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Math 307

Holomorphic functions

Lecture notes of Prof. Youssef Helou and Prof. Hicham [email protected]

Lebanese University, Fanar, Spring 2016http://hichamgebran.wordpress.com

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This is a second course on complex variables. First, we shall review what you learned inMath 209 but at a deeper and more rigorous level. Second, we shall prove the theorems that weaccepted in Math 209 as well as the fundamental theorem of algebra. And third, we shall learnnew topics such as the principle of analytic continuation, the maximum principle, the notionof simple connectedness and the homotopic version of Cauchy’s theorem. We shall also give ageneral solution to the problem of existence of antiderivatives.

References

Lars Ahlfors, Complex analysis (McGraw-Hill, 1979).

Henri Cartan, Theorie elementaire des fonctions analytiques d’une ou plusieurs variablescomplexes. (Hermann, 1995). Also available in English (Dover, 1995).

Walter Rudin, Real and complex analysis (McGraw-Hill, 1977).

Contents

1 Functions of a complex variable 51.1 Holomorphic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3 The exponential function and branches of the logarithm . . . . . . . . . . . . . . 141.4 Analytic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.5 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2 Complex integration 232.1 Line integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2 Analycity of holomorphic functions . . . . . . . . . . . . . . . . . . . . . . . . . . 292.3 Cauchy’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3 Laurent series and the residue theorem 393.1 Laurent series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.2 The residue theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

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4 CONTENTS

Chapter 1

Functions of a complex variable

1.1 Holomorphic functions

Recall the following. C denotes the set of complex numbers. It is a field (with the usual additionand multiplication). It is a vector space; as a vector space over IR it has dimension 2 and asa vector space over itself it has dimension 1. As a vector space over IR, C is isomorphic withIR2. C is equipped with the usual norm |z| =

√x2 + y2 (for z = x + iy). Then C is a Banach

space (over IR and over itself). As a topological space, C is homeomorphic to IR2. Therefore Cand IR2 can be identified algebraically and topologically. A complex number z = x + iy can bethought of as a couple (x, y) ∈ IR2.

Throughout this course, Ω will usually denote an open nonempty subset of C.

Definition 1.1 Let Ω be an open subset of C and let f : Ω → C. We say that f is holomorphicor C−differentiable at the point z0 ∈ Ω if the limit

limz→z0

f(z)− f(z0)z − z0

exists in C. In this case, the limit is denoted by f ′(z0) or dfdz (z0). If f is holomorphic at every

point of Ω, we say that f is holomorphic on Ω and we have a new function f ′ : Ω → C calledthe derivative of f .

Remark 1.1 Equivalently, f is holomorphic at z0 if there exists c ∈ C such that

∀ ε > 0, ∃α > 0 such that |z − z0| < α ⇒∣∣∣∣f(z)− f(z0)

z − z0− c

∣∣∣∣ ≤ ε

i.e.

∀ ε > 0, ∃α > 0 such that |z − z0| < α ⇒ |f(z)− f(z0)− c(z − z0)| ≤ ε|z − z0|.

This condition is usually written as

|f(z0 + h)− f(z0)− ch| = o(|h|) as h → 0.

A function f : Ω → C induces a function of two variables f : Ω ⊂ IR2 → IR2 when weidentify C with IR2. Indeed, let u and v be the real and imaginary parts of f , then we setf(x, y) = (u(x + iy), v(x + iy)). The function f is called a vector field. We usually still denotethis function by the same symbol f .

5

6 CHAPTER 1. FUNCTIONS OF A COMPLEX VARIABLE

Example. Let f(z) = z2. Then f(z) = (x+ iy)2 = x2−y2 +2ixy. So f(x, y) = (x2−y2, 2xy).

There is another concept of differentiability for this vector field. Recall from Math 302 (Dif-ferential calculus) that given two normed spaces X and Y and an open set U ⊂ X, a functionF : U → Y is said to be differentiable at a point x0 ∈ U if there exists a linear bounded mapL : X → Y such that

limh→0

||F (x0 + h)− F (x0)− Lh||||h||

= 0,

or equivalently

∀ε > 0, ∃δ > 0, ||h|| < δ ⇒ ||F (x0 + h)− F (x0)− Lh|| ≤ ε||h||.

This condition is written as

||F (x0 + h)− F (x0)− Lh|| = o(||h||) as h → 0.

In this case, the unique linear map L is called the Frechet derivative of F at (x0, y0) and isdenoted by DF (x0, y0).

Definition 1.2 A function f : Ω → C is called IR−differentiable at z0 = (x0, y0) ∈ Ω if thereexists a 2× 2−matrix L such that as h = (h1, h2) → 0, we have∣∣∣∣f(x0 + h1, y0 + h2)− f(x0, y0)− L

(h1

h2

)∣∣∣∣ = o(|h|).

Let u and v be the components of f . Recall from differential calculus that if f is IR−differentiable,then u and v have partial derivatives at (x0, y0) and

L = Df(x0, y0) =

(∂u∂x

∂u∂y

∂v∂x

∂v∂y .

)The relation between C−differentiability and IR−differentiability is given in the following propo-sition.

Proposition 1.1 Let Ω be an open subset of C, let f : Ω → C and let z0 = (x0, y0) ∈ Ω. Thenthe following conditions are equivalent

(i) f is C−differentiable (holomorphic) at z0.

(ii) f is IR−differentiable at (x0, y0) and Df(x0, y0) is of the form

(α −ββ α

).

(iii) f is IR−differentiable and the Cauchy-Riemann equations are satisfied.

Proof. (i)⇒(ii). Suppose first that f is holomorphic at z0 and let c = f ′(z0) = α+ iβ. Then,as h → 0,

|f(z0 + h)− f(z0)− ch| = o(|h|).Now observe that for h = (h1, h2) = h1 + ih2, we have

f(z0 + h)− f(z0)− ch = f(x0 + h1, y0 + h2)− f(x0, y0)− ch1 − ich2

= f(x0 + h1, y0 + h2)− f(x0, y0)− (α + iβ)h1 − i(α + iβ)h2

= f(x0 + h1, y0 + h2)− f(x0, y0)− (αh1 − βh2)− i(βh1 + αh2)

=(

u(x0 + h1, y0 + h2)− u(x0, y0)− (αh1 − βh2)v(x0 + h1, y0 + h2)− v(x0, y0)− (βh1 + αh2)

)= f(x0 + h1, y0 + h2)− f(x0, y0)−

(α −ββ α

)(h1

h2

)

1.1. HOLOMORPHIC FUNCTIONS 7

It follows that ∣∣∣∣f(x0 + h1, y0 + h2)− f(x0, y0)−(

α −ββ α

)(h1

h2

)∣∣∣∣ = o(|h|).

This means that f is IR−differentiable at (x0, y0) with Df(x0, y0) =(

α −ββ α

).

(ii)⇔(iii). If f = u + iv and Df(x0, y0) =(

α −ββ α

), then(

∂u∂x

∂u∂y

∂v∂x

∂v∂y

)=(

α −ββ α

).

This is equivalent to∂u

∂x=

∂v

∂yand

∂u

∂y= −∂v

∂x.

These relations are the well known Cauchy-Riemann equations.

(iii)⇒(i). The computations already made in the first step (i)⇒(ii) show that if f is IR−differentiableand the Cauchy-Riemann equations are satisfied at (x0, y0), then f is holomorphic at z0.

In summary,

C-differentiability ⇔ IR-differentiability + Cauchy-Riemann equations.

Thus, C−differentiability is a very restrictive condition.

Remark 1.2 It follows from the proof of the previous proposition that if f is holomorphic atz0 = (x0, y0) and f = u + iv then

f ′(z0) =∂u

∂x+ i

∂v

∂x=

∂u

∂x− i

∂u

∂y=

∂v

∂y+ i

∂v

∂x.

where the partial derivatives are evaluated at (x0, y0).

Remark 1.3 Let ||Df(x0, y0)|| = sup|h|≤1 ||Df(x0, y0)h|| be the Euclidean matrix norm ofDf(x0, y0). Then it can be shown that ||Df(x0, y0)|| = |f ′(z0)|.

Examples. (1) Every monomial azn is holomorphic on C and it derivative is anzn−1.(2) Every polynomial p(z) = a0 + a1z + · · ·+ anzn is holomorphic on C andp′(z) = a1 + 2a2z + · · ·+ nanzn−1.(3) The function z 7→ z is IR−differentiable on C but nowhere holomorphic.(4) The function z 7→ |z|2 is IR−differentiable on C and holomorphic only at z = 0.

The next two propositions were proved in Math 209.

Proposition 1.2 Let f, g : Ω → C be holomorphic at z0 ∈ Ω. Then f + g and fg areholomorphic at z0. If g(z0) 6= 0 then f

g is holomorphic at z0.

Corollary 1.1 Let H(Ω) denote the set of holomorphic functions on Ω. Then (H(Ω),+, ·) isan algebra on C. This means that (i) H(Ω) is a complex vector space when equipped with theusual addition and multiplication by scalars, (ii) (H(Ω),+, ·) is a ring and (iii) λ(fg) = (λf)gfor all λ ∈ C, f, g ∈ H(Ω).

Proposition 1.3 (The chain rule) Let f : Ω → C be holomorphic at z0 and h : Ω′ → C beholomorphic at w0 = f(z0) with f(Ω) ⊂ Ω′. Then h f is holomorphic at z0 and

(h f)′(z0) = h′(f(z0))f ′(z0).


Connectedness and path connectedness

Recall that given a topological space Ω, a separation of Ω is a couple O1, O2 of nonemptyopen subsets of Ω such that Ω = O1 ∪ O2 and O1 ∩ O2 = ∅. The space Ω is called connectedif there is no separation of Ω. This means that Ω is topologically ”one piece”. We clearly havethe following equivalence

(i) Ω is connected.

(ii) If A ⊂ Ω is closed and open in Ω, then either A = Ω are A = ∅.

Recall the following properties.

1. The image of a connected space under a continuous map is connected.

2. The only connected subsets of IR are the intervals.

A path in a topological space Ω is a continuous map γ : [α, β] → Ω (α < β). Note thatsince any compact interval [α, β] is homeomorphic to [0,1], we can always assume that a path isdefined on [0, 1]. γ(α) and γ(β) are called the endpoints of the path. A path joining two pointsa and b is a continuous function γ : [α, β] → Ω such that γ(α) = a and γ(β) = b. A topologicalspace Ω is called path connected if any two points of Ω can be joined by a path in Ω.

Note that a path in our terminology is a function γ. The image of the path, i.e. the setγ([0, 1]) will be refered to as a curve.

3. A path connected space is connected (the converse is generally not true, but see below).

Indeed, let Ω be path connected. If Ω is not connected, then there is a separation O1, O2 ofΩ. Let a ∈ O1 and b ∈ O2 and let γ : [0, 1] → Ω be a path joining a to b. Set A = ϕ([0, 1]).Then O1∩A,O2∩A is a separation of A = ϕ([0, 1]). But this set is connected as a continuousimage of [0,1]. This contradiction proves the claim.

Let γ1 : [0, 1] → Ω and γ2 : [0, 1] → Ω be two paths such that γ1(1) = γ2(0) (we say that γ1 andγ2 are juxtaposable). The product of γ1 and γ2 is the path denoted by γ1 ∗ γ2 and defined by

(γ1 ∗ γ2)(t) =

γ1(2t) if 0 ≤ t ≤ 1/2γ2(2t− 1) if 1/2 ≤ t ≤ 1.

This path is obtained by first going through γ1 and then going through γ2. It is also called thecomposition of γ1 and γ2.

Let now E be a normed space. The straight line path joining two points a, b ∈ E is thefunction γ : [0, 1] → E defined by γ(t) = (1− t)a + tb. It is clear that a straight line path is apath. A subset Ω ⊂ E is called convex if any two points in Ω can be joined by a straight linepath. Thus we have:

Convex ⇒ Path connected ⇒ Connected.

The reverse implications are not true in general. The image of the straight line path joining ato b will be denoted by [a, b], i.e. [a, b] = γ([0, 1]) and will be called a straight line segment.

A polygonal path joining two points a and b is a continuous function γ : [0, 1] → E suchthat γ(0) = a and γ(1) = b and there is a subdivision t0 = a < t1 < · · · < tn = b of [0,1] suchthat γ|[ti,ti+1] is a straight line path. Otherwise stated, a polygonal path is a product of straightline paths.

Proposition 1.4 Let Ω be an open set of a normed vector space E. Then the followingconditions are equivalent.

1.1. HOLOMORPHIC FUNCTIONS 9

(i) Every two points of Ω can be joined by a polygonal path in Ω.

(ii) Ω is path connected.

(iii) Ω is connected.

Proof. According to what we said above, it is clear that we always have (i)⇒(ii)⇒(iii) evenif Ω is not open. So now we need to prove (iii)⇒(i). Let a ∈ Ω be given and let A be theset of points in Ω that can be joined to a by a polygonal path in Ω. Observe first that A isnot empty since a can be joined to itself. Second, we claim that A is open. Indeed, let b ∈ A.Since Ω is open, there exists a ball B(b, r) centered at b and contained in Ω. Now every pointc ∈ B(b, r) can be joined to b by a straight line segment and therefore c can be joined to a by apolygonal path (by composition of paths; draw a figure). This means that B(b, r) ⊂ A. Third,,we claim that A is closed or that the complement Ac of A is open. Let therefore x ∈ Ac, then xcannot be joined to a by a polygonal path. Now there exists a ball B(x, δ) ⊂ Ω. If some pointy ∈ B(x, δ) can be joined to a by a polygonal path, then x could be joined to a by a polygonalpath (through y), contrary to the assumption. This means that B(x, δ) ⊂ Ac.

Since Ω is connected, it follows that A = Ω. This means that any point of Ω can be joinedto a by a polygonal path. But since a was arbitrary, this means that any two points of Ω canbe joined by a polygonal path.

Theorem 1.1 (The mean value theorem) Let X and Y be real Banach spaces. Let U ⊂ Xbe open and let F : U → Y be differentiable. Suppose that [a, b] ⊂ U , then

||F (b)− F (a)|| ≤ ||b− a|| supξ∈[a,b]

||DF (ξ)||.

In particular, if f : Ω → C is holomorphic and [a, b] ⊂ Ω, then

|f(b)− f(a)| ≤ |b− a| supξ∈[a,b]

|f ′(ξ)|.

Corollary 1.2 If U is connected and DF (x) = 0 for all x ∈ U , then F is constant.

Proof. We first claim that F is constant on any ball B ⊂ U . Indeed, let B be a ballcontained in U and let a, b ∈ U . Then [a, b] ⊂ B since B is convex. By the mean value theorem||F (b)−F (a)|| ≤ ||b−a||×0 = 0. Therefore F (b) = F (a). This means that F is constant on B.

Now we prove that F is constant on U . Let a ∈ U and consider the set A = x ∈ U |F (x) =F (a). Then, first, a ∈ A and so A 6= ∅. Second, continuity of F implies that A is closed.Third, we show that A is also open. Let x0 ∈ A. Then x0 ∈ U and since U is open, there is a ballB(x0, r) ⊂ U . By the claim, F is constant on B(x0, r). This implies that F (x) = F (x0) = F (a)for all x ∈ B(x0, r). Thus, B(x0, r) ⊂ A, proving that A is open. Connectedness of U impliesthat A = U . This means that F is constant on U .

Remark 1.4 The asumption of connectedness is essential. Indeed, consider for example thefunction f :]0, 1[∪]2, 3[→ IR defined by

f(x) =

1 if 0 < x < 12 if 2 < x < 3.

Then f ′ = 0 but f is not constant.

Here is another example. Consider the function f : IR∗ → IR defined by f(x) = arctanx +arctan 1

x . Then

f ′(x) =1

1 + x2+

11 + (1/x)2

(−1x2

) = 0.


However f(1) = 2 arctan 1 = π2 and f(−1) = 2 arctan(−1) = −π

2 . In fact

f(x) =

π/2 if x > 0−π/2 if x < 0.

Corollary 1.3 If DF (x) = 0 for all x ∈ U , then F is constant on the connected componentsof U .

Corollary 1.4 Let Ω ⊂ C be open and connected and let f : Ω → C be holomorphic. If f ′ = 0,then f is constant.

Proposition 1.5 Let Ω be an open and connected subset of C and f : Ω → C be holomorphic.Then the following conditions are equivalent.

(i) f is constant.

(ii) f ′ = 0.

(iii) Re(f) is constant.

(iv) Im(f) is constant.

(v) The function z 7→ f(z) is holomorphic on Ω.

(vi) |f | is constant

Proof. It is clear that condition (i) implies the remaining conditions.

(ii)⇒(i) is the previous corollary.

(iii)⇒(ii). Let f = u + iv. Then by assumption, u is constant and so ∂u∂x = ∂u

∂y = 0. It follows

that Df(x, y) =(

0 00 0

)or equivalently f ′(z) = 0 for all z = (x, y) ∈ U .

(iv)⇒(ii) is similar to (iii)⇒(ii).

(v)⇒(i). Since f and f are holomorphic, it follows that u = Re(f) = 12(f + f) is holomorphic.

But Im(u) = 0. Therefore u is constant by (iv)⇒(i). This means that Re(f) is constant. By(iii)⇒(i), f is constant.

(vi)⇒(i). Let |f | = C. If C = 0, then f = 0 and we are done. If not, then f never vanishes and

so 1f is holomorphic. Now, |f(z)|2 = C2 and so f(z)f(z) = C2. Therefore f(z) = C2

f(z) . Thisimplies that f is holomorphic. By (v)⇒(i), f is constant.

1.2 Power series

Recall first the following fundamental results from the theory of series.

The comparison test. Let (un) and (vn) be two sequences of real numbers satisfying0 ≤ un ≤ Mvn for some M > 0 and all n large enough. Then, the convergence of the series

∑vn

implies the convergence of the series∑

un.

The ratio test. Let (un) be a sequence of ]0,+∞[. Suppose lim un+1

un= `. Then the series∑

un is convergent if ` < 1 and divergent if ` > 1 (the test is inconclusive if ` = 1).

Absolute convergence. Let (an) be a sequence of complex numbers. We say that the series∑an is absolutely convergent if the series

∑|an| is convergent. Then,

Absolute convergence ⇒ Convergence.

1.2. POWER SERIES 11

This result is a consequence of the completeness of C.

Normal convergence. Let (fn) be a sequence of functions from a subset A ⊂ C to C.We say that the series

∑fn is normally convergent if the series (of norms)

∑supz∈A |fn(z)| is

convergent. Equivalently,∑

fn is normally convergent if there exists a convergent series∑

Mn

such that |fn(z)| ≤ Mn for all z ∈ A and all n large enough. A normally convergent series isabsolutely and uniformly convergent (Weierstrass M−test).

Uniform convergence. A uniform limit of continuous functions is continuous. In particular,the sum of a normally convergent series of continuous functions is continuous.

Now we move to power series. A power series about z = a is a series of the form∑

n≥0 an(z−a)n

where z is a complex variable and (an) is a sequence of complex numbers. It is enough to studyseries about the origin since we obtain the general series by a translation.

Lemma 1.1 (Abel) Let z0 ∈ C be given. If the sequence (anzn0 ) is bounded, then the series∑

n≥0 anzn is absolutely convergent for all z such that |z| < |z0|.

Proof. Let M be a bound for (anzn0 ). Then, |anzn| = |anz0|n |z|n

|z0|n ≤ M∣∣∣ zz0

∣∣∣n for all n ∈ IN.

Now by assumption∣∣∣ zz0

∣∣∣ < 1 and so the geometric series∑

n≥0

∣∣∣ zz0

∣∣∣n is convergent. By thecomparison test, it follows that the series

∑n≥0 |anzn| is convergent.

Theorem 1.2 (Radius of convergence) Let (an) be a sequence in C. There exists a uniquevalue R ∈ [0,+∞] such that the power series

∑n≥0 anzn is absolutely convergent if |z| < R and

divergent if |z| > R.

Proof. Uniqueness. Suppose that there are two values R1 < R2 satisfying the conclusion.Let R ∈]R1, R2[. Then

∑anRn should be convergent because R < R2 and divergent because

R > R1. Contradiction.

Existence. LetR = supr ≥ 0|(anrn) is bounded.

Let z ∈ C satisfy |z| < R. By a fundamental property of the sup, there exists a number ρsuch that |z| < ρ and (anρn) is bounded1. By the previous lemma,

∑n≥0 anzn is absolutely

convergent.Let now z ∈ C satisfy |z| > R. Then (anzn) is unbounded. This implies that (anzn) does

not converge to 0 and so in particular∑

n≥0 anzn is divergent.

Definition 1.3 The unique value R in the above proposition is called the radius of convergenceof the series

∑n≥0 anzn. The open disk B(0, R) is called the disk of convergence and the circle

|z| = R is called the circle of convergence.

Remark 1.5 The radius of convergence R of a power series∑

n≥0 anzn is also the uniquenumber such that

∑n≥0 |anzn| is convergent for |z| < R and divergent for |z| > R. We could

say that∑

n≥0 anzn is ”absolutely divergent” for |z| > R, but this terminology is not in use.

Remark 1.6 Note that the above theorem says nothing about the case |z| = R. In thiscase, the series

∑n≥0 anzn can be absolutely convergent, divergent or semi convergent (see the

examples below).

1|z| is not an upper bound of the set r ≥ 0|(anrn) is bounded.


Proposition 1.6 (Hadamard formula) If R is the radius of convergence of∑

n≥0 anzn, then

R =1

lim sup |an|1/n,

with the convention that 10 = +∞ and 1

+∞ = 0.

Proof. We distinguish between three cases. We shall prove that∑

n≥0 anzn is absolutelyconvergent if |z| < R and divergent if |z| > R.

Case 1. 0 < lim sup |an|1/n < +∞. Note that the series∑

n≥0 anzn always convergesfor z = 0. So let z satisfy 0 < |z| < 1

lim sup |an|1/n . Then there exists an number r such that

|z| < r < 1lim sup |an|1/n . Then lim sup |an|1/n < 1

r , i.e., infn supk≥n |ak|1/k < 1r . By a fundamental

property of the inf, there exists n0 such that supk≥n0|ak|1/k < 1

r . Therefore, |ak|1/k < 1r for

all k ≥ n0. Therefore |akrk| < 1 ∀k ≥ n0. It follows that |akz

k| = |akrk|∣∣ zr

∣∣k <∣∣ zr

∣∣k for allk ≥ n0. Since

∣∣ zr

∣∣ < 1, the geometric series∑

n≥0

∣∣ zr

∣∣k is convergent. By the comparison test,∑k≥0 |akz

k| is convergent.Let now z satisfy |z| > 1

lim sup |an|1/n . Then, lim sup |an|1/n > 1|z| , i.e., infn supk≥n |ak|1/k >

1|z| . It follows that supk≥n |ak|1/k > 1

|z| for all n. By a fundamental property of the sup,

∀n ∃m ≥ n |am|1/m > 1|z| , that is |amzm| > 1. This means that |amzm| > 1 for infintely

many indices m. In particular, the sequence (anzn) cannot converge to 0. This implies that∑n≥0 anzn is divergent.

Case 2. lim sup |an|1/n = +∞. In this case, infn supk≥n |ak|1/k = +∞. Then supk≥n |ak|1/k =+∞ for all n. This means that for all M > 0 ∀n ∃k ≥ n |ak|1/k > M . Let z 6= 0. TakingM = 1

|z| , we have ∀n ∃k ≥ n |ak|1/k > 1|z| , that is |akz

k| > 1 for infinitely many indices k. Asbefore, this implies that

∑n≥0 anzn is divergent. Therefore R = 0 = 1

lim sup |an|1/n .

Case 3. lim sup |an|1/n = 0. This imnplies that in fact lim |an|1/n = 0. This means thatfor every ε > 0, we have |an|1/n < ε for all n large enough. Now let z ∈ C be arbitrary.Choose a number M > |z|. Taking ε = 1

M , we have |an|1/n < 1M for all n large enough, i.e.,

|anMn| < 1. This means that the sequence (anMn) is bounded. Since |z| < M , it follows fromAbel’s lemma that

∑n≥0 anzn is absolutely convergent. Since z was arbitrary, this means that

R = +∞ = 1lim sup |an|1/n .

Corollary 1.5 (Cauchy’s test for the radius) If lim |an|1/n = ` then R = 1` .

Proposition 1.7 (D’Alembert test for the radius) If lim∣∣∣an+1

an

∣∣∣ = `, then R = 1` .

Proof. Suppose that lim∣∣∣an+1

an

∣∣∣ = `. Then lim∣∣∣an+1zn+1

anzn

∣∣∣ = `|z|. Therefore, first, if |z| < 1` ,

then `|z| < 1 and it follows from the ratio test that∑

n≥0 |anzn| is convergent, i.e.,∑

n≥0 anzn

is absolutely convergent. Second, if |z| > 1` , then `|z| > 1 and it follows from the ratio test that∑

n≥0 |anzn| is divergent. It follows from Remark 1.5 that R = 1` .

Exercise. Find the radius of convergence of the folllowing series and see what happens onthe circle of convergence.

∑n≥0 zn,

∑n≥1 nzn,

∑n≥1

1nzn,

∑n≥1

1n2 zn.

Proposition 1.8 Let R be the radius of convergence of the series∑

n≥0 anzn and let 0 ≤ r < R.Then

∑n≥0 anzn is uniformly convergent in the closed disk B′(0, r) = z ∈ C | |z| ≤ r.

1.2. POWER SERIES 13

Proof. We have ∀z ∈ B′(0, r), |anzn| ≤ |an|rn. Since∑

n≥0 |an|rn is convergent, we deducethat

∑n≥0 anzn is normally and hence uniformly convergent.

Consequence. The function S(z) =∑

n≥0 anzn is continuous on B′(0, r). But since rwas arbitrary, S is continuous on B(0, R). Actually S is much more than continuous: it isholomorphic.

Theorem 1.3 Let∑

n≥0 anzn be a power series with radius of convergence R. Then

1) The series∑

n≥0 nanzn−1 has the same radius of convergence.

2) The function S(z) =∑

n≥0 anzn is holomorphic on B(0, R) and S′(z) =∑

n≥1 nanzn−1.

Remark 1.7 This theorem states that we can differentiate term by term a convergent powerseries (and the radius of convergence of the series of derivatives is the same). This is a remarkableproperty of power series which is not true for arbitrary series of functions. For example, weknow from the theory of Fourier series that

∞∑n=1

1n

sin(nx) =π − x

2∀x ∈]0, 2π[.

However, if we differentiate term by, we get

∞∑n=1

cos(nx) =−12

∀x ∈]0, 2π[.

This is of course wrong because the series on the left is divergent.

Proof of Theorem 1.3. 1) Let R′ be the radius of convergence of∑

n≥0 nanzn−1. Notefirst that for n ≥ 1, |anzn| ≤ |nanzn| = |nanzn−1||z|.Claim 1. R′ ≤ R. If not, choose z such that R < |z| < R′. Then, first,

∑n≥0 |anzn| is

divergent. Second,∑

n≥0 |nanzn−1| is convergent. By the comparison test,∑

n≥0 |anzn| isconvergent; a contradiction.Claim 2. R ≤ R′. If not, choose z such that R′ < |z| < R and let r be such that |z| <r < R. Then

∑n≥0 anrn is convergent. In particular, (anrn) is bounded by some constant C.

Consequently,

|nanzn−1| = n|an|rn

|z|

∣∣∣zr

∣∣∣n ≤ C

|z|n∣∣∣zr

∣∣∣n .

Now the series∑

n≥1 n∣∣ zr

∣∣n is convergent because | zr | < 1. It follows from the comparisontest that

∑n≥1 |nanzn−1| is convergent. But it should be divergent beccause |z| > R′; a

contradiction.

2) Let z ∈ B(0, R). Let δ > 0 be such that |z| + δ < R. Then z + h ∈ B(0, R) for any h suchthat |h| < δ. Let r be such that |z|+ δ < r < R. Observe that for |h| < δ

S(z + h)− S(z)h

=

∑n≥0 an(z + h)n −

∑n≥0 anzn

h=

∑n≥1 an[(z + h)n − zn]

h.

But (z + h)n − zn = h[(z + h)n−1 + (z + h)n−2z + · · ·+ zn−1]. Therefore

S(z + h)− S(z)h

=∑n≥1

an[(z + h)n−1 + (z + h)n−2z + · · ·+ zn−1] =∑n≥1

ϕn(h).

Let ϕ(h) =∑

n≥1 ϕn(h) for |h| < δ. Now each ϕn is continuous with ϕn(0) = nanzn−1. Observethat |ϕn(h)| = |an||(z + h)n−1 + (z + h)n−2z + · · · + zn−1| ≤ n|an|rn−1. But

∑n|an|rn−1 is


convergent because r < R. Therefore∑

ϕn(h) is normally (and therefore uniformly) convergent.It follows that ϕ is continuous. In particular,

limh→0

ϕ(h) = ϕ(0) =∑n≥1

ϕn(0) =∑n≥1

nanzn−1.

This means thatlimh→0

S(z + h)− S(z)h

=∑n≥1

nanzn−1,

proving that S is holomorphic on B(0, R) with S′(z) =∑

n≥1 nanzn−1.

Corollary 1.6 1. Every power series S(z) =∑

n≥0 anzn is infinitely many time differen-tiable and

S(p)(z) =∑n≥p

n(n− 1) · · · (n + 1− p)anzn−p

for |z| < R and so S(p)(0) = p!ap.

2. If S(z) = 0 in a neighborhood of zero, then an = 0 ∀n ∈ IN.

3. If∑

n≥0 anzn =∑

n≥0 bnzn for all z in a neighborhood of zero, then an = bn ∀n ∈ IN.

Exercise. Find the sums of the series∑∞

n=1 nzn and∑∞

n=2 n(n− 1)zn.

1.3 The exponential function and branches of the logarithm

We set

ez = exp(z) =∞∑

n=0

1n!

zn.

Note that 1/(n+1)!1/n! = 1

n+1 → 0 as n → ∞ and so (by the d’Alembert test) the radius ofconvergence is R = +∞. Thus the series is absolutely convergent for any z ∈ C and ez is whatwe call an entire function. It should be clear that the exponential function on C is an extensionof the usual exponential function on IR.

Observe also that the derivative of the exponential is given by

d

dzez =

∞∑n=1

n

n!zn−1 =

∞∑n=1

1(n− 1)!

zn−1 =∞∑

n=0

1k!

zk = ez.

Proposition 1.9 The exponential function satisfies the following properties.

(i) eu+v = euev, ∀u, v ∈ C.

(ii) ez = ez ∀z ∈ C.

(iii) |eiy| = 1, ∀y ∈ IR.

Proof. (i) Let a ∈ C be given and set f(z) = ez+ae−z. Then f is holomorphic on C

and f ′(z) = ez+ae−z − ez+ae−z = 0. Therefore, f is constant since C is connected. Thusf(z) = f(0) = ea since e0 = 1, that is,

ez+ae−z = ea. (1.1)

But this relation is true for any a. In particular, letting a = 0, we get eze−z = 1, or, e−z = 1ez .

Multiplying relation (1.1) by ez we get ez+a = eaez.

1.3. THE EXPONENTIAL FUNCTION AND BRANCHES OF THE LOGARITHM 15

(ii) ez = limn→∞

n∑p=0

1p!

zp = limn→∞

n∑p=0

1p!

zp and since z → z is continuous (so lim xn = lim xn), we

get

ez = limn→∞

n∑p=0

1p!

zp =∞∑

p=0

1p!

zp = ez

(iii) |eiy|2 = eiyeiy = eiye−iy = 1 by part (i) and (ii).

Definition 1.4 We set cos z =eiz + e−iz

2and sin z =

eiz − e−iz

2i, for all z ∈ C.

It follows that eiz = cos z+ i sin z. Using the facts that (i)2p = (−i)2p = (−1)p, (i)2p+1 = (−1)piand (−i)2p+1 = −(−1)pi, we get

cos z =∑n≥0

in + (−i)n

2zn

n!=

∞∑p=0

i2p + (−i)2p

2z2p

(2p)!+

∞∑p=0

i2p+1 + (−i)2p+1

2z2p+1

(2p + 1)!

=∞∑

p=0

(−1)p

(2p)!z2p,

and

sin z =∑n≥0

in − (−i)n

2i

zn

n!=

∞∑p=0

i2p − (−i)2p

2i

z2p

(2p)!+

∞∑p=0

i2p+1 − (−i)2p+1

2i

z2p+1

(2p + 1)!

=∞∑

p=0

(−1)p

(2p + 1)!z2p+1,

The power series expansions of cos and sin show that these functions are the extension to C ofthe usual cos and sin function on IR. However, we can forget what we learned about the realtrigonometric functions and reprove everything starting from our definition of cosine and sine.As an exercise, prove the following identities

1. cos(a + b) = cos a cos b− sin a sin b for all a, b ∈ C.

2. cos2 z + sin2 z = 1 for all z ∈ C.

3. (cos z)′ = − sin z and (sin z)′ = cos z.

Also we can redefine the number π by

π = 2 infy ≥ 0 | cos y = 0.

With this definition, we can prove that ez = 1 ⇔ z = 2nπi and avoid the use of the inversetrigonometric functions in what follows. This approach is done in the appendix to this chapter.

Definition 1.5 Let S1 = u ∈ C , |u| = 1 be the unit circle. An argument of u ∈ S1 is anyreal number y satisfying eiy = u. More generally, if z ∈ C∗, then z

|z| ∈ S1 and an argument of

z is any number y satisfying eiy = z|z| .

An argument of a complex number z 6= 0 is the angle between the positive axis and the axisjoining the origin to z. But angles are defined modulo 2π; this is why we have several argumentsof the same number and sometimes one has to choose a determination of the argument.

Proposition 1.10 If y0 is an argument of z ∈ C∗, then the set of arguments of z is y0 + 2πZ.


Proof. y is an argument of z if and only if eiy = eiy0 , i.e., if and only if ei(y−y0) = 1. Noweib = 1 ⇔ cos b = 1 ⇔ b = 2kπ, k ∈ Z⇔ b ∈ 2πZ.

Definition 1.6 Let Ω be an open set of C∗. A continuous function g : Ω → IR is calleda determination or branch of the argument if eig(z) = z

|z| for all z ∈ Ω, that is, g(z) is anargument of z for all z ∈ Ω.

Remark 1.8 We shall see in the next chapter that there is no branch of the argument on C∗.There are however branches of the arguments on C\eiaIR+ (the set obatined by removing fromC a half line at the origin).

Fundamental example. Consider the map ϕ :]0, 2π[→ S1\1 defined by ϕ(y) = eiy.We can think of this map as a deformation that bends the straight line segment to a circlewithout a point. We claim that ϕ is a homeomorphism (a continuous bijection whose inverseis continuous). Continuity of ϕ is clear (ϕ is analytic). To prove that ϕ is a bijection wehave to show that the equation ϕ(t) = u is uniquely solvable for every u ∈ S1\1. Letu = a + ib. Then the equation ϕ(t) = u is equivalent to the system cos t = a; sin t = b. Weknow that this system has a unique solution t ∈]0, 2π[. Thus ϕ is a bijection. Let us findan expression of ϕ−1. Suppose first that b ≥ 0. We know that cos is a bijection from [0, π]to [−1, 1] whose inverse is denoted by Arccos. Now t ∈]0, π] because sin t ≥ 0. Thereforecos t = a ⇔ t = Arccosa. Suppose next that b ≤ 0. Then t ∈ [π, 2π[ and so t− π ∈]0, π[. Thencos t = a ⇔ cos(t− π) = −a ⇔ t = Arccos(−a) + π. Thus,

ϕ−1(u) =

Arccos(Re(u)) if Im(u) ≥ 0π + Arccos(−Re(u)) if Im(u) ≤ 0.

Now observe that when Im(u) = 0, we have u = −1 (since u ∈ S1). For this value of u,Arccos(Re(u)) = Arccos(−1) = π and π + Arccos(−Re(u)) = π + Arccos(1) = π. Thereforeϕ−1 is obtained by pasting together two continuous functions that agree on the intersection oftheir domains. By the pasting lemma in topology, ϕ−1 is continuous.

Remarks. 1) limu→1

Imu≥0

ϕ−1(u) = Arccos(1) = 0 and limu→1

Imu<0

ϕ−1(u) = π + Arccos(−1) = 2π.

Therefore ϕ−1 cannot be extended by continuity to the whole circle S1.

2) Arccos(−x) = π −Arccos(x) ∀x ∈ [−1, 1].

Consequence. Let Ω = C\IR+. Note that if z ∈ Ω then z|z| ∈ S1\1. Set g(z) = ϕ−1

(z|z|

).

Then g is a branch of the argument on Ω.Note however that g cannot be extended by continuity to C∗ (if not, then the extension will

have a jump of 2π at the positive real axis).

More generally, let α ∈ IR and consider the map ϕα :]α, α + 2π[→ S1\eiα defined byϕα(t) = eit.

Then ϕα(t) = eiαei(t−α) = eiαϕ(t− α). Therefore if we let rα(t) = t− α and sα(u) = eiαu,we can write ϕα = sα ϕ rα. Note that rα is a translation and sα is a rotation of angle α. Itfollows that ϕα is a homeomorphism.

Now let Vα := C\IR+eiα be the set obtained by removing from the plane the half line joining0 to eiα. Let

Aα :Vα →]α, α + 2π[

z 7→ ϕ−1

(z

|z|

).

1.3. THE EXPONENTIAL FUNCTION AND BRANCHES OF THE LOGARITHM 17

Then Aα is a branch of the argument on Vα.When α = −π, A−π is called the principal branch of the argument on C\IR−. It is sometimes

denoted by Arg.

Exercise. Compute A0(−1), A0(i), A0(−i), A−π(1), A−π(i), A−π(−i).

Definition 1.7 Let z ∈ C∗. A logarithm of z is any complex number w satisfying ew = z.

Proposition 1.11 Let z ∈ C∗ be given. Then the set of solutions to the equation ew = z isln |z|+ iv | v is an argument of z = ln |z|+ iv0 + 2πiZ. where v0 is some argument of z.

Proof. Let w = u + iv. Then

ew = z ⇔ eu = |z| and eiv =z

|z|⇔ u = ln |z| and v is an argument of z.

Definition 1.8 Let Ω be an open set of C∗. A continuous function f : Ω → C is called adetermination or branch of the logarithm if ef(z) = z for all z ∈ Ω.

Example 1.1 For any α ∈ IR,the function Lα : Vα → C defined by Lα(z) = ln |z| + iAα(z) isa branch of the logarithm. L−π is called the principal branch of the logarithm; it is defined onC\IR−. Observe that if x is real and > 0, then A−π(x) = 0 and so Lπ(x) = ln x. Thus L−π isan extension to C\IR− of the natural logarithm. Note that L−π/2 is also an extension of ln butto C\ − iIR+. However Lπ/2 is not an extension of ln because Lπ/2(x) = lnx + 2πi when x isreal and > 0.

Exercise. Compute L0(−1), L0(i), L0(−i), L−π(1), L−π(i), L−π(−i).

Proposition 1.12 Let Ω be open connected subset of C∗. Then two branches of the argumenton Ω, when they exist, differ by a constant of the form 2πk. Two branches of the logarithm onΩ, when they exist, differ by a constant of the form 2πki.

Proof. Let g1 and g2 be two branches of the argument on Ω , we know that ∀ z ∈ Ω, ∃ k ∈ Zsuch that g1(z)− g2(z) = 2πk, i.e., (g1 − g2)(Ω) ⊂ 2πZ. We need to show that (g1 − g2)(Ω) issingleton. Since Ω is connected and g1 − g2 is continuous, (g1 − g2)(Ω) is connected. But theonly connected subsets of 2πZ are singletons. The same reasoning establishes the result for thelogarithm.

Proposition 1.13 If L is a branch of the logarithm on Ω, then L is holomorphic and L′(z) = 1z

for all z ∈ Ω.

Proof. Note that L is continuous and eL(z) = z. Let z0 ∈ Ω, then

L(z)− L(z0)z − z0

=L(z)− L(z0)eL(z) − eL(z0)

=1

eL(z)−eL(z0)

L(z)−L(z0)

→ 1eL(z0)

=1z0

as z → z0.

Exercise. Let V−π = C\IR+e−iπ = C\IR−. Note that V−π is invariant under the symmetryz 7→ z.

1. Let g(z) = L−π(z). Show that g is holomorphic and compute g′(z).

2. Deduce that L−π(z) = L−π(z) for all z ∈ V−π.


1.4 Analytic functions

Definition 1.9 Let Ω be an open subset of C and f : Ω → C. We say that f is analyticat a point a ∈ Ω, if f has a power series expansion about a, that is, if there exists a powerseries

∑n≥0 anzn with radius of convergence r > 0 such that f(z) =

∑n≥0 an(z − a)n for all

z ∈ B(a, r). f is called analytic on Ω if it is analytic at every point of Ω.

Note that by Theorem 1.3, if a function is analytic at a point, it is holomorphic in a neighborhoodof that point. Conversely, as we shall see in chapter 2, a function which is holomorphic on anopen set is analytic there. Note however, that a function which is holomorphic at only onepoint need not be analytic at that point. For example, the function z 7→ |z|2 is holomorphic at0 but not analytic at 0.

Exercise. Show directly that the function f : C∗ → C defined by f(z) = 11−z is analytic on

C∗.

Definition 1.10 Let f : Ω → C be analytic and let a ∈ Ω. The germ of f at a is the set

Germ(f, a) = f(a), f ′(a), . . . , f (n)(a), . . ..

It follows that Germ(f, a) = 0 ⇔ f (k)(a) = 0 ∀ k ∈ IN ⇔ f vanishes in a neighborhood of a

(since an = f (n)(a)

n! ).

Proposition 1.14 (Principle of isolated zeros) Let f : Ω → C be analytic and let a ∈ Ωbe such that f(a) = 0 and Germ(f, a) 6= 0. Then there exists r > 0 such that f(z) 6= 0 for allz ∈ B(a, r)\a. Otherwise stated, a is an isolated zero of f .

Proof. Since Germ(f, a) 6= 0, there exists k ∈ IN∗ such that fk(a) 6= 0, that is, the setk ≥ 1 | f (k)(a) 6= 0 is not empty. Let then p = mink ≥ 1 | f (k)(a) 6= 0 (every nonemptysubset of IN has a smallest element). It follows that fk(a) = 0 for all k < p. Now since f isanalytic, there exists a disk B(a, r1) such that f(z) =

∑∞k=0

f (k)(a)k! (z − a)k for all z ∈ B(a, r1).

Therefore, f can be written in the form

f(z) =∞∑

k=p

f (k)(a)k!

(z − a)k = (z − a)p

( ∞∑k=0

bk(z − a)k

)= (z − a)pf1(p)

with b0 = f (p)(a)p! 6= 0. Since f1(a) = b0 6= 0, and f1 is continuous, it follows that f1 does not

vanish on a neighborhood say B(a, r) of a.

Proposition 1.15 Let f : Ω → C be analytic and let a ∈ Ω. Then, the following conditionsare equivalent.

(i) There exists a subset E ⊂ Ω having a as a limit point in Ω and such that f |E = 0.

(ii) Germ(f, a) = 0.

(iii) f ≡ 0 in a neighborhood of a.

Proof. (i)⇒(ii). The continuity of f implies that f |E = 0 and so in particular, f(a) = 0.Suppose that Germ(f, a) 6= 0, then, by the previous proposition, there exists r > 0 suchthat f(z) 6= 0 for all z ∈ B(a, r)\a. Since a is an accumulation point of E, there existsz0 ∈ E ∩ B(a, r)\a. Then, on the one hand, f(z0) 6= 0 because z0 ∈ B(a, r)\a; and on theother hand f(z0) = 0 because z0 ∈ E; a contradiction.

1.4. ANALYTIC FUNCTIONS 19

(ii)⇒(iii). Since f is analytic, then f(z) =∑

n≥0f (n)(a)

n! (z − a)n for all z in a neighborhood ofa. But f (n)(a) = 0 for all n ∈ IN, hence f(z) = 0 for all z in a neighborhood of a.

(iii)⇒(i). Let B(a, r) ⊂ Ω be a disk on which f vanishes. Note that a is a limit point of the setE := B(a, r)\a.

Theorem 1.4 (Principle of analytic continuation) Let Ω be an open and connected sub-set of C and let f : Ω → C be analytic. If there exists a ∈ Ω such that Germ(f, a) = 0, thenf ≡ 0 in Ω.

Proof. Let S = z ∈ Ω |Germ(f, z) = 0. Then

1) S 6= ∅ because a ∈ S.

2) S is open, for if z0 ∈ S, then f = 0 in a neighborhood U of z0. But then Germ(f, z) = 0 forall z ∈ U .

3) S is closed in Ω. Indeed, let z ∈ SΩ = S ∩ Ω. Then, there exists a sequence (zn) of

S converging to z . If z /∈ S then z is a limit point of the set E = zn|n ∈ IN. ButGerm(f, zn) = 0 and so in particular f(zn) = 0 so that f |E = 0. By the previous proposition,Germ(f, z) = 0 and so z ∈ S, contradicting our assumption. Therefore z ∈ S. This meansthat S ∩ Ω ⊂ S.

Since Ω is connected, it follows that S = Ω and so f(z) = 0 ∀ z ∈ Ω.

Said differently, if an analytic function vanishes on an open set, then it vanishes on the connectedcomponent contaning that set. Observe also the following. Suppose that f : Ω → C is analyticand let Ω′ be an open and connected set containing Ω. If g : Ω′ → C and h : Ω′ → C aretwo analytic extensions (continuations) of f then g = h. Otherwise stated, there is at mostone analytic continuation of an analytic function to a connected set. This is why the previoustheorem is called the principle of analytic continuation.

Also due to the previous proposition, if an analytic function vanishes on set E having alimit point a, then it vanishes on the connected component containing a.

The principle of analytic continuation is a remarkable property which does not hold forarbitrary smooth functions f : IR → IR. For example consider the function f defined by

f(t) =

0 if t ≤ 0e−1/t if t > 0.

It is not difficult to see that f is differentiable and even C∞. However f vanishes on the openset ]−∞, 0[ without being identically zero.

Corollary 1.7 Let Ω be open and connected and f : Ω → C be analytic and not identicallyzero. Then the set of zeros of f is closed in Ω, discrete and therefore countable.

Proof. Let Z(f) = f−1(0). Z(f) is closed in Ω because f is continuous. If Z(f) were notdiscrete then it would have a limit point a ∈ Z(f). Since f vanishes at Z(f), it follows fromthe previous theorem that f vanishes on Ω, contrary to our assumption.

Now for every n ∈ IN∗, set Kn = z ∈ C | |z| ≤ n and dist (z,Ωc) ≥ 1n. Then Kn is a closed

and bounded set contained in Ω; therefore it is compact. Note also that Ω = ∪nKn. Now, forevery n, the set Z(f)∩Kn is discrete and compact; therefore it is finite. But since we can writeZ(f) =

⋃(Kn ∩ Z(f) we see that Z(f) is countable as a countable union of finite sets.


Remark 1.9 Z(f) as a subset of C may posses limit points outside Ω. For example let f :C∗ → C be defined by f(z) = sin(1

z ). Then Z(f) = 1nπ |n ∈ Z∗. Z(f) is indeed discrete,

closed in C∗ (but not in C) and countable. It has however 0 as a limit point (which is not inC∗).

Note the following. Let X be a topological space and let A ⊂ X. Then A is discrete (in thesubspace topology) if and only if no limit point of A (in X) belongs to A.

1.5 Appendix

In this appendix, we assume no knowledge about the real trigonometric functions. We shall givea definition of π and prove (without the use of inverse trigonometric functions) that the mapϕ :]0, 2π[→ S1\1 defined by ϕ(t) = eit is an homeomorphism. The appendix is important tothose interested in the foundations of Analysis.

Proposition 1.16 There is a smallest real number y0 > 0 such that cos y0 = 0. We set π = 2y0.

Proof. It is clear from the definition that cos 0 = 1. Now cos 2 is equal to the alternatingseries

∑∞p=0

(−1)p22p

(2p)! = 1− 2+ 23 −

64720 + · · · = −1

3 −64720 + · · · < −1

3 < 0. Since cos is continuous,there exists by the intermediate value theorem a number y between 0 and 2 such that cos y = 0.Therefore the set A := y ≥ 0 | cos y = 0 is not empty. Continuity of cos implies that this setis closed in [0,∞[ and therefore also closed in IR. Therefore, inf A ∈ A. Note that since 0 /∈ A,we have inf A > 0. Setting π = 2 inf A, we see that cos(π

2 ) = 0. So π2 is the first positive zero

of cos.

Corollary 1.8 (i) The function sin is strictly increasing on [0, π2 ] and sin π

2 = 1.

(ii) The function cos is strictly decreasing on [0, π2 ].

Proof. (i) We know that sin′ = cos. But by definition of π2 , cos(y) > 0 for y ∈ [0, π

2 [.Therefore sin is strictly increasing on [0, π

2 ]. In particular, sin π2 > sin 0 = 0. On the other

hand, we have sin2 π2 + cos2 π

2 = 1. It follows that sin2 π2 = 1. Since sin π

2 > 0, we get sin π2 = 1.

Therefore ei π2 = i.

(ii) We have cos′(y) = − sin y < 0 for y ∈]0, π2 ]. It follows that cos is strictly decreasing [0, π

2 ].

Let us divide the unit circle S1 into 4 equal parts

Γ1 = u ∈ S1 |Re(u) ≥ 0 and Im(u) ≥ 0 Γ2 = u ∈ S1 |Re(u) ≤ 0 and Im(u) ≥ 0Γ3 = u ∈ S1 |Re(u) ≤ 0 and Im(u) ≤ 0 Γ4 = u ∈ S1 |Re(u) ≥ 0 and Im(u) ≤ 0.

Note that Γ2 = iΓ1, Γ3 = iΓ2 and Γ4 = iΓ3 (each quarter of the circle is obtained by rotatingthe previous quarter by an angle of π

2 ).

Proposition 1.17 Let ϕ(y) = eiy for y ∈ IR. Then

(i) ϕ : [0, π2 ] → Γ1 is a homeomorphism.

(ii) ϕ : [π2 , π] → Γ2 is a homeomorphism.

(iii) ϕ : [π, 3π2 ] → Γ3 is a homeomorphism.

(iv) ϕ : [3π2 , 2π] → Γ4 is a homeomorphism.

1.5. APPENDIX 21

Proof. (i). It follows from the previous proposition that cos : [0, π2 ] → [0, 1] is a drecreasing

bijection and sin : [0, π2 ] → [0, 1] is an increasing bijection. Let now u = α + iβ ∈ Γ1. The

equation ϕ(y) = u is equivalent to the system cos y = α; sin y = β. This system is uniquelysolvable in [0, π

2 ] since α, β ∈ [0, 1]. This means that ϕ : [0, π2 ] → Γ1 is a bijection. It is clearly

continuous. Since [0, π2 ] is compact, it follows from a theorem of topology that ϕ−1 is also

continuous.

(ii) Observe thatϕ(y) = eiy = ei(y−π

2+π

2) = ei π

2 ei(y−π2) = iϕ(y − π

2).

Therefore if y ∈ [π2 , π], then y− π

2 ∈ [0, π2 ]. It follows from (i) that ϕ is a homeomorphism from

[π2 , π] to iΓ1 = Γ2. Therefore it is also a homeomorphism from [π, 3π

2 ] to iΓ2 = Γ3. We alsodeduce that ϕ is a homeomorphism from [3π

2 , 2π] to iΓ3 = Γ4.

Remark 1.10 Using the relation ϕ(y) = iϕ(y − π2 ), we see that eiπ = iei π

2 = i.i = −1. Itfollows also that ei 3π

2 = ieiπ = −i and so e2πi = iei 3π2 = i(−i) = 1.

Corollary 1.9 2π is the smallest positive number y such that eiy = 1.

Proof. According to what we said, points of ]0, π2 ] are maped under ϕ onto Γ1\1; points of

[π2 , π] are mapped onto Γ2 and therefore are distinct from 1; points of [π, 3π

2 ] are mapped ontoΓ3 and therefore are distinct from 1; points of [3π

2 , 2π[ are mapped onto Γ3\1 and thereforeare distinct from 1. Thus 2π the first positive number whose image is 1.

Corollary 1.10 ϕ : [0, 2π[→ S1 is a bijection.

Proposition 1.18 ϕ :]0, 2π[→ S1\1 is a homeomorphism.

Proof. We already know that ϕ is a continuous bijection from ]0, 2π[ to S1\1. It remains toshow that ϕ−1 is continuous. So let (un) be a sequence of S1\1 that converges to u ∈ S1\1.Claim. There exists ε > 0 such that ϕ−1(un) ∈ [ε, 2π − ε] for all n large enough.Proof of the claim. If not, we have the following condition

∀ε > 0 ∀n ∈ IN∗ ∃m ≥ n ϕ−1(um) < ε or ϕ−1(um) > 2π − ε.

Letting ε = 1, 12 , 1

3 , . . . , 1k , . . ., we see that there exists a increasing sequence nk of integers such

that for each k either ϕ−1(unk) < 1

k or ϕ−1(unk) > 2π − 1

k .Now consider the set E = k ∈ IN∗ |ϕ−1(unk

) < 1k. If it is infinite, then there exists a

subsequence of (unk) denoted by (unk`

) such that ϕ−1(unk`) < 1

k`. This means that ϕ−1(unk`

) →0 and therefore unk`

→ ϕ(0) = 1, contrary to our assumption.If E is finite then, ϕ−1(unk

) > 2π− 1k for all k large enough. This means that ϕ−1(unk

) → 2πand therefore unk`

→ ϕ(2π) = 1 contrary to our assumption. This proves the claim.

Since ϕ is continuous and [ε, 2π − ε] is compact, it follows that ϕ−1 is continuous fromϕ([ε, 2π − ε]) to [ε, 2π − ε]. Therefore (ϕ−1(un)) converges to ϕ−1(u) ∈ [ε, 2π − ε] ⊂]0, 2π[.

Proposition 1.19 ϕ : (IR,+) → (S1, ·) is a group homomorphism whose kernel is 2πZ.

Proof. We clearly have

ϕ(y1 + y2) = ei(y1+y2) = eiy1eiy2 = ϕ(y1)ϕ(y2).

So ϕ is a homomorphism. Next let n ∈ Z. Then ϕ(2πn) = ei2πn = (e2πi)n = 1n = 1.Conversely let ϕ(y) = 1 and let n be the floor function of y

2π . Then, 2πn ≤ y < 2π(n + 1)and so 0 ≤ y − 2πn < 2π. Since ei(y−2πn) = eiye−2πin) = 1, it follows from Corollary 1.9 thaty − 2πn = 0 and so y ∈ 2πZ.

Exercise. Solve in C the equation cos z = 0 and sin z = 0.

Chapter 2

Complex integration

2.1 Line integrals

Definition 2.1 Let Ω ⊂ C. A path in Ω is a continuous function γ : [α, β] → Ω. The pointsγ(α) and γ(β) are called the endpoints of the path; γ(α) is called the starting point and γ(β)is called the ending point. The image of [α, β] under a path γ will be denoted by γ∗ and wewill call it a curve.1 A piecewise smooth path is a path γ such that there exists a subdivisionof [α, β], t0 = α < t1 < · · · < tn−1 < tn = β for which γ|[tk,tk+1] is C1. This means that γ′ exists(and is continuous) except at finitely many points tk but γ′ has left and right limits at tk. Apath γ is called a loop or a closed path if γ(α) = γ(β).

Remark 2.1 A curve can be the image of different paths, i.e., it can be described by differentparamerizations. For example let C = (x, y) |x2 + y2 = 1 and − 1√

2≤ x ≤ 1√

2. Let

γ : [π4 , 3π

4 ] → C be given by γ(t) = eit. Then C = γ∗. Also if δ : [− 1√2, 1√

2] → C is given by

δ(t) = t + i√

1− t2, then C = δ∗.

Definition 2.2 Let γ be a piecewise smooth path and f : γ∗ → C be continuous. The lineintegral of f over γ is ∫

γf(z) dz =

n−1∑k=0

∫ tk+1

tk

f(γ(t))γ′(t) dt

where t0 = α < t1 < · · · < tn−1 < tn = β and γ|[tk,tk+1] is C1.

Actually, the integral depends only on the oriented curve and not on the parametrization.

Proposition 2.1 (Invariance under reparametrization) Let ϕ : [0, 1] → [0, 1] be an in-creasing diffeomorphism. Let γ be piecewise smooth path and γ = γ ϕ. Let f : γ∗ → C becontinuous. Then ∫

γf(z) dz =

∫γf(z) dz.

Proof. We have∫γf(z) dz =

n−1∑k=0

∫ tk+1

tk

f(γ(t))γ′(t) dt =n−1∑k=0

∫ tk+1

tk

f(γ(ϕ(t))γ′(ϕ(t))ϕ′(t) dt

=n−1∑k=0

∫ ϕ(tk+1)

ϕ(tk)f(γ(s))γ′(s) ds =

∫γf(z) dz.

1Therefore, a curve is a subset of C, whereas a path is a function. A path is also called a parametrization ofthe curve and it gives an orientation to it.

23

24 CHAPTER 2. COMPLEX INTEGRATION

Remark 2.2 If ϕ is decreasing, then∫γ f(z) dz = −

∫γϕ f(z) dz.

Definition 2.3 The length of a path γ is the integral

L(γ) =∫ 1

0|γ′(t)|dt =

n−1∑i=0

∫ ti+1

ti

|γ′(t)|dt.

The next proposition was proved in Math 209.

Proposition 2.2∣∣∣∣∫

γf(z) dz

∣∣∣∣ ≤ L(γ) supz∈γ([0,1])

|f(z)|.

Corollary 2.1 Let γ be a path in C.

(a) (Interchange of lim and∫

). Let (fn) be a sequence of continuous functions that convergesuniformly to f on γ∗. Then

limn→∞

∫γfn(z) dz =

∫γf(z) dz.

(b) (Integration term by term). If∑

fn is a series of continuous functions that convergesuniformly on γ∗, then ∫

γ

( ∞∑n=1

fn(z)

)dz =

∞∑n=1

∫γfn(z) dz.

Proof. (a) follows from∣∣∣∣∫γfn(z) dz −

∫γf(z) dz

∣∣∣∣ ≤ supz∈γ∗

|fn(z)− f(z)|L(γ).

(b) follows from (a) and the linearity of the integral.

Let γ1 : [0, 1] → C and γ2 : [0, 1] → C be two paths such that γ1(1) = γ2(0) (we say that γ1

and γ2 are juxtaposable). The product of γ1 and γ2 is the path

(γ1 ∗ γ2)(t) =

γ1(2t) if 0 ≤ t ≤ 1/2γ2(2t− 1) if 1/2 ≤ t ≤ 1.

Proposition 2.3 Let γ1 and γ2 be two juxtaposable piecewise smooth paths and let f : γ∗1 ∪γ∗2 → C be continuous. Then∫

γ1∗γ2

f(z) dz =∫

γ1

f(z) dz +∫

γ2

f(z) dz.

If γ : [α, β] → C is a path, the opposite path of γ is the path γ− defined by γ−(t) =γ(α + β − t).

Proposition 2.4 Let γ be a piecewise smooth path and f : γ∗ → C be continuous. Then∫γ−

f(z) dz = −∫

γf(z) dz.

Proof. Straightforward.

2.1. LINE INTEGRALS 25

Definition 2.4 An antiderivative of a function f : Ω → C is a holomorphic function F : Ω → C

such that F ′(z) = f(z) for all z ∈ Ω. If f has an antiderivative, we say that the differentialform f(z) dz is exact.

Remark 2.3 We shall see shortly that a continuous or even analytic function need not have anantiderivative. One of our targets in this chapter is to give conditions under which a functionhas an antiderivative.

Theorem 2.1 (Barrow’s rule) If f : Ω → C has an antiderivative F and γ : [α, β] → Ω is apiecewise smooth path, then ∫

γf(z) dz = F (γ(β))− F (γ(α)).

Proof. Let t0 = α < t1 < · · · < tn−1 < tn = β be a subdivision of [α, β] such that γ|[tk,tk+1]

is C1. Then∫γf(z) dz =

n−1∑k=0

∫ tk+1

tk

f(γ(t))γ′(t) dt =n−1∑k=0

∫ tk+1

tk

F ′(γ(t))γ′(t) dt =n−1∑k=0

∫ tk+1

tk

(F γ)′(t) dt

=n−1∑k=0

[(F γ)(tk+1)− (F γ)(tk)] = (F γ)(β)− (F γ)(α).

Exercise. Compute∫

γz2 dz where γ(t) = et cos3 t + iet sin3 t for 0 ≤ t ≤ 2π.

Theorem 2.2 Let Ω be a connected open set of C and let f : Ω → C be continuous. Then thefollowing conditions are equivalent.

(i) f has an antiderivative.

(ii)

∫γf(z) dz = 0 for every piecewise smooth loop γ.

(iii)

∫γf(z) dz = 0 for every polygonal loop γ.

If moreover Ω is convex, then the above conditions are equivalent to

(iv)

∫γf(z) dz = 0 for every triangular loop γ.

Proof. (i)⇒(ii). Let F be an antiderivative of f and let γ be a piecewise smooth loop. Bythe previous theorem, ∫

γf(z) dz = F (γ(1))− F (γ(0)) = 0

because γ is closed.

(ii)⇒(iii). A polygonal path is piecewise smooth.

(iii)⇒(i). Let a ∈ Ω be given and let z vary in Ω. We claim first that if γ and µ are twopolygonal paths joining a to z then

∫γ f(z) dz =

∫µ f(z) dz. Indeed, the path γ ∗ µ− is a closed


polygonal path and so∫γ∗µ− f(z) dz = 0 by assumption. But it follows from Proposition 2.3

and Proposition 2.4 that∫γ∗µ−

f(z) dz =∫

γf(z) dz +

∫µ−

f(z) dz =∫

γf(z) dz −

∫µ

f(z) dz.

Hence the claim.This claim permits to define the function F by

F (z) =∫

γf(u) du

where γ is a polygonal path joining a to z. We will show that F is holomorphic on Ω and thatF ′ = f .

Consider now a point z ∈ Ω. Since Ω is open, there exists a disc B(z, r) ⊂ Ω. Then[z, z + h] ⊂ Ω whenever |h| < r. Then

F (z + h)− F (z)− hf(z) =∫

[z,z+h]f(u) du− hf(z) = h

∫ 1

0[f(z + th)− f(z)] dt.

and so

|F (z + h)− F (z)− hf(z)| ≤ |h|∫ 1

0|(z + th)− f(z)|dt.

Let now ε > 0 be given. Since f is continuous at z, there exits α > 0 that we can take < rsuch that |f(z + k)− f(z)| < ε whenever |k| < α. Let h satisfy |h| < α. Then for all t ∈ [0, 1],|th| < α. Therefore |f(z + th) − f(z)| < ε for all t ∈ [0, 1] and so

∫ 10 |(z + th) − f(z)|dt ≤ ε.

Thus,|F (z + h)− F (z)− hf(z)| ≤ ε|h|

whenever |h| < α. This means that

limh→0

F (z + h)− F (z)h

= f(z).

Suppose now that Ω is convex. Then (iii)⇒ (iv) because a triangular path is a closed polygonalpath.

(iv)⇒ (i) Let a ∈ Ω be fixed. For z ∈ Ω, set G(z) =∫[a,z] f(u) du. This makes sense because

now [a, z] ⊂ Ω. The proof (iii)⇒(i) can be repeated verbatim to show that G′ = f .

Consequence. The fnction z 7→ 1z has no antidrivative on C∗. Indeed, take γ(t) = eit,

t ∈ [0, 2π]. Then ∫γ

1z

dz =∫ 2π

0

1γ(t)

γ′(t) dt =∫ 2π

0

ieit

eitdt = 2πi 6= 0.

It follows that there is no branch of the logarithm on C∗; (L−π is an antideravative of 1z on

C\IR− and not on C∗).

Exercise. Explain why there is no branch of the argument on C∗.

Remark 2.4 By a triangular region, we mean a triangle together with its interior, that is, thesmallest convex set containing the three vertices of a triangle. If T is a triangular region, itsboundary (which is a triangle) is denoted by ∂T .

Theorem 2.3 (Goursat-Cauchy version 1) Let f : Ω → C be holomorphic and let T be atriangular region in Ω. Then ∫

∂Tf(z) dz = 0.

2.1. LINE INTEGRALS 27

Proof. Let I =∫∂T f(z) dz. By means of the midpoints of the edges T , divide T into four

equal triangular regions T 1, T 2, T 3 and T 4 (draw a figure). Then

I =∫

∂Tf(z) dz =

4∑i=1

∫∂T i

f(z) dz.

Among the T i, there is a triangular region T1 such that

14|I| ≤

∣∣∣∣∫∂T1

f(z) dz

∣∣∣∣(otherwise, we would have |I| > |I|). By induction, we construct a decreasing sequence (Tn) oftriangular regions such that

14n|I| ≤

∣∣∣∣∫∂Tn

f(z) dz

∣∣∣∣ = |In|. (2.1)

The sequence (Tn) is such that diam(Tn) = diamT2n → 0 as n → ∞. By a theorem of topology,

there exits a ∈ Ω such that∞⋂

n=1

Tn = a.

At the point a, f(z) = f(a) + (z − a)f ′(a) + (z − a)ε(z) where limz→a ε(z) = 0. Then

In =∫

∂Tn

f(z) dz =∫

∂Tn

[f(a) + (z − a)f ′(a)] dz +∫

∂Tn

(z − a)ε(z) dz.

The first integral on the right hand side vanishes because the function z 7→ f(a) + (z − a)f ′(a)has an antiderivative (which is z 7→ f(a)z + f ′(a)

2 (z − a)2) and ∂Tn is a closed curve. Therefore∣∣∣∣∫∂Tn

f(z) dz

∣∣∣∣ ≤ ∣∣∣∣∫∂Tn

(z − a)ε(z) dz

∣∣∣∣≤ L(∂Tn) sup

z∈∂Tn

|z − a| supz∈∂Tn

|ε(z)|

≤ L(∂T )2n

diam(T )2n

supz∈Tn

|ε(z)|

By inequality (2.1), we get

|I|4n

≤ L(∂T )2n

diam(T )2n

supz∈Tn

|ε(z)|.

Consequently,|I| ≤ L(∂T )diam(∂T ) sup

z∈Tn

|ε(z)|.

Letting n →∞, we get I = 0.

Now we have a first answer to the problem of existence of antiderivatives.

Corollary 2.2 Let Ω be open and convex and f : Ω → C be holomorphic. Then f has anantiderivative on Ω.

Proof. Let γ be a triangular loop in Ω. Then γ∗ = ∂T where T is a triangular region.

Since Ω is convex, T ⊂ Ω. By the previous theorem∫

γf(z) dz = 0. By Theorem 2.2, f has an

antiderivative because Ω is convex.


Corollary 2.3 Let f : Ω → C be holomorphic. Then every z ∈ Ω has a neighborhood on whichf has an antiderivative. Otherwise stated, a holomorphic function has locally an antiderivative.

Proof. A ball B(z, r) is convex.

In the next section, we need a slightly different version of the above theorem.

Theorem 2.4 (Goursat-Cauchy version 2) Let Ω ⊂ C be open and a ∈ Ω. Let f : Ω → C

be continuous and holomorphic on Ω\a. If T is a triangular region in Ω, then∫∂T

f(z) dz = 0.

Proof. If a /∈ T , then T ⊂ Ω\a and the result follows from version 1 of this theorem. Sowe assume that a ∈ T . Let v1, v2 and v3 be the vertices of T .

Step 1. a is one of the vertices say v1. Let ε > 0 be given. Choose a2 ∈ [v1, v2] anda3 ∈ [v1, v3] such that |v1 − a2| + |a2 − a3| + |a3 − v1| < ε (that is, a2 and a3 are sufficientlyclose to v1 so that the perimeter of the triangle v1a2a3 is small; draw a figure). Then∫

∂Tf(z) dz =

∫a2a3v2

f(z) dz +∫

a3v3v2

f(z) dz +∫

v1a2a3

f(z) dz.

The first two intergals are zero by version 1 of this theorem. Let M be a bound of |f(z)| on T .Then, ∣∣∣∣∫

∂Tf(z) dz

∣∣∣∣ = ∣∣∣∣∫v1a2a3

f(z) dz

∣∣∣∣ ≤ L(v1a2a3)M ≤ εM.

Since ε was arbitrary, we conclude that∫∂T f(z) dz = 0.

Step 2. a is in the interior of T or one some side. Then∫∂T

f(z) dz =∫

av1v2

f(z) dz +∫

av2v3

f(z) dz +∫

av1v3

f(z) dz.

By step 1, each of these integrals is zero.

Remark 2.5 The assumptions of this theorem imply that actually f is holomorphic at a (seethe next section or chapter 3).

We shall deal later with integrals depending on a complex parameter. The following resultis fundamental and will be useful.

Theorem 2.5 (Differentiation under the integral sign) Let ϕ : [a, b] × Ω → C be con-tinuous, holomorphic with respect to the second variable and such that ∂2ϕ is continuous on[a, b]× Ω. We set

f(z) =∫ b

aϕ(t, z) dt.

Then f is holomorphic and

f ′(z) =∫ b

a∂2ϕ(t, z) dt.

2.2. ANALYCITY OF HOLOMORPHIC FUNCTIONS 29

Proof. Let z ∈ Ω. Then [z, z + h] ⊂ Ω for all h whose modulus is small enough. Letg(z) =

∫ ba ∂2ϕ(t, z) dt. Then g is continuous (prove this). Now,∫ z+h

zg(u) du =

∫ z+h

z

∫ b

a∂2ϕ(t, u) dt du

=∫ b

a

∫ z+h

z∂2ϕ(t, u) du dt by Fubini’s theorem

=∫ b

a[ϕ(t, z + h)− ϕ(t, z)] dt

= f(z + h)− f(z).

Therefore,f(z + h)− f(z)

h=

1h

∫ z+h

zg(u) du.

Now we claim that limh→01h

∫ z+h

zg(u) du = g(z). Indeed, let ε be given. Since g is continuous

at z, there is δ > 0 such that |g(u)− g(z)| < ε for all u such that |u− z| < δ. Then for |h| < δ,we have∣∣∣∣1h∫ z+h

zg(u) du− g(z)

∣∣∣∣ = ∣∣∣∣1h∫ z+h

zg(u) du− 1

h

∫ z+h

zg(z) dz

∣∣∣∣ = ∣∣∣∣1h∫ z+h

z[g(u)− g(z)] dz

∣∣∣∣ ≤ ε.

It follows fro the claim that f is holomorphic at z and f ′(z) = g(z).

2.2 Analycity of holomorphic functions

Definition 2.5 Let γ : [α, β] → C be a piecewise smooth loop and let a /∈ γ∗. Then index orthe winding number of γ with respect to a is the number

I(γ, a) =1

2πi

∫γ

dz

z − a=

12πi

∫ β

α

γ′(t)γ(t)− a

dt.

Example 2.1 Let γ(t) = a + reit, 0 ≤ t ≤ 2π. Then I(γ, a) =1

2πi

∫ 2π

0

rieit

reit= 1.

Proposition 2.5 I(γ, a) ∈ Z.

Proof. Let γ : [0, 1] → C be a piecewise smooth loop not passing through a. Set

g(t) = exp(∫ t

0

γ′(s)γ(s)− a

ds

), 0 ≤ t ≤ 1.

Then, except at a finite number of points where γ is not differentiable, we have by the funda-mental theorem of calculus

g′(t) = exp(∫ t

0

γ′(s)γ(s)− a

ds

)γ′(t)

γ(t)− a= g(t)

γ′(t)γ(t)− a

.

Let ϕ(t) =g(t)

γ(t)− a. Then ϕ′(t) =

g′(t)[γ(t)− a]− γ′(t)g(t)(γ(t)− a)2

= 0. It follows that ϕ is piecewise

constant. But since it is continuous, it is constant and so ϕ(1) = ϕ(0). This means that

exp(2πiI(γ, a))γ(1)− a

=1

γ(0)− a.

Consequently, (since γ(0) = γ(1)), we get exp(2πiI(γ, a)) = 1 and therefore I(γ, a) ∈ Z.


Remark 2.6 I(γ, a) is the number of times γ ”winds” around a in the counterclockwise direc-tion. Let us see why. Set

f(t) =∫ t

0

γ′(s)γ(s)− a

ds.

By the fundamental theorem of calculus, f is piecewise smooth and f ′(t) = γ′(t)γ(t)−a except at

finitely many points. Keeping the same notation as in the above proof, we have

ef(t) = g(t) =γ(t)− a

γ(0)− a.

Now, there exists a complex number C such that eC = 1γ(0)−a . Therefore ef(t)−C = γ(t) − a.

Setting h(t) = f(t)− C, we get

(1) h′(t) = γ′(t)γ(t)−a and

(2) eh(t) = γ(t)− a.

It follows from the first point that

I(γ, a) =1

2πi

∫ 1

0f ′(t) dt =

12πi

(f(1)− f(0)).

Next, let α(t) and θ(t) denote respectively the real and imaginary part of h(t). It follows from thesecond point that α(t) = ln |γ(t)−a| and so α(1) = α(0). Therefore h(1)−h(0) = i(θ(1)−θ(0))and so

I(γ, a) =12π

(θ(1)− θ(0)).

But θ(t) is the angle that γ(t)− a makes with the positive x-axis. Therefore θ(1)− θ(0) is thevariation of this angle as γ goes from 0 to 1. Therefore I(γ, a) is the oriented number of timesγ winds around a (I(γ, a) is positive if γ winds counterclockwise, negative it it winds clockwise,and 0 if a is outside γ).

Exercise. a) Check that I(γ−, a) = −I(γ, a).b) Let γ(t) = eint for t ∈ [0, 2π] and n ∈ Z. Find I(γ, 0).b) More generally, let γ : [0, 1] → C be a loop. Extend γ by periodicity to IR. Let µ(t) = γ(nt)for t ∈ [0, 1]. Show that I(µ, a) = nI(γ, a).

Proposition 2.6 Let γ be a piecewise smooth loop in C. Then the function ζ 7→ I(γ, ζ) fromC\γ∗ → Z is constant on the connected component of C\γ∗.

Proof. (First method). The function f is continuous and takes values in Z. Thereforethe image of every connected component is a connected subset of Z. But the only connectedsubsets of Z are singletons.

(Second method). Let f(ζ) = I(γ, ζ). Then, f(ζ) =1

2πi

∫γ

dz

z − ζ=

12πi

∫ 1

0

γ′(t)γ(t)− ζ

dt. It

follows from Theorem 2.5, that f is holomorphic on C\γ∗ and that

f ′(ζ) =1

2πi

∫γ

dz

(z − ζ)2.

Now the function z 7→ 1(z−ζ)2

has an antiderivative on C\ζ which is z 7→ −1z−ζ . Consequently,∫

γ

dz

(z − ζ)2= 0. It follows that f ′(ζ) = 0 for all ζ ∈ C\γ∗. This implies that f is constant on

the connected components of C\γ∗ (draw a figure).


Proposition 2.7 I(γ, a) = 0 whenever a belongs to the unbounded connected component ofC\γ∗.

Proof. Since γ∗ is bounded, it is contained in some ball B(0, r). Take a ∈ C such that |a| > r(draw a figure). Then a belongs to the unbounded component of C\γ∗. Then, for z ∈ γ∗,∣∣∣ 1z−a

∣∣∣ ≤ 1||z|−|a|| ≤

1|a|−r

|I(γ, a) =∣∣∣∣∫

γ

dz

z − a

∣∣∣∣ ≤ L(γ)|a| − r

.

This estimate is true for any a outside C, and so letting |a| → ∞ we get I(γ, a) = 0.

Proposition 2.8 (Cauchy’s integral formula in a special case) Let f : B(z0, r) → C beholomorphic. Let γ(t) = z0 + r0e

it where r0 < r and t ∈ [0, 2π]. Then

f(z) =1

2πi

∫γ

f(u)u− z

du

for all z ∈ B(z0, r0).

Proof. Fix z ∈ B(z0, r0) and define g : B(z0, r) → C by

g(u) =

f(u)−f(z)

u−z if u 6= z

f ′(z) if u = z.

Then g is continuous on B(z0, r) and holomorphic on B(z0, r)\z. Now by Theorem 2.4,∫µ g(u) du = 0 for every triangular path µ in B(z0, r)2. But it follows from Theorem 2.2 that∫γ g(u) du = 0. This means that

∫γ

f(u)−f(z)u−z du = 0 since z /∈ γ∗. Therefore

12πi

∫γ

f(u)u− z

du =1

2πi

∫γ

f(z)u− z

du = f(z)∫

γ

du

u− z= f(z)I(γ, z).

Now, since z and z0 are in the same connected component of C\γ∗ (both are inside γ), it followsthat

I(γ, z) = I(γ, z0) =1

2πi

∫γ

du

u− z0=

12πi

∫ 2π

0

γ′(t)γ(t)− z0

dt =1

2πi

∫ 2π

0

r0ieit

r0eitdt = 1.

Exercise. a) Compute∫

γ

ez

zdz where γ is a parametrization of the unit circle (positively

oriented).

b) Compute∫

γ

1z2 + 1

dz where γ is a parametrization of the circle |z| = 2 (positively oriented).

Theorem 2.6 (Analycity of holomorphic functions)Let Ω ⊂ C be an open set and let f : Ω → C be holomorphic. Then f is analytic on Ω.

Proof. Let z0 ∈ Ω. Then, there exists r > 0 such that B(z0, r) ⊂ Ω. Let now r0 satisfy0 < r0 < r. Let γ(t) = z0 + r0e

it for 0 ≤ t ≤ 2π. Let z ∈ B(z0, r0). By Cauchy’s integralformula,

f(z) =1

2πi

∫γ

f(u)u− z

du.

2Since B(z0, r) is convex, the interior of every tringular path in B(z0, r) is contained in B(z0, r); so everytriangular path is the boundary of a triangular region contained in B(z0, r).


Now observe that1

u− z=

1(u− z0)− (z − z0)

=1

u− z0

11− z−z0

u−z0

and that∣∣∣ z−z0u−z0

∣∣∣ = |z−z0|r0

< 1 for u ∈ γ∗. Therefore

1u− z

=∞∑

n=0

(z − z0)n

(u− z0)n+1.

Consequently,f(u)u− z

=∞∑

n=0

f(u)(u− z0)n+1

(z − z0)n.

Therefore,

f(z) =1

2πi

∫γ

f(u)u− z

du =1

2πi

∫γ

[ ∞∑n=0

f(u)(u− z0)n+1

(z − z0)n

]du.

Now let M be a bound of |f | on γ∗. Then,∣∣∣∣ f(u)(u− z0)n+1

(z − z0)n

∣∣∣∣ ≤ M

r1

(|z − z0|

r0

)n

for all u ∈ γ∗. This means that the series∑∞

n=0f(u)

(u−z0)n+1 (z − z0)n is normally (and thereforeuniformly) convergent. Thus, we can interchange the operations of integration and summation.Therefore,

f(z) =∞∑

n=0

[1

2πi

∫γ

f(u)(u− z0)n+1

du

](z − z0)n =

∞∑n=0

an(z − z0)n ∀z ∈ B(z0, r0)

wherean =

12πi

∫γ

f(u)(u− z0)n+1

du

for all n ∈ IN.

Remark 2.7 If f is analytic at a point z0, then it is analytic in a neighborhood of z0. Indeed,we now that f is holomorphic on a neighborhood U of z0. Let z1 ∈ U , then there exists adisk B(z1, r1) ⊂ U . The function f is holomorphic on B(z1, r1), and according to the previoustheorem, it is analytic at z1.

Remark 2.8 Suppose that f cannot be extended to a holomorphic function outside Ω (i.e., Ωis the biggest open set on which f is holomorphic). Let z0 ∈ Ω. We know that

f(z) =∞∑

n=0

an(z − z0)n.

We claim that the radius of convergence of this series is R = dist (z0,Ωc) (if Ω = C, R = ∞).Points in Ω are called regular points of f and points in Ωc are called singular points. So inwords, the claim says that the radius of convergence of the above series is the distance from z0

to the set of singular points of f .Indeed, let us go back to the proof of the previous theorem. We know that there exists

r > 0 such that B(z0, r) ⊂ Ω. Take r as large as possible, that is, take r = dist (z0,Ωc) withthe convention that dist (z0, ∅) = inf ∅ = +∞. Let 0 < r0 < r. Our proof shows that theseries

∑∞n=0 an(z − z0)n converges absolutely in B(z0, r0) so R ≥ r0. Since r0 was an arbitrary


number less than r, it follows that R ≥ r. Next, if R > r, then by a property of the infinimum,B(z0, R) would contain a point in Ωc; however f is holomorphic on B(z0, R). Thus R = r.

Let us illustrate the use of this fact in practise. Let f(z) = ez

1−z for z 6= ±i. Here Ω = C\1.We know that f(z) =

∑∞n=0 anzn for z ∈ B(0, R). Even without computing the an, we know

that R = dist (0, 1) = 1.

Exercise. a) Compute the an in the previous example and check directly that R = 1.

b) Show in the context of the previous remark, that dist (z0,Ωc) = dist (z0, ∂Ω).

Corollary 2.4 Let f : Ω → C be holomorphic. Then f has derivatives of all orders given by

f (n)(z0) =n!2πi

∫γ

f(u)(u− z0)n+1

du

where γ is as in the proof of the above theorem.

Remark 2.9 In particular, the derivative of a holomorphic function is holomorphic. Thereforeif a function has an antiderivative, it is necessarily holomorphic. Thus for example z 7→ z andz 7→ |z|2 have no antiderivatives.

Theorem 2.7 (Morera; converse to Goursat-Cauchy) Let f : Ω → C be continuous andsatisfy

∫∂T f(z) dz = 0 for every triangular region T ⊂ Ω. Then f is holomorphic on Ω.

Proof. Let z ∈ Ω be given and let U a disk containing z and contained in Ω. Note that Uis convex. By Theorem 2.2 (iv), f has an antiderivative F on U . By the previous remark, f isholomorphic on U and in particular at z. Since z was arbitrary, it follows that f is holomorphicon Ω.

Corollary 2.5 Let f : Ω → C be continuous and holomorphic on Ω\a for some a ∈ Ω. Thenf is holomorphic on Ω.

Proof. By Theorem 2.4,∫∂T f(z) dz = 0 for every triangular region T ⊂ Ω. By Morera’s

theorem, f is holomorphic on Ω.

Proposition 2.9 (Cauchy’s inequality) Let f(z) =∑

n≥0 an(z − z0)n be holomorphic on

a disk B(z0, R). Let 0 < r < R and γ(t) = z0 + reit for 0 ≤ t ≤ 2π. Finally, let M(r) =sup|u−z0|=r |f(u)| = supu∈γ∗ |f(u)|. Then

|an| ≤M(r)

rn

for all n ∈ IN.

Proof. We know from Theorem 2.6 that an =1

2πi

∫γ

f(u)(u− z0)n+1

du. Therefore,

|an| ≤12π

L(γ) supu∈γ∗

|f(u)||u− z0|n+1

=12π

2πrM(r)rn+1

=M(r)

rn.

Definition 2.6 A entire function is a holomorphic function f : C→ C.

We know that if f is holomorphic function then f is the sum of a power series∑

n≥0 anzn withinfinite radius of convergence.


Corollary 2.6 (Liouville’s theorem) A bounded entire function is constant.

Proof. Let f(z) =∑

anzn for all z ∈ C. Let r > 0 and let A be a bound for f . By Cauchy’sinequality, |an| ≤ M(r)

rn ≤ Arn . Letting r → ∞, we get an = 0 for all n ≥ 1. Consequently,

f(z) = a0 for all z ∈ C.

Corollary 2.7 (Fundamental theorem of Algebra) Every nonconstant polynomial withcomplex coefficients has a root in C.

Proof. Let P (z) = a0+a1z+· · ·+anzn with an 6= 0. By writing P (z) = zn[an+ an−1

z +· · ·+ a0zn ],

we see that lim|z|→∞ |P (z)| = ∞. Suppose that P has no roots in C. Then z 7→ 1P (z) is

holomorphic on C. Since lim|z|→∞ | 1P (z) | = 0, there is a number λ > 0 such that

∣∣∣ 1P (z)

∣∣∣ ≤ 1 for

all z such that |z| > λ. On the other hand, the map z 7→ 1P (z) is continuous on the compact set

B′(0, λ) and therefore it is bounded there. Consequently, 1P is bounded on C. It follows from

Liouville theorem that 1P is constant and therefore P is constant, contradicting our assumption.

Theorem 2.8 (Strong maximum principle) Let Ω ⊂ C be open and connected and letf : Ω → C be holomorphic. If |f | has a relative maximum at a point a ∈ Ω, then f is constant.

Proof. By definition, there exists a disk B(a, ρ) ⊂ Ω such that |f(z)| ≤ |f(a)| for allz ∈ B(a, ρ). Let 0 < r < ρ and let γ(t) = a + reit, t ∈ [0, 2π]. By Cauchy’s integral formula,

f(a) =1

2πi

∫γ

f(u)u− a

du =1

2πi

∫ 2π

0

f(a + reit)rieit

reitdt =

12π

∫ 2π

0f(a + reit) dt.

Therefore

|f(a)| = 12π

∣∣∣∣∫ 2π

0f(a + reit) dt

∣∣∣∣ ≤ 12π

∫ 2π

0|f(a + reit)|dt ≤ 1

2π

∫ 2π

0|f(a)|dt = |f(a)|.

It follows that ∫ 2π

0

[|f(a)| − |f(a + reit)|

]dt = 0.

Since the integrand is nonnegative and continuous, it is zero. That is, |f(a + reit)| = |f(a)| forall t ∈ [0, 2π]. But r was an arbitrary number in ]0, ρ[. This means that |f(z)| = |f(a)| forall z ∈ B(a, ρ), i.e., |f | is contant on B(a, ρ) (which is open and connected). It follows fromProposition 1.5 that f is constant on B(a, ρ). Since Ω is connected, the principle of analyticcontinuation implies that f is constant on Ω.

Corollary 2.8 (Weak maximum principle) Let Ω ⊂ C be open, bounded and connected.Let f : Ω → C be continuous and holomorphic on Ω. Then

supz∈Ω

|f(z)| = supz∈Ω

|f(z)| = supz∈∂Ω

|f(z)|.

Proof. We prove first that supz∈Ω |f(z)| = supz∈Ω |f(z)|. This is true under only thecontinuity assumption. Let m = supz∈Ω |f(z)| and M = supz∈Ω |f(z)|. It is clear that m ≤ Msince Ω ⊂ Ω. Conversely, let A = z ∈ Ω||f(z)| ≤ m. Then A is closed in Ω and therefore inC. Note that Ω ⊂ A and so Ω ⊂ A since A is closed. It follows that A = Ω. This means thatfor all z ∈ Ω, |f(z)| ≤ m and so M ≤ m.

Next, we prove the remaining part of the equality. Since Ω is compact and |f | is continuous,it attains its maximum at a point a ∈ Ω = Ω∪ ∂Ω. If a ∈ Ω, then f is constant by the previoustheorem and so the equality holds. If a ∈ ∂Ω, then supz∈∂Ω |f(z)| ≤ supz∈Ω |f(z)| = |f(a)| ≤supz∈∂Ω |f(z)|. Hence the equality.

2.3. CAUCHY’S THEOREM 35

2.3 Cauchy’s theorem

Definition 2.7 Let Ω ⊂ C be open. Let γ0 and γ1 be two loops in Ω. We say that γ0 ishomotopic (or Ω−homotopic) to γ1 if there exists a continuous function H : [0, 1]× [0, 1] → Ωsuch that

(i) ∀t ∈ [0, 1], H(t, 0) = γ0(t) and H(t, 1) = γ1(t).

(ii) ∀s ∈ [0, 1], H(0, s) = H(1, s).

Let us explain this definition. Set γs(t) = H(t, s). Then, for each s ∈ [0, 1], γs is a loop in Ωsince γs(0) = γs(1). We can think of a homotopy as a continuous deformation that takes γ0 toγ1. We can also think of a homotopy as a path in the space of loops.

Remark 2.10 The homotopy is an equivalence relation between loops.

Examples. 1) An two loops γ0, γ1 : [0, 1] → C in C are C−homotopic. Indeed, let H(t, s) =(1− s)γ0(t) + sγ1(t). H is called the straight line homotopy. It is clear that H is a homotopybetween γ0 and γ1.

2) More generally if Ω is convex, then any two loops γ0, γ1 : [0, 1] → Ω are Ω−homotopic.

3) Let γ0(t) = eit and γ1(t) = 2eit for 0 ≤ t ≤ 2π. Then γ0 and γ1 are C∗−homotopic.However, if γ2(t) = e2it for t ∈ [0, 2π], then γ0 and γ1 which parametrize the same circle are notC∗−homotopic; γ0 runs one time through the circle, whereas γ2 runs twice.

4) Let γ0(t) = eit and γ1(t) = 2 + eit for t ∈ [0, 2π]. We shall see later that γ0 and γ1 are notC∗−homotopic. The intuitive idea behind this is that if we want to deform continuously γ0 toγ1, we have at some point to pass by 0. So there is an obstruction in the domain that preventsthe existence of a homotopy.

Theorem 2.9 (Cauchy) Let f : Ω → C be holomorphic and let γ0 and γ1 be two piecewisesmooth loops that are homotopic in Ω. Then∫

γ0

f(z) dz =∫

γ1

f(z) dz.

To prove Cauchy’s theorem, we need some preliminary results and concepts.

Lemma 2.1 Let Ω ⊂ C be open and let γ : [0, 1] → Ω be a path. Then there exists asubdivision t0 = 0 < t1 < · · · < tn−1 < tn = 1 of [0,1] and open disks D0, . . . , Dn−1 containedin Ω such that γ([tk, tk+1]) ⊂ Dk for all k = 0, . . . , n− 1.

The subdivision t0, . . . , tn and the collection of disks D0, . . . , Dn−1 are called a chaincovering of γ.

Proof. Note that γ is uniformly continuous on the compact space [0,1]; this means thatfor every ε > 0, there exists η > 0 such that |γ(t) − γ(t′)| < ε whenever |t − t′| < η. Sinceγ∗ is compact and C\Ω is closed, the distance d(γ∗,C\Ω) between γ∗ and C\Ω is > 0. Takeε < d(γ∗,C\Ω) and η as in the uniform continuity condition. Next let n ∈ IN∗ be such that1n < η. Define a subdivision t0, . . . , tn by t0 = 0 and tk+1 = tk + 1

n for k = 0, . . . , n − 1.Finally, let Dk = B(γ(tk), ε) for k = 0, . . . , n − 1. Now one can easily check that Dk ⊂ Ω andγ([tk, tk+1]) ⊂ Dk.


Definition 2.8 Let f : Ω → C be holomorphic. Let γ, t0, . . . , tn and D0, . . . , Dn−1 be asin the above lemma. Let finally γ(tk) = zk. Since each Dk is convex, there exists on Dk anantiderivative Fk of f . The number

n−1∑k=0

[Fk(zk+1)− Fk(zk)]

is called the variation of antiderivatives of f along γ.

Lemma 2.2 If γ is a piecewise smooth path, then the variation of antiderivatives of f along γis equal to

∫γ f(z) dz.

Proof. Let γk denote the restriction of γ to [tk, tk+1]. Then by Barrow’s rule, Fk(zk+1) −Fk(zk) =

∫γk

f(z) dz. The result follows by summation over k.

Lemma 2.3 The variation of antiderivatives along a path is independent of the choice of an-tiderivative.

Proof. For each k = 0, . . . , n, let Fk and Gk be two antiderivatives of f on Dk. Thenthey differ by a constant ck. But then Fk(zk+1) − Fk(zk) = Gk(zk+1) + ck − (Fk(zk) + ck) =Gk(zk+1)−Gk(zk).

Lemma 2.4 Let γ and µ be two piecewise smooth loops in Ω. If we can associate to them thesame chain covering, then γ and µ have the same variation of antiderivatives.

Proof. Let t0, . . . , tn be a subdivision of [0,1] and D0, . . . , Dn−1 be disks such thatγ([tk, tk+1]) ⊂ Dk ⊂ Ω and µ([tk, tk+1] ⊂ Dk ⊂ Ω. Let F0, . . . Fn−1 be antiderivatives of frespectively on D0, . . . , Dn−1. Finally let zk = γ(tk) and wk = µ(tk). Note that z0 = zn sinceγ(0) = γ(1) and similarly w0 = wn. Let

∆ =n−1∑k=0

[Fk(zk+1)− Fk(zk)]−n−1∑k=0

[Fk(wk+1)− Fk(wk)]

Then

∆ =n−1∑k=0

[Fk(zk+1)− Fk(zk)]− [Fk(wk+1)− Fk(wk)]

=

n−1∑k=0

[Fk(zk+1)− Fk(wk+1)]− [Fk(zk)− Fk(wk)]

= [Fn−1(zn)− Fn−1(wn)]− [Fn−1(zn−1)− Fn−1(wn−1)]+

+ [Fn−2(zn−1)− Fn−2(wn−1)]− [Fn−2(zn−2)− Fn−2(wn−2)] + · · ·+ [F0(z1)− F0(w1)]− [F0(z0)− F0(w0)] .

Now observe that zn−1 ∈ Dn−1 ∩ Dn−2 and that wn−1 ∈ Dn−1 ∩ Dn−2. On the convex setDn−1 ∩Dn−2, the antiderivatives Fn−1 and Fn−2 differ by a constant. Thus, we see that ∆ = 0(the underlined terms cancel each other).

Proof of Cauchy’s theorem. Let H : [0, 1]× [0, 1] → Ω be a homotopy between γ0 and γ1.Note that H is uniformly continuous on the compact set [0, 1]× [0, 1]. Therefore for any ε > 0,there exists α > 0 such that |H(t, s)−H(t′, s′)| < ε whenever |t− t′| < α and |s− s′| < α.

Let δ = dist(H([0, 1]×[0, 1]),C\Ω). Let 0 < ε < δ. Take two subdivisions t0, t1, . . . , tn ands0, s1, . . . , sn such that max |ti+1− ti| < α and max |si+1−si| < α. Let Di,j = B(H(ti, sj), ε).

2.3. CAUCHY’S THEOREM 37

It follows from the uniform continuity condition that H([ti, ti+1]× [sj , sj+1]) ⊂ Di,j . Set αj(t) =H(t, sj) for all t ∈ [0, 1]. Then αj is a loop and αj([ti, ti+1]) ⊂ Di,j and αj+1([ti, ti+1]) ⊂ Di,j .This means that αj and αj+1 have the same chain covering. By Lemma 2.4, the variation ofantiderivatives of f along αj is equal to the variation of antiderivatives of f along αj+1. Itfollows from Lemma 2.2 that

∫αj

f(z) dz =∫αj+1

f(z) dz for every j = 0, . . . , n− 1. Therefore∫γ0

f(z) dz =∫

α0

f(z) dz =∫

α1

f(z) dz =∫

α2

f(z) dz = · · · =∫

αn

f(z) dz =∫

γ1

f(z) dz.

Corollary 2.9 Let Ω be an open set and a ∈ Ω. If γ and µ are two piecewise smooth loopsthat are homotopic in Ω\a, then I(γ, a) = I(µ, a).

Examples. 1) Let γ0(t) = eit and γ1(t) = e2it for 0 ≤ t ≤ 2π. Then γ0 and γ1 are notC∗−homotopic. Indeed, I(γ0, 0) = 1 whereas I(γ1, 0) = 2.

2) Let γ0(t) = eit and γ1(t) = e−it for 0 ≤ t ≤ 2π. Then γ0 and γ1 are not C∗−homotopic.Indeed, I(γ0, 0) = 1 whereas I(γ1, 0) = −1.

3) Let γ0(t) = eit and γ1(t) = 3 + eit for 0 ≤ t ≤ 2π. Then γ0 and γ1 are not C∗−homotopic.Indeed, I(γ0, 0) = 1 whereas I(γ1, 0) = 0 (since 0 is in the unbounded component of C\γ∗1).

Definition 2.9 Let Ω ⊂ C be an open and connected. We say that Ω is simply connected ifevery loop is homotopic to a point in Ω.

Intuitively, a simply connected domain is a domain without ”holes”. Every convex set is simplyconnected. On the other hand C∗ is not simply connected. The annulus 1 < |z| < 2 is notsimply connected.

Now we have a very general answer to the problem of existence of antiderivatives.

Corollary 2.10 Let Ω be simply connected and f : Ω → C be holomorphic. Then f has anantiderivative on Ω.

Proof. Let γ be a loop in Ω. Since Ω is simply connected, γ is homotopic to a pointz0. Therefore

∫γ f(z) dz =

∫z0 f(z) dz = 0. It follows from Theorem 2.2 that f has an

antiderivative on Ω.

Remark 2.11 We have therefore a topological condition on the domain of a holomorphicfunction that guarantees the existence of a global antidervative. This condition is optimal:if every holomorphic function f : Ω → C has an antiderivative, then Ω is necessarily simplyconnected; see Theorem 13.11 in the book of Rudin.

Definition 2.10 An open set Ω is called star convex with respect to a point a ∈ Ω if [a, z] ⊂ Ωfor any z ∈ Ω.

It is clear that a convex set is star convex (with respect to any of its points). The set C\IR− isstar convex (with respect to the point 1 for example) but not convex. More generally if D is ahalf line then C\D is star convex.

Proposition 2.10 Every open star convex set is simply connected.

Proof. Let Ω be a star convex set. It should be clear that Ω is path connected and thereforeconnected. Next, let γ be a closed path in Ω and let a be a point with respect to which Ω isstar convex. Let H(t, s) = (1 − s)γ(t) + sa. Then it is easy to check that H is a homotopy inΩ joining γ to a.


Theorem 2.10 (Cauchy’s integral formula) Let f : Ω → C be holomorphic and a ∈ Ω.

1) If γ and µ are two piecewise smooth loops that are homotopic in Ω and not passing througha, then

12πi

∫γ

f(u)u− a

du− 12πi

∫µ

f(u)u− a

du = f(a)[I(γ, a)− I(µ, a)].

2) If γ is a piecewise smooth loop not passing through a and homotopic to a point in Ω, then

12πi

∫γ

f(u)u− a

du = I(γ, a)f(a).

Proof. Set

g(z) =

f(z)−f(a)

z−a if z 6= a

f ′(a) if z = a.

It follows from Corollary 2.5 that g is is holomorphic in Ω. By Cauchy’s theorem,∫γ g(u) du =∫

µ g(u) du. This implies that∫γ

f(u)u− a

du− f(a)∫

γ

du

u− a=∫

µ

f(u)u− a

du− f(a)∫

µ

du

u− a.

Dividing by 12πi and recalling that I(γ, a) = 1

2πi

∫γ

duu−a , we get the result.

The proof of 2) is similar.

By differentiating under the integral, we obtain the following.

Corollary 2.11 (Cauchy’s integral formula of order n) Let f : Ω → C be holomorphicand a ∈ Ω. If γ is a piecewise smooth loop not passing through a and homotopic to a point inΩ, then

I(γ, a)f (n)(a) =n!2πi

∫γ

f(u)(u− a)n+1

du.

Exercise. Let R > 1 and let CR be the upper half circle of center 0 and radius R (orientedpositively). Set ΓR = [−R,R] ∪ CR.

(a) Compute∫

ΓR

eiz

1 + z2dz by using properly Cauchy’s formula.

(b) Deduce the value of∫ ∞

0

cos x

1 + x2dx.

Chapter 3

Laurent series and the residuetheorem

3.1 Laurent series

Definition 3.1 A Laurent series is a series of the form+∞∑

n=−∞an(z − a)n. Such series is con-

vergent if the series

+∞∑n=0

an(z − a)n and

−1∑n=−∞

an(z − a)n =+∞∑p=1

a−p

(z − a)pare both convergent.

Same definition for absolutely and uniformly convergent.

Proposition 3.1 Let∑

n∈Z an(z−a)n be a Laurent series. Then there exist two values p1 andp2 in [0,∞] such that the Laurent series is absolutely convergent in the annulus p2 < |z−a| < p1.

Proof. Let p1 denote the radius of convergence of+∞∑n=0

an(z−a)n. Then this series is absolutely

convergent for |z − a| < p1 and divergent for |z − a| > p1.

The series−1∑

n=−∞an(z− a)n can be written as

+∞∑p=1

bpup where bp = a−p and u = 1

z−a . Let R2

be the radius of convergence of the series+∞∑p=1

bpup. Then this series is absolutely convergent for

|u| < R2 and divergent for |u| > R2. Let p2 = 1R2

. Then the series−1∑

n=−∞an(z−a)n is absolutely

convergent for |z − a| > p2. Hence the conclusion.

Remark 3.1 Since limu→0

+∞∑p=1

bpup = 0, it follows that lim|z−a|→∞

−1∑n=−∞

an(z − a)n = 0.

Remark 3.2 A Laurent series need not be convergent. For example the series∑

n∈Z zn isnowhere convergent (it is convergent on the empty annulus 1 < |z| < 1).

Proposition 3.2 If two Laurent series∑

n∈Z an(z − a)n and∑

n∈Z bn(z − a)n coincide on anonempty annulus p2 < |z − a| < p1 then an = bn for all n ∈ Z.

Proof. Let S(z) denote the common value of the two series. A complex number z in theannulus can be written in the form z = a + reiθ with p2 < r < p1. Then S(a + reiθ) =

39

40 CHAPTER 3. LAURENT SERIES AND THE RESIDUE THEOREM

∑n∈Z anrneinθ and this series is normally convergent with respect to θ ∈ [0, 2π]. Therefore, we

can multiply by e−ipθ and integrate term by term:∫ 2π

0S(a + reiθ)e−ipθ dθ =

∫ 2π

0

∑n∈Z

anrnei(n−p)θ dθ =∑n∈Z

anrn

∫ 2π

0ei(n−p)θ dθ = 2πapr

p

since ∫ 2π

0ei(n−p)θ dθ =

0 if n 6= p

2π if n = p.

Repeating the same reasoning with∑

n∈Z anrneinθ, we get∫ 2π

0S(a + reiθ)e−ipθ dθ = 2πbpr

p.

Hence ap = bp for all p ∈ Z.

Proposition 3.3 Let R > 0 and f be holomorphic on z ∈ C | |z − a| > R and satisfylim|z|→∞ f(z) = 0. Then there exists a Laurent series

∑p≥1

a−p

(z−a)p which converges to f(z) on

z ∈ C | |z − a| > R.

Proof. Set u = 1z−a for |z − a| > R and g(u) = f( 1

u + a) = f(z). Then g is holomorphic onthe set u ∈ C | 0 < |u| < 1

R. Note that limu→0 g(u) = lim|z−a|→∞ f(z) = 0 (by assumption).So letting g(0) = 0, we see that g is continuous on u ∈ C | |u| < 1

R. By Corollary 2.5, g isholomorphic B(0, 1

R). Therefore, there exists a power series∑∞

p=1 bpup that converges to g(u)

on the above disk. Hence f(z) =∑∞

p=1bp

(z−a)p for all |z − a| > R.

Theorem 3.1 Let f be a holomorphic function on the annulus p2 < |z − a| < p1. Then thereexists a Laurent series

∑n∈Z an(z − a)n that converges to f(z) on the annulus.

Proof. Let z belong to the annulus A. Then there exists r1 and r2 such that p2 < r2 <|z − a| < r1 < p1. Letγ1(t) = a + r1e

it and γ2(t) = a + r2eit. Then γ1 and γ2 are homotopic in

the annulus A (for example by the homotopy H(t, s) = (1−s)γ2(t)+sγ1(t)). Then by Cauchy’sintegral formula,

12πi

∫γ1

f(u)u− z

du− 12πi

∫γ2

f(u)u− z

du = f(z)[I(γ1, z)− I(γ2, z)].

Now note that I(γ1, z) = 1 since z is inside γ1 and I(γ2, z) = 0 since z is outside γ2. Therefore,

f(z) =1

2πi

∫γ1

f(u)u− z

du− 12πi

∫γ2

f(u)u− z

du = f1(z) + f2(z).

It follows from the theorem of differentiation under the integral (Theorem 2.5), that f1 isholomorphic on C\γ∗1 and f2 is holomorphic on C\γ∗2 . In particular f1 is holomorphic onthe disk B(a, r1). Therefore f1 has a power series expansion f1(z) =

∑n≥0 an(z − a)n for

|z − a| < r1. Also f2 is holomorphic outside the closed disk B′(a, r2) with lim|z|→∞ |f2(z)| = 0.By the previous proposition, f2(z) =

∑p≥1

a−p

(z−a)p =∑

n≤−1 an(z − a)n for |z − a| > r2.Consequently,

f(z) = f1(z) + f2(z) =∑n≥0

an(z − a)n +∑

n≤−1

an(z − a)n =∑n∈Z

an(z − a)n.

3.1. LAURENT SERIES 41

Remark 3.3 This Laurent series depends a priori on r1 and r2. But according to the uniquenessof a Laurent series expansion, this series does not depend on r1 and r2.

Definition 3.2 A point a ∈ C is called an isolated singular point of a function f if f isholomorphic on the annulus 0 < |z − a| < R. The function f then has a Laurent seriesexpansion on this annulus:

f(z) =∞∑

n=1

a−n

(z − a)n+

∞∑n=0

an(z − a)n

for 0 < |z − a| < R.

Let A = n ∈ IN∗ | a−n 6= 0. There are three possibilities.

1. A = ∅.

2. A 6= ∅ and is finite.

3. A is infinite.

In the first case, f extends to a holomorphic function at a and we say that a is a removablesingularity of f .In case 2, f(z) = a−p

(z−a)p + · · ·+ a−1

z−a + a0 + a1(z − a) + · · · . We say that a is pole of order p forf .In case 3, we say that a is an essential singularity for f .

Proposition 3.4 Let f : B(a,R)\a → C be holomorphic. Then the following conditions areequivalent.

(i) a is a removable singularity for f .

(ii) f has a limit as z → a.

(iii) |f | has a limit as z → a

(iv) |f | is bounded on some annulus C = z ∈ C | 0 < |z − a| < R.

Proof. (i)⇒(ii). f extends to a holomorphic and therefore continuous function f : Ω → C.Therefore f has a limit as z → a.

(ii)⇒(iii) is clear.

(iii)⇒(iv). Since |f | has a limit as z → a, then |f | extends to a continuous function g : B(a, δ) →C. Let 0 < R < δ. The function g is continuous on the compact set B′(a,R), and therefore isbounded there.

(iv)⇒(i). There exists R > 0 such that Let f(z) =∑

n∈Z an(z − a)n for 0 < |z − a| < R. Let0 < r < R. Then, it follows from the proof of Proposition 3.2 that∫ 2π

0f(a + reiθ)e−ipθ dθ = 2πapr

p

for all p ∈ Z. Therefore |ap|rp ≤ sup0≤θ≤2π |f(a + reiθ)|.Since f is bounded on C, there exists M > 0 such that

sup0≤θ≤2π

|f(a + reiθ)| ≤ M

and therefore |ap| ≤ Mr−p. If p < 0, then −p > 0 and limr→0 r−p → 0. Thus letting r → 0, weget ap = 0. This means that A = ∅ and so a is a removable singularity for f .


Proposition 3.5 Let f : B(a,R)\a → C be holomorphic. Then, the following conditionsare equivalent.

(i) a is a pole of order p for f .

(ii) There exists δ > 0 and a holomorphic function g : B(a, δ) → C such that g(a) 6= 0 and

f(z) =g(z)

(z − a)p

for all 0 < |z − a| < δ.

(iii) limz→a |f(z)| = ∞.

Proof. (i)⇒(ii). We have

f(z) =a−p

(z − a)p+ · · ·+ a−1

z − a+ a0 + a1(z − a) + · · ·+ an(z − a)n + · · ·

for all z ∈ B(a,R)\a, with a−p 6= 0. Reducing to the same denominator, we get

f(z) =a−p + · · ·+ a−1(z − a)p−1 + a0(z − a)p + · · ·

(z − a)p=

g(z)(z − a)p

and g(a) = a−p 6= 0.

(ii)⇒(i). Let f(z) = g(z)(z−a)p for all z ∈ B(a, δ)\a, where g is holomorphic on B(a, δ) and

g(a) 6= 0. Then g(z) =∑∞

n=0 bn(z − a)n for all z ∈ B(a, δ) and therefore

f(z) =b0

(z − a)p+ · · ·+ bp−1

z − a+ b0 + bp+1(z − a) + · · ·+ an(z − a)n + · · · .

This means that a is a pole of order p for f since b0 = g(a) 6= 0.

(ii)⇒(iii). We have

limz→a

|f(z)| = limz→a

|g(z)||z − a|p

= |g(a)| limz→a

1|z − a|p

= ∞.

(iii)⇒(ii). Since limz→a |f(z)| = ∞, there exists δ ∈]0, R[ such that |f(z)| > 1 for all z ∈B(a, δ)\a. Let h(z) = 1/f(z). Then h is holomorphic on B(a, δ)\a and |h(z)| < 1 there.This means that |h| is bounded on B(a, δ)\a. By the previous proposition, h extends to aholomorphic function H : B(a, δ) → C. Moreover, limz→a H(z) = limz→a 1/f(z) = 0, so thatH(a) = 0. Therefore there exist p ≥ 1 such that the Taylor series expansion of H is

H(z) =∞∑

k=p

bk(z − a)k = (z − a)pG(z)

where bp 6= 0 and G is holomorphic on B(a, δ). Moreover, G(a) = bp 6= 0 and G does not vanishon B(a, δ). Letting g = 1/G, we see that g is holomorphic on B(a, δ) and that

f(z) =1

h(z)=

1H(z)

=1

(z − a)pG(z)=

g(z)(z − a)p

for all z ∈ B(a, δ)\a.

Proposition 3.6 Let f : B(a,R)\a → C be holomorphic. Then the following conditions areequivalent.

3.2. THE RESIDUE THEOREM 43

(i) a is an essential singularity of f .

(ii) For any ε ∈]0, R], the image of B(a, ε)\a under f is dense in C (Cassorati-Weierstrasstheorem).

(iii) |f | has no limit as z → a.

Proof. Note that the following conditions are equivalent to (ii)

1. For all ε ∈]0, R] and for all w ∈ C, there exists a sequence (zn) ⊂ B(a, ε)\a such thatf(zn) → w as n →∞. We can take zn → a.

2. For every ε ∈]0, R], every nonempty open set meets f(B(a, ε)\a).

3. For every ε ∈]0, R], for every w ∈ C and every δ > 0, the disc B(w, δ) meets f(B(a, ε)\a).

To prove that (i)⇒(ii), we reason by contradiction. Let ε ∈]0, R] be given. Assume that thereexists w ∈ C, and δ > 0 such that B(w, δ) does not meet f(B(a, ε)\a). Then for everyz ∈ B(a, ε)\a, |f(z)− w| ≥ δ. Accordingly, let for z ∈ B(a, ε)\a,

g(z) =1

f(z)− w.

Then g is holomorphic on B(a, ε)\a and bounded there. It follows from Proposition 3.4that g has a limit L ∈ C as z → a. Moreover, L 6= 0 because otherwise we would havelim |f(z) − w| = ∞ so lim |f(z)| = ∞ and this implies by the previous proposition that ais a pole of f . Setting g(a) = L, we see that g is holomorphic on B(a, ε) with g(a) 6= 0.Therefore G := 1/g is holomorphic on B(a, ε) (because g does not vanish there). In particular,lim G(z) = G(a) = 1/g(a) = 1

L . But lim G(z) = lim[f(z)−w]. Therefore lim f(z) = w + 1L . By

Proposition 3.4, this means that a is a removable singularity for f , contrary to the assumption.

(ii)⇒(iii). Let w = 1, then there exists a sequence (zn) converging to a such that f(zn) → 1and so |f(zn)| → 1. Letting, w = 2, there exists a sequence (tn) converging to a such thatf(tn) → 2 and so |f(tn)| → 2. Thus |f | has no limit as z → a.

(iii)⇒(i). By the previous two propositions, a cannot be neither a removable singularity nor apole. Therefore a is an essential singularity.

3.2 The residue theorem

Definition 3.3 If f is holomorphic on B(a,R)\a, then f is expandable in a Laurent series

f(z) =∞∑

n=1

a−n

(z − a)n+

∞∑n=0

an(z − a)n

on B(a,R)\a. The residue of f at a denoted by Res(f, a) is the coefficient a−1 in the Laurentseries.

Theorem 3.2 (Residue) Let let Ω be an open subset of C and A be a discrete and closedsubset of Ω. Let f : Ω\A → C be holomorphic. Let γ be a piecewise smooth loop not intersectingA and homotopic to a point in Ω. Then∫

γf(z) dz = 2πi

∑a∈A

I(γ, a)Res(f, a).

where I(γ, a) 6= 0 for finitely many elements a ∈ A.


Proof. Let B = a ∈ A|I(γ, a) 6= 0. Let H : [0, 1]× [0, 1] → Ω be a homotopy taking γ to apoint z0 and let K = H([0, 1]× [0, 1]). Then K is compact (as the image of a compact set by acontinuous map). We proved in an exercise of the last chapter that B ⊂ C := A ∩K and thatC is finite.

Let for c ∈ C, fc denote the singular part of f . Then fc is holomorphic on C\c. Letg = f −

∑c∈C fc. Then g is holomorphic on the set Ω\(A\C) = (Ω\A) ∪ C1. Also γ is

homotopic to z0 on Ω\(A\C) since K ⊂ Ω\(A\C). It follows from Cauchy’s theorem that∫γ g(z) dz = 0. Therefore ∫

γf(z) dz =

∑c∈C

∫γfc(z) dz.

So let us compute∫γ fc(z) dz. We know that fc(z) =

∑∞n=1

b−n

(z−c)n and the series convergesuniformly on γ (because it converges on the compact subsets of C\c). Now observe that forn 6= 1, the function z 7→ (z − c)−n has antiderivative on C\c and therefore

∫γ

dz(z−c)n = 0 for

n 6= 1 since γ is a loop. Accordingly,∫γfc(z) dz =

∞∑n=1

b−n

∫γ

dz

(z − c)n= b−1

∫γ

dz

z − c= b−12πiI(γ, c) = 2πiRes(f, c)I(γ, c).

1Note that A\C is closed in A because A is discrete; it is therefore closed in Ω and so Ω\(A\C) is open in Ωand therefore in C.

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