Proof For each b ∈ f (D), there exists a ∈ D such that b = f (a)

Open Mapping Theorem Let D ⊂ C be an open subset of C. If f : D → C is a non-constant holomorphic function in D, then f (D) is an open subset in C, i.e. a non-constant holomorphic function is an open mapping.

Proof For each b ∈ f (D), there exists a ∈ D such that b = f (a). Since f is a non-constant holomorphic function in D and by the Uniqueness Theorem, there exists r > 0 such that

f (z) 6= f (a) for all z ∈ ¯D(a; r) = {z | |z − a| ≤ r} ⊂ D.

Let

C = {z ∈ D | |z − a| = r} and 2ε = min

z∈C |f (z) − f (a)| > 0.

For each w ∈ D(b, ε), consider the holomorphic function f (z) − w defined on ¯D(a; r).

Since

|f (a) − w| = |b − w| < ε and |f (z) − f (a)| ≥ 2ε for each z ∈ C, we have

|f (z) − w| = |f (z) − f (a) − (w − f (a))| ≥ |f (z) − f (a)| − |f (a) − w|

≥ 2ε − ε = ε> |f (a) − w| for all z ∈ C.

This implies that min

z∈ ¯D(a;r)

|f (z) − w| is achieved at an interior point of ¯D(a; r), and by the Minimum Modulus Theorem

min

z∈ ¯D(a;r)

|f (z) − w| = 0 ⇐⇒ ∃ z₀ ∈ D(a, r) such that w = f (z₀) ∈ f ( ¯D(a; r)) ⊂ f (D).

Hence

D(b; ε) ⊂ f (D) ⇐⇒ b is an interior point of f (D).

This proves that f (D) is an open subset of C.

The Maximum-Modulus Theorem can also be used in conjunction with other given information about a function to obtain stronger estimates for the modulus of f in its domain of analyticity.

The following example is typical.

Schwarz Lemma Let D be the unit disk. If f : D → D is holomorphic (extending continuously to the boundary) and if f (0) = 0, then

(i) |f (z)| ≤ |z| for all z ∈ D, (ii) |f⁰(0)| ≤ 1.

and equality holds in either [(i)] or [(ii)] if and only if f (z) = e^iθz for some θ ∈ R.

Proof Let g : D → C be defined by

g(z) =





 f (z)

z = f (z) − f (0)

z − 0 if z 6= 0,

f⁰(0) if z = 0.

For each 0 < r < 1, since g is holomorhic, |f (z)| ≤ 1 for all z ∈ D, we have max

|z|≤r|g(z)|= max

|z|=r|g(z)| = max

|z|=r

|f (z)|

|z| ≤ 1

r by the Maximum Modulus Theorem.

By letting r → 1, we obtain that

|g(z)| ≤ 1 ∀ |z| ≤ 1 ⇐⇒







|f (z)|

|z| = |g(z)| ≤ 1 ∀ z 6= 0 ∈ D

|f⁰(0)| = |g(0)| ≤ 1

⇐⇒

(|f (z)| ≤ |z| ∀ z ∈ D

|f⁰(0)| ≤ 1 If the equality occurs in either case, i.e.

• either |g(z₀)| = 1 for some 0 < |z₀| < 1

• or |g(0)| = |f⁰(0)| = 1,

then g is a constant function on D with modulus equals to 1, and by the Maximum Modulus Theorem there exists a θ ∈ R such that

g(z) = e^iθ ∀z ∈ D =⇒ f(z) = e^iθz ∀z 6= 0 ∈ D =⇒ f (z) = e^iθz ∀z ∈ D since f (0) = 0.

Examples

(a) Let a, b, c, d ∈ C such that deta b c d



= ad − bc 6= 0. We define a bilinear transformation (or a M¨obius transformation) T : C ∪ {∞} → C ∪ {∞} as follows.

If c 6= 0,

T (z) =

















 az + b

cz + d if z ∈ C \ {−d c} a

c if z = ∞

∞ if z = −d

c If c = 0,

T (z) =









 az + b

d if z ∈ C

∞ if z = ∞

When c 6= 0, since

T (z) = az + b cz + d = 1

c



a − ad − bc cz + d



= T₃◦ T₂◦ T₁(z), where

T₁(z) = cz + d a dilation and a translation T₂(z) = 1

z an inversion T₃(z) = a

c − ad − bc

c
z a dilation and a translation maps {circles and lines} to {circles and lines}, so we have

• T : {circles and lines} → {circles and lines}

• T is bijective and invertible such that T (z) = az + b

cz + d ⇐⇒ T⁻¹(z) = dz − b

−cz + a.

(b) For each a ∈ C such that |a| < 1, let Ba(z) : ¯D → C be the bilinear transformation defined by

Ba(z) = z − a

1 − ¯az for |z| ≤ 1.

This implies that B0(z) = z, and for 0 < |a| < 1, we have

• B_a(z) is holomorphic throughout |z| ≤ 1 since 1

|¯a| = 1

|a| > 1,

• Ba(z) maps the unit circle |z| = 1 onto |z| = 1 since

|B_a(z)|² = z − a 1 − ¯az

¯ z − ¯a

1 − a¯z = |z|²− a¯z − ¯az + |a|²

1 − a¯z − ¯az + |a|²|z|² = 1 for |z| = 1, this implies that B_a(z) : ¯D →D, i.e. |B¯ a(z)| ≤ 1 for all z ∈ ¯D.

• B_a(a) = 0, B_a⁰(a) = 1

1 − |a|² and B_a⁻¹(z) = z + a

1 + ¯az = B−a(z).

Corollary (Generalizied Schwarz Lemma) Let D be the unit disk. If f : D → D is holo- morphic (extending continuously to the boundary) and f (a) = 0 for some |a| < 1, then

(i) |f (z)| ≤ |B_a(z)| for all z ∈ D, (ii) |f⁰(a)| ≤ |B_a⁰(a)| = 1

1 − |a|².

and equality holds in either [(i)] or [(ii)] if and only if f (z) = e^iθB_a(z) for some θ ∈ R.

Proof Let g : D → C be defined by

g(z) =













 f (z)

Ba(z) = f (z) − f (a)

Ba(z) − Ba(a) if z 6= a, f⁰(a)

B_a⁰(a) = f⁰(a)(1 − |a|²) if z = a.

For each 0 < r < 1, since g is holomorhic, |f (z)| ≤ 1 for all z ∈ D and since |Ba(z)| = 1 for all

|z| = 1, we have, by the Maximum Modulus Theorem, max|z|≤r|g(z)| = max

z=r |g(z)| = max

|z|=r

|f (z)|

|B_a(z)| ≤ max

|z|=r

1

|B_a(z)|. By letting r → 1, we obtain that

|g(z)| ≤ lim

r→1max

|z|=r

1

|B_a(z)| = 1 ∀ |z| ≤ 1 which implies that







|f (z)| ≤ |B_a(z)| ∀ z ∈ D

|f⁰(a)| ≤ |B⁰_a(a)| = 1 1 − |a|² If the equality occurs in either case, i.e.

• either |g(z₀)| = 1 for some z₀ ∈ D, z0 6= a,

• or |g(a)| = |f⁰(a)| = 1 1 − |a|²,

then g is a constant function on D with modulus equals to 1, and there exists a θ ∈ R such that g(z) = e^iθ ∀z ∈ D =⇒ f(z) = e^iθBa(z) ∀z 6= 0 ∈ D =⇒ f (z) = e^iθBa(z) ∀z ∈ D since f (a) = 0.

Example Let H = {f | f : D → D is anaytic on ¯D}. Find max

f ∈H |f⁰(¹₃)| . Suppose f (¹₃) = 0, we have

|f⁰(¹₃)| ≤ |B_1/3⁰ (¹₃)| by the Generalized Schwarz Lemma.

Suppose f (¹₃) 6= 0, we consider the map g defined by

g(z) = f (z) − f (¹₃)

1 − f (¹₃)f (z) = B_{f (1/3)}(f (z)) for |z| ≤ 1.

Since |f (z)| < 1 is holmorphic for all |z| < 1, and B_{f (1/3)}(w) is holomorphic with |B_{f (1/3)}(w)| < 1 for all |w| < 1,

g(z) = B_{f (1/3)}(f (z)) : D → D is holomorphic on D satisfying g(1/3) = 0.

Thus, by the above case, we have

|g⁰(¹₃)| ≤ |B_1/3⁰ (¹₃)|.

Since

g⁰(1/3) = f⁰(¹₃) 1 −

f (¹₃) 

2 and 0 < 1 − f (¹₃)

2 < 1 this implies that

f⁰(¹₃)

< |g⁰(¹₃)| ≤ |B⁰_1/3(¹₃)| =⇒ max

f ∈H |f⁰(¹₃)| = |B_1/3⁰ (¹₃)|.

Example If f is entire satisfying

|f (z)| ≤ 1

|Im z| ∀ z ∈ C \ R, then f ≡ 0.

Proof _y

x

0 R

z

θ

For any R > 0 and for each z ∈ C \ R satisfying that |z| = R, note that if Re z ≥ 0 =⇒ |(z − R)f (z)| ≤ |z − R|

|Im z| = sec θ ≤√

2 for some θ ∈ [0,π 4], Re z ≤ 0 =⇒ |(z + R)f (z)| ≤ |z + R|

|Im z| = sec θ ≤√

2 for some θ ∈ [0,π 4].

Thus the entire function g defined by g(z) = (z²− R²)f (z) satisfies that

|g(z)| = |z + R| |z − R| |f (z)| ≤ 3R ∀ z ∈ C with |z| = R.

By the Maximum-Modulus Theorem,

|g(z)| = |z²− R²| |f (z)| ≤ 3R ∀ |z| ≤ R =⇒ |f (z)| ≤ 3R

|z²− R²| ∀ |z| ≤ R, ∀R > 0.

By letting R → ∞, we obtain that f (z) = 0 for each (fixed) z ∈ C.

Morera’s Theorem If f is continuous on a region D and Z

Γ

f (z)dz = 0 whenever Γ is the boundary of a closed triangle ¯T in D, then f is holomorphic in D.

Proof

For each z₀ ∈ D, there exists a r > 0 such that the disc D_r(z₀) ⊂ D. Since Z

Γ

f (z)dz = 0 whenever Γ is the boundary of a closed triangle ¯T in Dr(z0), we can define

F (z) = Z z

z0

f (ζ)dζ

where the path of integration is the horizontal followed by the vertical segments from z₀ to z.

Since

h→0lim

F (z + h) − F (z)

h = lim

h→0

1 h

Z z+h z

f (ζ)dζ = f (z) ∀ z ∈ Dr(z0)

F is analytic on D_r(z₀). Since analytic functions are infinitely differentiable and F⁰(z) = f (z), f is analytic at z₀. Finally, since z₀ is arbitrary, f is analytic on D.

Corollary If f is continuous on D and f is holomorphic on D except at α ∈ D, then f is holomorphic on D.

Proof Let Γ = ∂T. Note that if α /∈ T, then Z

Γ

f (z) dz = 0.

If α ∈ T, we split the contour so that α ∈ T_{`</sub> and assume that the boundary Γ<sub>`} = ∂T satisfies that lim

`→∞|Γ<sub>`</sub>| = 0, i.e. the perimeter of Γ<sub>`</sub> decreases to 0 as ` → ∞. Then 0 = lim

`→∞

Z

Γ`

f (z) dz = Z

Γ

f (z) dz by continuity of f.

Proposition Suppose {f_n} is holomorphic in an open set D, and f_n→ f uniformly on comapct subsets of D. Then f = lim

n→∞f_n is holomorphic in D.

Proof For each z₀ ∈ D and for any D_r(z₀) ⊂ D, since D_r/2(z₀) is compact, so by assumption fn → f uniformly there. Note that fn → f uniformly on D_r/2(z0) implies f is continuous on D_r/2(z₀). Also

Z

Γ

f (z)dz = Z

Γ

n→∞lim f_n(z)dz = lim

n→∞

Z

Γ

f_n(z)dz = 0 whenever Γ is a triangle in D_r/2(z₀), f is holomorphic on D_r/2(z₀) by Morera’a Theorem. Since z₀ is arbitrary, f is holomorphic on D.

Examples (a) Let f (z) =

Z ∞ 0

e^zt

t + 1dt for z ∈ C with Re z < 0.

For each z ∈ C with Re z = x < 0, since

|f (z)| ≤ Z ∞

0

|e^zt| t + 1dt =

Z ∞ 0

e^xt t + 1dt <

Z ∞ 0

e^xtdt = −1 x,

the integral is absolutely convergent and f (z) is well defined and bounded by 1

|x|. Let Γ be the boundary of a closed triangle in D = {z ∈ C | Re z < 0}, since

Z

Γ

Z ∞ 0

|e^zt| t + 1dt dz

converges, we can interchange the order of integration; hence

Z

Γ

f (z)dz = Z

Γ

Z ∞ 0

e^zt

t + 1dt dz = Z ∞

0

Z

Γ

e^zt

t + 1dz dt = Z ∞

0

0 dt = 0 by the analyticity of e^zt

t + 1 as a function of z. By Moreras Theorem, then, f is analytic in D.

(b) Show that the zeta function ζ(z) ζ(z) =

∞

X

n=1

1

n^z is holomorphic for Re z > 1.

For each N ∈ N, R > 1 and z ∈ C with Re z ≥ R > 1, define

f_N(z) =

N

X

n=1

1 n^z. Then f_N(z) is holomorphic on Re z ≥ R. Since

|

N

X

n=1

1

n^z| = |

N

X

n=1

e^{−(log n)z}| ≤

N

X

n=1

|e^{−(log n)z}| =

N

X

n=1

|e−(log n)Rez|

≤

N

X

n=1

e^{−R(log n)} =

N

X

n=1

1 n^R ≤

∞

X

n=1

1

n^R < ∞ ∀N ∈ N, ∀ Re z ≥ R,

f_N(z) converges uniformly to ζ(z) on Re z ≥ R > 1. Hence ζ(z) converges for Re z > 1 and ζ(z) is holomorphic by Morera’s Theorem.

(c) Show that the Gamma function Γ(z) Γ(z) =

Z ∞ 0

e^−tt^z−1dt is holomorphic for Rez > 0.

To show that Γ(z) is holomorphic on Re z > 0 is equivalent to show that f (z) = Z ∞

0

e^−tt^zdt is holomorphic on Re z > −1.

For each n ∈ N and z ∈ C with Re z > −1, define f_n(z) :=

Z 1 1/n

e^−tt^zdt + Z n

1

e^−tt^zdt.

Then fn(z) is holomorphic for Re z > −1. To show that fn converges uniformly on any compact subset of Re z > −1, we observe that if K is a compact subset of Re z > −1, then there exist R₁, R₂ such that −1 < R₁ ≤ Re z ≤ R₂ for all z ∈ K.

Claim f_n(z) converges uniformly on K to f.

For t ≥ 1, since

|e^−tt^z| = |e^−te^{z log t}| = |e^−tex log t+iy log t| = |e^−te^{x log t}| = |e^−tt^{Re z}| ≤ e^−tt^R2, we have

| Z n

1

e^−tt^zdt − Z ∞

1

e^−tt^zdt| ≤ Z ∞

n

|e^−tt^z| dt ≤ Z ∞

n

e^−tt^R2dt

for all n ∈ N. For each ε > 0, since Z ∞

1

e^−tt^R2dt converges, there exists N ∈ N such that if n ≥ N, then

| Z n

1

e^−tt^zdt − Z ∞

1

e^−tt^zdt| < ε 2. For t ≤ 1, since

|e^−tt^z| = |e^−tt^{Re z}| ≤ e^−tt^R1, we have

| Z 1

1/n

e^−tt^zdt − Z 1

0

e^−tt^zdt| ≤

Z 1/n 0

|e^−tt^z| dt ≤ Z 1/n

0

e^−tt^R1dt

= Z ∞

n

e^−1/ss^−R1−2ds ≤ n^−R1−1 R₁+ 1.

For each ε > 0, since −R₁− 1 < 0 and

n→∞lim

n^−R1−1 R₁+ 1 = 0, there exists N ∈ N such that if n ≥ N, then

| Z 1

1/n

e^−tt^zdt − Z 1

0

e^−tt^zdt| < ε 2. This proves that f_n converges uniformly on K to f.

Singularities of functions

Singularities of functions will be important when trying to access their behaviors around certain points so that we can define things like Laurent series or perform residue calculus. Needless to say, an understanding of the different types of singularities and how they pertain to classes of functions is important for complex anaysis in general.

Definition Given α ∈ C, a deleted neighborhood D of α is a neighborhood not containing α, e.g. D = {z | 0 < |z − α| < ε}. f is said to have an isolated singularityat α if f is defined and holomorphic on a deleted neighborhood of α.

Types of singularities. For D a deleted neighborhood of α, we have the following types of singularities:

1. Removable singularity: ∃ g(z) on D ∪ {α} holomorphic with g(z) = f (z) on D.

2. Pole of order k: ∃ A(z), B(z) holomorphic on D ∪ {α} such that A(α) 6= 0, B(α) = 0, f (z) = A(z)

B(z) and B(z) has a zero of order k at α. Recall that we can expand B(z) into a power series as B(z) = c_k(z − α)^k+ c_k+1(z − α)^k+1+ · · · .

3. Essential singularity: Neither of the above.

The following proposition allows us to recognize if something has a removable singularity.

Proposition Suppose f (z) is holomorphic in a deleted neighborhood D of α and if

z→αlimf (z)(z − α) = 0, then f (z) has a removable singularity at α.

Proof Define h(z) as

h(z) =

(f (z)(z − α) if z 6= α,

0 if z = α.

Since h is continuous on D ∪ {α} and holomorphic on D, h(z) is holomorphic on D ∪ {α} by the corollary to Morera’s theorem.

Moreover, since h(α) = 0, the function g(z) = h(z)

z − α is holomorphic on D ∪ {α} (by using power series of h at α) and that g(z) = f (z) for all z ∈ D, f (z) has a removable singularity at α.

Likewise, the following proposition allows us to recognize a pole of f.

Proposition Suppose f (z) is holomorphic in a deleted neighborhood D of α and ∃ n ∈ N such that

z→αlimf (z)(z − α)ⁿ= 0.

If k + 1 is the least such n, then f (z) has a pole of order k at α.

Proof Let

h(z) =

(f (z)(z − α)^k+1 if z 6= α,

0 if z = α.

Since h is continuous on D ∪ {α} and holomorphic on D, h(z) is holomorphic on D ∪ {α} by the corollary to Morera’s theorem.

Since h(α) = 0, the function g(z) = h(z)

z − α is holomorphic on D ∪ {α} (by using power series of h at α) and that lim

z→αg(z) = lim

z→αf (z)(z − α)^k 6= 0 (by assumption), f (z) = g(z)

(z − α)^k has a pole of order k at α.

Examples

• f (z) = sin z

z =

∞

X

k=0

(−1)^kz^2k

(2k + 1)! has a removable singularity at z = 0.

• f (z) = sin z

(z − 1)² has a pole of order 2 at z = 1.

• g(z) = 1

sin z has countably infinite poles of order 1 (simple poles) at z = nπ.

For essential singularities, we’ve shown that it must be the case that lim

z→αf (z)(z − α)ⁿ does not exist for all n ∈ N. Notice as well the following:

• If f is bounded in a deleted neighborhood, then f has a removable singularity.

• If |f (z)| ≤

N

X

j=0

|c_j|

|x − α|^j in a neighborhood of α, then f has a pole of order ≤ N.

We have a theorem for essential singularities, which tells us that the image of any deleted neighborhood of an essential singularity under a holomorphic function is necessarily dense in the complex plane:

Casorati-Weierstrass Theorem If f has an essential singularity at α and if D is a deleted neighborhood, then f (D) = {f (z) | z ∈ D} is dense in C, i.e. for any ε > 0 and for any w ∈ C,

B_ε(w) ∩ f (D) 6= ∅.

Proof Suppose not. Then ∃B_ε(w) with B_ε(w) ∩ f (D) = ∅. This means that

|f (z) − w| > ε ⇐⇒ 1

|f (z) − w| < 1

ε ∀ z ∈ D.

Thus 1

f (z) − w is bounded holomorphic on a deleted neighborhood D of α which implies that α is a removable singularity of 1

f (z) − w and the function g defined by g(z) = 1

f (z) − w is holomorphic on the neighborhood D ∪ {α}.

Since f (z) = 1 + wg(z)

g(z) is a ratio of two holomorphic functions, by using the Taylor series of g(z) at α, we see that α is

• either a removable singularity of f (z) when g(α) 6= 0,

• or a pole of order k of f (z) when g^(k)(α) 6= 0 and g^(j)(α) = 0 for each 0 ≤ j ≤ k − 1.

Example Consider f (z) = e^1/z for z ∈ C^∗ = C \ {0}, which has an essential singularity at z = 0 ∈ C. Show that f (Bε(0) \ {0}) = C^∗.

−π π 3π

ε 1/ε 0

Proof Note that f (z) is the composition of 1

z and then e^z. Under 1

z, the set B_ε(0) \ {0} gets inverted outside the ball B_1/ε(0).

For each n ∈ Z, let

H_n= {z = x + iy | x ∈ R, π + 2(n − 1)π ≤ y < π + 2nπ} the horizontal strip in C.

Since e^z is 2πi periodic, so

e^z : H_n→ C \ {0} is an onto map for all n ∈ Z.

By choosing n so that H_n lies outside B_1/ε(0), we have shown that f (B_ε(0) \ {0}) = C^∗.

Laurent expansions We say

∞

X

k=−∞

µk= L if both

∞

X

k=0

µk and

∞

X

k=1

µ−k converge and if the sum of their sums is L.

Definition A Laurent expansions of a function f (z) about an isolated singularity α is a series of the form

f (z) =

∞

X

k=−∞

c_k(z − α)^k.

Proposition If

1 lim

k→∞|c_k|^1/k ≥ R₂ and lim

k→∞|c−k|^1/k ≤ R₁, then f (z) =

∞

X

k=−∞

c_kz^k converges for all z ∈ A(R₁, R₂).

Proof Write

f (z) =

∞

X

k=0

c_kz^k+

∞

X

k=1

c_−k 1 z

k

. Note that

• the first sum converges and is holormorphic for |z| < 1 lim

k→∞|c_k|^1/k,

• the second sum converges and is holomorphic for 1

|z| < 1 lim

k→∞|c_−k|^1/k =⇒ |z| > lim

k→∞|c−k|^1/k,

• the second sum is holomorphic for |z| > R₁ since it is a compsition of holomorphic functions of 1/z for z 6= 0 and g(w) =

∞

X

k=1

c−kw^k for |w| < 1 R₁. So f is holomorphic on z ∈ A(R₁, R₂).

Theorem For 0 ≤ R₁ < R₂ ≤ ∞, let A(R₁, R₂) be the annulus defined by A(R₁, R₂) := {z ∈ C | R1 < |z| < R₂}.

If f (z) is holomorphic on A(R₁, R₂), then f has a Laurent expansion f (z) =

∞

X

k=−∞

c_kz^k, where c_k = 1 2πi

Z

C(0;R)

f (w)

w^k+1 dw for k ∈ Z, converges for all z ∈ A(R₁, R₂).

z

R1 rR R2

Proof For each z ∈ A(R₁, R₂), consider the function g : A(R₁, R₂) → C defined by

g(w) =







f (w) − f (z)

w − z if w 6= z, f⁰(z) if w = z.

Since

• g is continuous on A(R₁, R₂) and

• Z

Γ

g(w) dw = 0 whenever Γ is the boundary of a closed triangle ¯T in D, there exists a holormorphic function G(w) (an antiderivative of g) such that G⁰(w) = g(w) for each w ∈ A(R₁, R₂) =⇒

Z

C

g(w) dw = 0 for any closed curve C ∈ A(R₁, R₂).

For each z ∈ A(R₁, R₂) and for any r, R such that R₁ < r < |z| < R < R₂, let

C(0; R) = {w ∈ A(R₁, R₂) | |w| = R} and C(0; r) = {w ∈ A(R₁, R₂) | |w| = r}

be two positively oriented concentric circles in A(R₁, R₂). Since

• Z

C(0;R)S

−C(0;r)
g(w) dw = 0,

• if w ∈ C(0; R), 0 < r < |z| < |w| =⇒ |z|

|w| < 1, then

∞

X

k=0

z^k

w^k+1 = 1

w · 1

1 − (z/w)= 1 w − z converges uniformly on C(0; R),

• if w ∈ C(0; r), 0 < |w| < |z| =⇒ |w|

|z| < 1, then

−

−∞

X

k=−1

z^k w^k+1 =−

∞

X

k=0

w^k

z^k+1 = −1

z · 1

1 − (w/z)= −1 z − w converges uniformly on C(0; r),

we have

f (z) = 1 2πi

Z

C(0;R)

f (w)

w − zdw − 1 2πi

Z

C(0;r)

f (w) w − zdw

= 1

2πi Z

C(0;R)

f (w)

∞

X

k=0

z^k

w^k+1 dw + 1 2πi

Z

C(0;r)

f (w)

−∞

X

k=−1

z^k w^k+1 dw

=

∞

X

k=0

 1 2πi

Z

C(0;R)

f (w) w^k+1 dw

 z^k+

−∞

X

k=−1

 1 2πi

Z

C(0;r)

f (w) w^k+1 dw

 z^k

converges for all z ∈ A(R₁, R₂).

Examples

(a) The following Laurent expansion converges on A(0, ∞).

e^1/z =

∞

X

k=0

1

k! z^k = 1 + 1 z + 1

2 z² + 1

3! z³ + · · · . (b) Near z = 0 (pole of order 1), we have

1

z(1 + z)² = 1 z

1

(1 + z)² = 1

z 1 − 2z + 3z²− 4z³+ · · ·
because ∂

∂z 1

1 + z
= − 1

(1 + z)² and 1

1 + z = 1 − z + z² − z³+ z⁴− · · · . We can do the same near z = −1.

Remark If f (z) is holomorphic on the annulus R1 < |z − α| < R2, then f has a unique repre- sentation

f (z) =

∞

X

k=−∞

c_k(z − α)^k where c_k = 1 2πi

Z

C

f (z) (z − α)^k+1dz

and C = C(α; R) with R₁ < R < R₂. Hence Laurent series are unique to every function. We define the principal part of f (z) =

∞

X

k=−∞

c_k(z − α)^k near α as the sum

−1

X

k=−∞

c_k(z − α)^k, the

analytic part of f (z) =

∞

X

k=−∞

c_k(z − α)^k near α as the sum

∞

X

k=0

c_k(z − α)^k and observe that

• α is a removable singularity iff c_k = 0 for k < 0, e.g. f (z) = sin z

z =

∞

X

k=0

(−1)^kz^2k

(2k + 1)! has a removable singularity at z = 0.

• α is a pole of order n iff c_k = 0 for k < −n, and c_−n 6= 0.

• α is an essential singularity iff ck 6= 0 for infinitely many negative k.

Here is an idea: at a pole α, we can make the function f (z) still be holomorphic if we expand its range to ˆC, so that f (z) = ∞ there and holomorphically so.

Proposition If f (z) has a pole of order k at α, then 1

f (z) is holomorphic near α and has a zero of order k at α.

Proof Write

f (z) = g(z) (z − α)^k.

Then g(z) is holomorphic and g(z) 6= 0 on a neighborhood of α, so 1

f (z) = (z − α)^k g(z)

is holomorphic on a neighborhood of α and has a zero of order k at α.

Definition f is meromorphic on D if f (z) is holomorphic on D except at isolated singularities at which f haspoles.

Theorem Any proper rational function R(z) = P (z)

Q(z) = P (z)

(z − z1)^k1(z − z2)^k2· · · (z − zn)^kn,

where P and Q are polynomials with deg P < deg Q, can be expanded as the sum of polynomials in 1

z − z_k, k = 1, 2, . . . , n.

Proof Since R has a pole of order k₁ at z₁,

R(z) = P1

 1

z − z₁



+ A1(z),

where P₁

 1

z − z₁



is the principal part of R around z₁ and A₁ is the analytic part.

Furthermore

A₁(z) = R(z) − P₁

 1

z − z₁



has a removable singularity at z1 and the same principal parts as R at z2, . . . , zn. Thus if we take P₂

 1

z − z₂



to be the principal part of R around z₂ and proceed inductively, we find

A_n(z) = R(z) −

 P₁

 1

z − z₁

 + P₂

 1

z − z₂



+ · · · + P_n

 1

z − z_n



is an entire function (once it is defined “correctly” at z1, z2, . . . , zn). Furthermore An is bounded since R and all its principal parts approach 0 as z → ∞. Thus, by Liouville’s Theorem, A_n is constant; indeed A_n ≡ 0. Hence

R(z) =

 P1

 1

z − z₁

 + P2

 1

z − z₂



+ · · · + Pn

 1

z − z_n



.

Remark Let f be meromorphic on D with poles at points α₁, α₂, · · · without limit point in D, let D^∗ = D \ {α₁, α₂, · · · }, and let ˆC = C ∪ {∞} be the extended complex plane. Then we can think of meromorphic functions

f : D^∗ → C as functions

f : D → ˆC by defining

f (α_j) = ∞ for all poles p⁰_js.

Similarly, in studying meromorphic functions on C, it is also useful to consider the extension of the function themselves to ˆC. We say that z = ∞ is a pole (resp. removable or essential singularity) of f (z) if z = 0 is a pole (resp. removable or essential singularity) for the function

f (z) = f (1/z).ˆ

We then say that a meromorphic function on C is meromorphic on the extended plane, if it does not have an essential singularity at z = ∞. It turns out that meromorphic functions on ˆC can be classified. The proposition above can then be restated as: a function f being meromorphic in D really means that f : D → ˆC is holomorphic.

TheoremLet f be meromorphic in ˆC = C ∪ {∞}. Suppose that lim_z→∞f (z) = ∞. Then f (z) is a rational function, i.e. f (z) = P (z)

Q(z), where P and Q are polynomials.

Proof Since lim

z→∞f (z) = ∞ ⇐⇒ lim

z→0f (1/z) = ∞, this implies that

• f (1/z) has a pole at z = 0,

• there exists a small neighborhood B_ε(0) = {z ∈ C | |z| < ε} on which f (1/z) has no other pole ⇐⇒ f (z) has no other pole in C \ B1/ε(0).

Since ¯B1/ε(0) = {z ∈ C | |z| ≤ 1

ε} is compact and all poles are isolated, there are only finitely many poles z₁, z₂, . . . , z_n of f (z) in ¯B_1/ε(0) (and in C). Since

• for each pole z_k∈ C,

f (z) = P_k

 1

z − zk



+ G_k(z),

where P_k

 1

z − z_k



is the principal part of f around z_k and G_k is holomorphic on a neigh- borhood B_rk(z_k) of z_k.

• at z = ∞,

f 1 z



= P∞

 1 z



+ G∞(z),

where as before, G∞(z) is holomorphic in a neighborhood B_ε(0) of z = 0.

Since the function

H(z) = f (z) − P∞(z) −

n

X

k=1

Pk

 1

z − z_k



is entire and bounded, so, by the Liouville’s Theorem, H(z) is a constant and f (z) is a rational function.

Theorem If f : ˆC →C is holomorphic, then f is a rational function.ˆ