therefore ∫ ∞−∞

1

(x2 + 1)2dx = lim

r→∞

∫ r

−r

1

(x2 + 1)2dx

= limr→∞

(∫γr

f −∫Cr

f

)=π

2.

Example 16.6. We can evaluate ∫ ∞−∞

cosx

x2 + 1dx

by applying the calculus of residues to the meromorphic function f(z) = eiz/(z2 + 1). Notethat the rational function 1/(x2 + 1) has a zero of order 2 at∞, and since eix is bounded forx ∈ R the integral

∫∞−∞ f(x) dx converges, with∫ ∞

−∞f(x) dx = lim

r→∞

∫ r

−rf(x) dx.

As in Example 16.5, for r > 0 let γr be the path consisting of the interval [−r, r] and thesemicircle Cr in the upper half-plane, oriented counterclockwise. Let r > 1.

For z ∈ C∗r , we have

|f(z)| = |eiz||z2 + 1|

=e−y

|z2 + 1|≤ 1

|z|2 − 1=

1

r2 − 1.



Thus ∣∣∣∣∫Cr

f

∣∣∣∣ ≤ L(Cr)

r2 − 1=

πr

r2 − 1

r→∞−−−→ 0.

γr encloses the pole i of f . The pole is simple, so

Res(f, i) = limz→i

(z − i)f(z) = limz→i

eiz

z + i=ei

2

2i= −e

−1

2i.

Since we are only considering r > 1, γr encloses the pole i, so by the Residue Formula∫γr

f = 2πiRes(f, i) =π

e.

Therefore

limr→∞

(∫ r

−r

cosx

x2 + 1dx+ i

∫ r

−r

sinx

x2 + 1dx

)= lim

r→∞

∫ r

−rf(x) dx

= limr→∞

(∫γr

f −∫Cr

f

)=π

e.

Taking the real part, we get∫ ∞−∞

cosx

x2 + 1dx = lim

r→∞

∫ r

−r

cosx

x2 + 1dx =

π

e.

Example 16.7. Let’s evaluate ∫ ∞0

x sinx

x2 + 1dx. (11)

There is a subtlety: since x/(x2 + 1) only has a zero of order 1 at∞, it is not obvious at theoutset that the improper integral converges, so we need to compute directly

limb→∞

∫ b

0

x sinx

x2 + 1dx.

In fact, we will compute ∫ ∞−∞

x sinx

x2 + 1dx = lim

a,b→∞

∫ ∞−a

x sinx

x2 + 1dx,

then note that the integrand is an even function, so we can finish by dividing by 2. Letg(z) = z/(z2 + 1) and f(z) = g(z)eiz (chosen because when z = x is real the functiong(x) sinx is the imaginary part of f(x)). This time, instead of semicircles, we will userectangles. Note that there exist M,K > 0 such that

|g(z)| ≤ M

|z|if |z| > K.



Of course, we must take K > 2 since the poles of f are ±2i, but other than that the particularchoices of M,K do not matter. For a, b > K and c = a+ b, let

γ1 = [−a, b], γ2 = [b, b+ ic], γ3 = [b+ ic,−a+ ic], γ4 = [−a+ ic,−a],

and let γ =∑4

j=1 γj.

First consider γ2: we have z = b+ iy with 0 ≤ y ≤ c, so

|g(z)| ≤ M

|z|≤ M

b,

and|eiz| = e−y.

Parameterizing by z = b+ iy for 0 ≤ y ≤ c, we have∣∣∣∣∫γ2

f

∣∣∣∣ =

∣∣∣∣∫ c

0

f(b+ iy)i dy

∣∣∣∣≤∫ c

0

M

be−y dy

≤ M

b

(since

∫ c

0

e−y dy ≤∫ ∞

0

e−y dy = 1

)b→∞−−−→ 0.

Similarly, ∣∣∣∣∫γ4

f

∣∣∣∣ ≤ M

a

a→∞−−−→ 0.

For γ3, we can integrate from left to right since we will take the absolute value. We havez = x+ ic with −a ≤ x ≤ b,

|g(z)| ≤ M

|z|≤ M

c,

and|eiz| = e−c,

so ∣∣∣∣∫γ3

f

∣∣∣∣ =

∣∣∣∣∫ b

−af(x+ ic) dx

∣∣∣∣≤ Me−c(a+ b)

ca,b→∞−−−−→ 0 (since c = a+ b).

γ encloses the pole i of f . Since this pole is simple, we have


(z − i)zeiz

z2 + 1=iei

2

2i=e−1

2=

1

2e,



and so by the Residue Formula ∫γ

f = 2πiRes(f, i) =πi

e.

Therefore

lima,b→∞

(∫ b

−a

x cosx

x2 + 1dx+ i

∫ b

−a

x sinx

x2 + 1dx

)= lim

a,b→∞

∫ b

−af(x) dx

= lima,b→∞

(∫γ

f −4∑j=2

∫γj

f

)

=πi

e.

Taking imaginary parts gives∫ ∞0

x sinx

x2 + 1dx =

1

2

∫ ∞−∞

x sinx

x2 + 1dx =

π

e.

Example 16.8. Examples 16.6 and 16.7 involved integrating a rational function R(x) timescosx or sinx, where R had no zeros on the real axis. We can in fact allow simple poles atthe zeros of the trig function. The Dirichlet integral is an example:∫ ∞

0

sinx

xdx.

As in Example 16.7 it is not obvious a priori that the integral converges, and this time weshould avoid 0, so we need to compute directly

limr↓0,R→∞

∫ R

r

sinx

xdx.

Let f(z) = eiz/z. Although we could modify the technique of Example 16.7 by introducinga small semicircle avoiding 0 (for example, see [Ahl78, Section 5.3]), we will instead use thesmall semicircle trick with the method of Example 16.6: for 0 < r < R let

γ = [−R,−r] + Cr + [r, R] + CR,

where Cr is the upper semicircle reit oriented clockwise and CR is the upper semicircle Reit

oriented counterclockwise. The function f is meromorphic in C with a simple pole at 0.Since γ does not enclose the pole, by Cauchy’s Theorem

0 =

∫γ

f =

∫[−R,−r]+[r,R]

f +

∫Cr+CR

f.

Since f has odd real part and even imaginary part, we get

2i

∫ R

r

sinx

xdx = −

∫Cr+CR

f.



We claim that∫CRf → 0 as R → ∞, and this takes a bit of calculus. On CR we have

z = Reit for 0 ≤ t ≤ π, so ∣∣∣∣∫CR

f

∣∣∣∣ =

∣∣∣∣∣∫ π

0

eiReit

ReitiReit dt

∣∣∣∣∣=

∣∣∣∣∫ π

0

eiR(cos t+i sin t) dt

∣∣∣∣≤∫ π

0

e−R sin t dt

= 2

∫ π/2

0

e−R sin t dt.

Now, on the interval [0, π/2], we have sin t ≥ t/2, so∫ π/2

0

e−R sin t dt ≤∫ π/2

0

e−Rt/2 dt

= − 4

Re−Rt/2

∣∣π/20

=4

R(1− e−Rπ/4)

R→∞−−−→ 0,

verifying the claim.

Next, the residue of f at 0 is 1, so there is a holomorphic function g such that

f(z) =1

z+ g(z),

and then ∫Cr

f =

∫Cr

dz

z+

∫Cr

g.

Now, on Cr we have z = re−it for 0 ≤ t ≤ π, and the curve is oriented clockwise, so∫Cr

dz

z= −

∫ π

0

ireit

reitdt = −πi.

On the other hand, ∫Cr

gr→0−−→ 0

because g is holomorphic and L(Cr)→ 0. Thus∫Cr

fr→0−−→ −πi.



Thus

2i

∫ R

r

sinx

xdx = −

(∫Cr

f +

∫CR

f

)r↓0,R→∞−−−−−→ πi,

so ∫ ∞0

sinx

xdx = lim

r↓0,R→∞

∫ R

r

sinx

xdx =

π

2.

Example 16.9. Consider ∫ ∞0

x1/3

x2 + 1dx

We want to use the function z1/3(z2+1), and for this we must choose a branch of the functionz1/3. Since

z1/3 = e(1/3) log z,

we must choose a branch of log. We use the slit plane C \ [0,∞), and use the branch

log z = log |z|+ i arg z with 0 < arg a < 2π,

with the convention that since |z| > 0 the value of log |z| is the real logarithm. Thus, lettingθ = arg z, we take the branch

z1/3 = |z|1/3eθi/3,

where |z|1/3 is positive cube root.

For our contours, we use something that requires some explanation. We want to define

γ = L+ + CR + L− + Cr,

where CR is the circle Reit oriented counterclockwise, Cr is the circle reit oriented clockwise,L+ is the interval [r, R] oriented from left to right as usual, and L− is the same interval butoriented from right to left. However, the new wrinkle is that we parameterize L+ in theusual way, but for L− we let arg z = 2π, so the parameterization is z = xe2πi for x goingfrom R to r. Of course, since e2πi = 1, this looks like we have not really changed anythingwith this business about the argument. But it will make a difference when we compute z1/3:for z = x in L− we take

z1/3 = x1/3e2πi/3.

Thus ∫L−

f = −∫ R

r

x1/3e2πi/3

x2 + 1dx = −e2πi/3

∫ R

r

x1/3

x2 + 1dx = −e2πi/3

∫L+

f.

On Cr, we have

|f(z)| = |z1/3||z2 + 1|



=r1/3

|z2 + 1|

≤ r1/3

1− r2(we will have r < 1),

so ∣∣∣∣∫Cr

f

∣∣∣∣ ≤ L(Cr)r1/3

1− r2=

πr4/3

1− r2

r↓0−−→ 0.

On CR, we have

|f(z)| = |z1/3||z2 + 1|

=R1/3

|z2 + 1|

≤ R1/3

R2 − 1(we will have R > 1),

so ∣∣∣∣∫CR

f

∣∣∣∣ ≤ L(CR)R1/3

R2 − 1=

πR4/3

R2 − 1

R→∞−−−→ 0.

For small r and large R, γ will enclose the poles ±i, which are both simple. Note thatthe other singularity of f is 0, which is not enclosed by γ. The residues are


z1/3

z + i

=i1/3

2i

=eπi/6

2i

and

Res(f,−i) = limz→−i

z1/3

z − i

=(−i)1/3

−2i

=eπi/2

−2iusing −i = e3πi/2)

=−eπi/2

2i.

We wrote the last value without simplifying eπi/2 because it will be convenient. It will alsobe convenient to let α = eπi/6, so that

Res(f, i) =α

2iand Res(f,−i) =

−α3

2i.



Thus by the Residue Formula ∫γ

f = 2πi

(α

2i− α3

2i

)= πα(1− α2).

Therefore ∫ ∞0

x1/3

x2 + 1dx = lim

r↓0,R→∞

∫ R

r

x1/3

x2 + 1dx

= limr↓0,R→∞

1

1− e2πi/3

∫L++L−

f

=1

1− α4lim

r↓0,R→∞

(∫γ

f −∫Cr

f −∫CR

f

)=πα(1− α2)

1− α4

=πα

1 + α2

=π

α + α(α ∈ T)

=π√3

(since α =√

3/2 + i/2).

Theorem 16.10 (Argument Principle). Let f be meromorphic on U , and let Z and P bethe sets of zeros and poles, respectively, of f . Then for any cycle γ in U \ (Z ∪ P ) that ishomologous to 0 in U ,

1

2πi

∫γ

f ′

f=∑a∈Z

Ord(f, a)Wγ(a)−∑a∈P

Ord(f, a)Wγ(a).

Proof. By Problems 32 in this section, and 8 in Section 11, both sums are only finitelynonzero.

If a is a zero of f of order m, we can write f(z) = (z− a)mg(z), where g is meromorphicon U and holomorphic at a, and g(a) 6= 0. Then

f ′(z)

f(z)=

m

z − a+g′(z)

g(z)

has a simple pole at a, and Res(f ′/f, a) = m.

Similarly, if a is a pole of f of order m, then f(z) = (z−a)−mg(z) with g(a) 6= 0, and theabove computation with −m replacing m shows that Res(f ′/f, a) = −m. The result nowfollows from the Residue Formula.



The above theorem is called the Argument Principle because the left-hand side is Wf◦γ(0),which can be interpreted as the net change in the argument of the image cycle f ◦ γ.

Corollary 16.11 (Rouche’s Theorem). Let γ be homologous to 0 in U , and suppose thatWγ(a) = 0 or 1 for every a ∈ U \ γ∗. If f, g ∈ H(U) and

|f(z)− g(z)| < |f(z)| for all z ∈ γ∗,

then f and g have the same number of zeros enclosed by γ, counted according to multiplicity.

Proof. The hypotheses imply that f and g have no zeros in γ∗. Let h = g/f ∈ H(U). Then|1− h| < 1 on γ∗. Thus h ◦ γ is a closed path in D1(1), so

0 = Wh◦γ(0)

=1

2πi

∫γ

h′

h(as in proof of Corollary 11.13)

=1

2πi

∫γ

(g′

g− f ′

f

)= number of zeros of g enclosed by γ

− number of zeros of f enclosed by γ,

counting multiplicity, where we used the Argument Principle at the last step.

Exercises

1. How many roots does z5 − 7z3 + z2 + 2z + 1 = 0 have in the open disk |z| < 1?

2. Prove that if f ∈ H(U) and a ∈ U then

Res

(f(z)

z − a, a

)= f(a).

3. Prove that if f and g are holomorphic in a neighborhood of a, f(a) 6= 0, and g has a simple0 at a, then

Res

(f

g, a

)=f(a)

g′(a).

4. Let U be open, a ∈ U , f ∈ H(U), and g ∈ H(U \ {a}), and assume that g has a simple poleat a.

(a) Prove that if f(a) 6= 0 then

Res(fg, a) = f(a) Res(g, a).



(b) Prove that if f(a) = 0 then fg has a removable singularity at a.

5. Find the singularities, classify them (removable, etc.), and find the residues:

f(z) =z2

(z2 + a2)2if a > 0.

6. Find the singularities and classify them :

f(z) =1

sin z


f(z) =1

sin2 z


f(z) =1

(z − a)m(z − b)na 6= b,m, n ∈ N


f(z) =1

ez − 1


f(z) =z3

sin z


f(z) = cos1

z

12. Find using the calculus of residues: ∫|z−i|=1/2

1

z4 − 1dz

13. Find using the calculus of residues:∫ 2π

0

cos t

a+ cos tdt if a > 1.



14. Find using the calculus of residues:∫ π/2

0

1

a+ sin2 tdt if a > 1.

15. Find using the calculus of residues:∫ 2π

0

1

1− 2a cos θ + a2dθ if 0 < a < 1.

16. Find using the calculus of residues (not the integral formula from calculus!):∫ ∞0

1

x2 + 1dx

17. Find using the calculus of residues: ∫ ∞0

x2

x4 + 1dx

18. Find using the calculus of residues: ∫ ∞−∞

cosx

(x2 + 1)2dx

19. Find using the calculus of residues: ∫ ∞−∞

cosx

x4 + 1dx


x3 sinx

x4 + 1dx


1− cosx

x2dx

using f(z) = (1− eiz)/z2.

22. Let n ∈ {2, 3, 4, . . . }, and find ∫ ∞0

1

xn + 1dx

using paths of the following form: from 0 to r, then along a circular arc to re2πi/n, then backto 0.



23. Find the Fresnel integrals ∫ ∞0

cos(x2) dx and

∫ ∞0

sin(x2) dx

by integrating eiz2

over paths of the following form: [0, r], then along a circular arc to reπi/4,then back to 0. You may use the identity

∫∞−∞ e

−x2 dx =√π.


√x

x2 + 1dx

25. Show that the function∞∑n=1

1

z2 + n2(12)

is meromorphic and find the poles together with their orders.

26. Prove that if f is holomorphic on a punctured disk D = Dr(a) \ {a}, then f has a primitiveon D if and only if Res(f, a) = 0.

27. Prove that if f is entire and has a zero at 0 of order at least 2, then∫|z|=1

f(1/z) dz = 0.

28. Let f be holomorphic in a region containing the closed unit disk D, and suppose that|f(z)| < 1 for all z ∈ T. Prove that there is a unique z ∈ D such that f(z) = z.

29. Let f be a nonconstant continuous function on the closed disk |z| ≤ 1 that is holomorphicon the interior.

(a) Prove that if |f | is constant on T, then f has at least one zero in the disk.

(b) Prove that if |f(z)| = r for all z ∈ T, then the image of the disk D is the disk Dr. Hint:you could use (a) and a fractional linear transformation.

30. Let f, g be holomorphic on |z| < 2, and assume that |f(z)| = |g(z)| for all |z| = 1.

(a) Prove that if neither f nor g has a zero in the closed disk |z| ≤ 1, then there exists c ∈ Tsuch that f = cg.

(b) Prove that if neither f nor g has a zero in the disk |z| < 1, then there exists c ∈ T suchthat f = cg. Hint: show that any zero of either one of f or g on the circle T is also a zerofor the other, with the same order, and in fact there exist holomorphic functions h, φ, ψ suchthat f = hφ, g = hψ, and φ, ψ satisfy the hypotheses of part (a).



31. Let f be a polynomial of degree 8. Prove that∫|z|=r

f ′

f= 16πi for all sufficiently large r.

32. Let f be holomorphic in a region U except for a set A of isolated singularities, and let γ bea cycle in U \ A that is homologous to 0 in U . Let B = {a ∈ A : Wγ(a) 6= 0}. Show that Bis finite.

References

[Ahl78] L. V. Ahlfors. Complex analysis. McGraw-Hill Book Co., New York, third edition,1978. An introduction to the theory of analytic functions of one complex variable,International Series in Pure and Applied Mathematics.

[Con78] J. B. Conway. Functions of one complex variable, volume 11 of Graduate Texts inMathematics. Springer-Verlag, New York-Berlin, second edition, 1978.

[Lan85] S. Lang. Complex analysis, volume 103 of Graduate Texts in Mathematics. Springer-Verlag, New York, second edition, 1985.

[Rud87] W. Rudin. Real and complex analysis. McGraw-Hill Book Co., New York, thirdedition, 1987.

[Spi] J. Spielberg. MAT 571 Notes. (unpublished course notes, 2010).


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