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MATH 5120 - COMPLEX ANALYSIS- LECTURE NOTES

1/20/2015

• C = a+ bi | a, b ∈ R where i =√−1. This is a eld.

Addition: (a+ bi) + (c+ di) = (a+ c) + (b+ d) i. Multiplication: (a+ bi) · (c+ di) = (ac− bd) + (ad+ bc) i.

• Remark: We identify

a+ bi←→[

a b−b a

]we can identify Q with the set of matrices with matrix addition and multiplication.

• Denitions/Facts: z = a+ bi, and we let a = Rez (real part) and b = Imz (imaginary part). z = a− bi , the conjugate of z. We can write a, b in terms of z, z and i:

a =z + z

2b =

z − z2i

.

Here is another easy formula: If z, w ∈ C then

z + w = z + w

zw = z · w.

z is real (Im(z) = 0) i z = z. z is imaginary (Rez = 0) i z = −z. The modulous (or norm, or absolute value ) of z is

|z| =√a2 + b2.

∗ It follows that |z| = |z| and zz = |z|2. The distance between z and w is dened by |z − w|. Which is the familiar Euclidean distance. Easy to verify:

|z · w| = |z| |w|∣∣∣ zw

∣∣∣ =|z||w|

if w 6= 0

|Rez| ≤ |z| ,|Imz| ≤ |z| .

Triangle Inequality: z, w ∈ C then

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

which implies

|z + w| ≤ |z|+ |w| .1

MATH 5120 - COMPLEX ANALYSIS- LECTURE NOTES 2

Proof. We have

|z + w|2 = (z + w) · (z + w)

= (z + w) · (z + w)

= zz + ww + zw + wz

≥ |z|2 + |w|2

as needed.

• We can represent z by polar coordinates (r, θ):

z = reiθ, where r = |z| , θ = argz, argument of z.

If z1 = r1eiθ1 and z2 = r2e

iθ2 then

z1z2 = r1r2ei(θ1+θ2).

Note: eiθ = cos θ + i sin θ.• C is not compact. We dene it's compactication:

C = C ∪ ∞ . Riemann sphere.

Dene:

a+∞ = ∞+ a =∞,∀a ∈ Cb · ∞ = ∞ · b =∞

a

0= ∞,∀a ∈ (C ∪ ∞) \ 0

b

∞= 0,∀b ∈ C.

See Figure 1. We use stereogrpahic projection and use

S2 =

(x1, x2, x3) | x21 + x2

2 + x23 = 1

.

Where ϕ : S2\ 0, 0, 1 → C where

ϕ(x1, x2, x3) =x1 + ix2

1− x3= z

and

|z|2 =x2

1 + x22

(1− x3)2 =

1− x23

(1− x3)2 =

1 + x3

1− x3

and

x1 =z + z

|z|2 + 1x2 =

z − z

i(|z|2 + 1

) x3 =|z|2 − 1

|z|2 + 1.

• Computations:

|z|2 + 1 =1 + x3 + 1− x3

1− x3=

2

1− x3

which implies

1− x3 =2

1 + |z|2


and implies

x3 = 1− 2

1 + |z|2=|z|2 − 1

|z|2 + 1.

• Now ϕ cna be exended to S2 → C by sending (0, 0, 1) to ∞. So they have the same topology.

Proposition. The subspace topology of C from C is the same as the metric topology on C (from |z − w|).

Proof. It suces to show that we can dene a metric d (z, w) on C and the metric d (z, w) is equivalent to|z − w| for all z, w ∈ C in the following sense: Given any real number c > 0, there exist α, β > 0 such thatif |z| ≤ c, |w| ≤ c then

α |z − w| ≤ d (z, w) ≤ β |z − w| .For any z, w ∈ C, then ϕ−1(z)− ϕ−1(w) ∈ R3 so dene

d (z, w) =∣∣ϕ−1(z)− ϕ−1(w)

∣∣ ,where |·| = |·|R3 and say ϕ−1(z) = (x1, x2, x3) and ϕ−1(w) = (y1, y2, y3). Then

d (z, w)2

= |(x1, x2, x3)− (y1, y2, y3)|2

= |(x1, x2, x3)|2 + |(y1, y2, y3)|2

−2 (x1y1 + x2y2 + x3y3) , dot product

= 2− 2 (x1y1 + x2y2 + x3y3) , norm of 1

= 2− 2

(1 + |z|2

)(1 + |w|2

)− 2 |z − w|2(

1 + |z|2)(

1 + |w|2)

=4 |z − w|2(

1 + |z|2)(

1 + |w|2) .

Thusd (z, w) ≤ 2 |z − w| ,∀z, w ∈ C.

If |z| , |w| ≤ c, then2

1 + c2|z − w| ≤ d (z, w) .

• Now f : C→ C dened by w = f(z) where z, w ∈ C.• Dene

limz→a

f(z) = A where a,A,∈ C.

For any ε > 0, there exists δ > 0 such that 0 < |z − a| < δ implies |f(z)−A| < ε.• Remark:

(1) The denition canbe extended to the case a or A is ∞. (2) In the case of R, there are +∞ and −∞, but in C, we only have ∞.

• Note thatlimz→a

f(z) = A is equivalent to limz→a

f(z) = A

andis also equivalent to lim

z→aRef(z) = ReA and lim

z→aImf(z) = ImA.


• Also f is continuous at a ilimz→a

f(z) = f(a).

• We say f is dierentiable at z0 if

f ′(z0 = limh→0

f (z0 + h)− f (z0)

hexists.

(Note that h ∈ C.)• We say f is holomorphic (or analytic) at z0 if there exists a neighborhood V of z0 such that f isdierentiable at every point of V .

• We say f is holomorphic (or analytic) in a domain (connected open set in C) if f is holomorphicat every point of the domain.

Example: f(z) = (Rez)2is dierentiable at z = 0 but not holomorphic at z = 0.

Let's compute the derivative ∣∣∣∣f(h)− f(0)

h

∣∣∣∣ =|Reh|2

|h|≤ |h|

2

|h|= |h| → 0

as |h| → 0. This shows f ′(0) = 0. Let z0 6= 0. Write z0 = x0 + iy0. Let h ∈ R. Then

f (z0 + h)− f(z0)

h=

(x0 + h)2 − x2

0

h= 2x0 + h→ 2x0

as h→ 0. Let h ∈ R thenf (z0 + ih)− f(z0)

ih=x2

0 − x20

ih0→ 0,

so we have dierent limits. This means f ′(z0) doesn't exist.

1/22/2015

Denition. f : C→ C f is dierentiable at z0 if

limh→0

f (z0 + h)− f (z0)

h= f ′(z0) (h ∈ C)

exists. f is holomorphic at z0 if ∃Ω s.t. f is dierentiable at every point of Ω.

Remark. In fact, any real-valued function of a complex variable has the derivative zero, or the derivativedoes not exist.

Note that if h ∈ R thenf (z0 + h)− f (z0)

h∈ R

whilef (z0 + ih)− f (z0)

ihis purely imaginary.

So it must be zero.

Theorem. Let f(z) = u(z) + iv(z) be a function dened in a domain Ω ⊂ C where u and v are real-valued.Then f is dierentiable at z0 ∈ Ω i u, v are dienretiable at z0, when viewed as function of x and y. i.e.u(z) = u(x, y) and they satisfy the Cauchy-Riemann equations,

ux = vy

uy = −vx.

Remark. The Cauchy-Riemann equations alone is not sucient. Say f(z) = f(x+ iy) =√|xy| ???


Proof. ( =⇒ ) Assume f is dierentiable at z0. Then

f ′(z0) = A+Bi, A,B ∈ R.

Write z0 = x0 + iy0and∣∣∣∣f(z0 + h)− f(z0)

h− f ′(z0)

∣∣∣∣ =

∣∣∣∣u (x0 + h, y0) + iv(x0 + h, y0)− (u(x0, y0) + iv(x0, y0))

h− (A+Bi)

∣∣∣∣→ 0

implies that ∣∣∣∣u (x0 + h, y0)− u(x0, y0)

h−A

∣∣∣∣→ 0 and

∣∣∣∣v(x0 + h, y0)− v(x0, y0)

h−B

∣∣∣∣→ 0

which implies that

ux = A, vx = B.

Now do the same thing but replace h with ih: Write∣∣∣∣f(z0 + ih)− f(z0)

ih− f ′(z0)

∣∣∣∣ =

∣∣∣∣u (x0, y0 + h) + iv(x0, y0 + h)− (u(x0, y0) + iv(x0, y0))

ih− (A+Bi)

∣∣∣∣→ 0

which implies that

vy = A and − uy = B.

Remark:

f ′(z) = ux + ivx = vy − iuy.(⇐=) Assume u, v diernetiable as functions of x, y and satisfy the Cauchy-Riemann equations. Let

z0 = x0 + iy0. Denote

A = ux(z0) = vy(z0)

B = uy(z0) = −vx(z0)

and the same computation is similar except its in the opposite direction. Compute:

lim|h+ik|→0

∣∣∣∣u (x0 + h, y0 + k) + iv(x0 + h, y0 + k)− (u(x0, y0) + iv(x0, y0))

h+ ik− (A+Bi)

∣∣∣∣= lim|h+ik|→0

∣∣∣∣u (x0 + h, y0 + k) + iv(x0 + h, y0 + k)− (u(x0, y0) + iv(x0, y0))

h+ ik− Ah−Bk + (Ak +Bh) i

h+ ik

∣∣∣∣≤ lim|h+ik|→0

|u (x0 + h, y0 + k)− u(x0, y0)− (Ah−Bk)|√h2 + k2

+ lim|h+ik|→0

|v (x0 + h, y0 + k)− v(x0, y0)− (Ak +Bh)|√h2 + k2

→ 0.

Corollary. f(z) = u(z) + iv(z) is Holomorphic in a domain i u, v re dierentiable in Ω and satisfy theCauchy-Riemann equations.

• We have f ′′(z) = (f ′(z))′similarly f (k) is dened analogously.

Theorem. (2) If f is holomorphic in a domain Ω , then f (k) exists for all k ∈ N. Assume the theorem.

Corollary. f is holomorphic in a domain Ω i u = Ref and v = Imf are both C1 and satisfy the C-Requations.


Proof. If f is holomorphic, then write

f ′(z) = ux + ivx = vy − iuy.By Theorem 2, f ′′(z) exists so tht

f ′′(z) = uxx + ivxx = vxy − iuxy= uyx − ivyx = −uyy − ivyy.

Which implies u, v ∈ C1 as needed.

Corollary. f = u+ iv holomorphic in Ω. Then u, v are harmonic i.e.

∆u := uxx + uyy = 0

and∆y := vxx + vyy = 0.

Proof. By Cauchy Riemann, we have that ux = vy and uy = −vx. Which implies that

uxx = vyy = vxy = −uyybecause u, v ∈ C2 by theorem 2.

Denition. If u, v are harmonic and satisfy the C-R equations

ux = vy

uy = −vx,then we call v is the conjugate harmonic of u.

(−u is the conjugate harmonic of v)

Corollary. Suppose Ω is simply-connected. (π1(Ω) = 0 e.g. a unit disk). Let u be harmonic in Ω. Thenthere exists a conjuagte harmonic v in Ω. Consequently, f = u+ iv is holomorphic in Ω.

Proof. Recall in Calculus: F = (P,Q) is conservative vector eld if F = ∆f . A theorem says if Py = Qx ina simply connected domain,then F = ∇f . Dene F = (−uy, ux) then

(−uy)y = (ux)x .

But since Ω is simply connected =⇒ F = ∇V = (vx, vy). Which implies u, v satisfy the Cauchy Riemannequations.

• Now z = x+ iy and z = x− iy so that

x =z + z

2and y =

z − z2i

,−iz − z2

.

• We compute∂f

∂z=∂f

∂x

∂x

∂z+∂f

∂y

∂y

∂z=

1

2

(∂f

∂x− i∂f

∂y

)and

∂f

∂z=∂f

∂x

∂x

∂z+∂f

∂y

∂y

∂z=

1

2

(∂f

∂x+ i

∂f

∂y

)with

∂f

∂z=

1

2(ux + ivx + i (uy + ivy))

=1

2((uy − vy) + i (vx + uy)) ,


So that

C-R equations ⇐⇒ ∂f

∂z= 0

f is holomorphic implies f is independent of z.

Denition. f ∈ Ck (Ω) if u, v ∈ Ck (Ω) where u = Ref and v = Imf .

Corollary. f is holomorphic in Ω i f ∈ C1 (Ω) and ∂f∂z = 0 in Ω.

Example. Take

f(z) = (Rez)2

=(z + z)

2

4we have that

∂f

∂z=

(z + z)

2We have f is not holomorphic.

Example. f(z) = log |z| is not Holomorphic: Compute

f(z) = log |z|

=1

2log |z|2

=1

2log (zz) .

In C\ 0 we have∂f

∂z=

z

2zz=

1

2z.

Denition. f is harmonic if

∂2f

∂z∂z= 0

(∂2f

∂z∂z=

1

4(fxx + ifyy)no i?

).

Remark. A holomorphic function is harmonic.

∂2f

∂z∂z= 0

in C\ 0.

1/29/2015

Denition. A complex-valued function f is holomorphic in a domain Ω if f ∈ C1 (Ω) and

∂f

∂z= 0.

• Sequence/Series of complex numbers• Sequence/Series of complex valued function.• power series:

∑∞n=1 anz

n

• Polylonmial: P (z) = anzn + · · ·+ a1z.

• Rational Functions: R(z) = P (z)

Q(z) where P,Q are polynomials. is holomoprihc on the domain z ∈ C | Q(z) 6= 0.


• an sequence of complex numbers is said to have a limt A ∈ C if

limn→∞

an = A.

Given ε > 0, there exists N ∈ N such that if n ≥ N then

|an −A| < ε.

• Cauchy criterion an coverges if and only if given ε > 0 there exists N ∈ N such that if n,m ≥ N then

|an − am| < ε.

Proof. Assume limn→∞ = A then

|an − am| ≤ |an − L|+ |L− am| < 2ε

for large enough n and m.Now assume an = βn + γni where βn, γn ∈ R. Then

|an − am| = |βn − βm + (γn − γm) i|

=

√(βn − βm)

2+ (γn − γm)

2< ε

which implies βn and γn are Cauchy sequewnces (of real nuimbers).This implies that ,βn → B and γn → C, so that an → B + iC.

• Innite Series:∑∞n=1 an where partial sum is Sn =

∑nk=1 an.

Denition.∑an is said to be convergent if the sequence Sn converges.

Proposition. A series∑an converges i given ε > 0 there esists N ∈ N such that if n ≥ N and p ≥ 0 then

|an + an+1 + · · ·+ an+p| < ε.

Corollary. If a series∑an convefrges, then limn→∞ an = 0.

Denition.∑an is absolutely covergent if

∑|an| converges.

Corollary. If a series∑an converges absolutely, thenit converges.

Proof. Triangle inequality:|an + · · · an+p| ≤ |an|+ · · ·+ |an+p| .

• fn a sequence of complex-valued functions dened on E ⊂ C.• fn converges (pointwise) to a function f on E, if limn→∞ fn(z) = f(z) for every z ∈ E.• fn converges uniformly to f . if given ε > 0, there exists N ∈ N such that if n ≥ N , then

‖fn − f‖E = supz∈E|fn(z)− f(z)| < ε.

Proposition. If fn is a sequene of continus functions on E ⊂ C and dn convefrges uniformly to fon E,then f is continuous.

Proof. Take z, w ∈ C then

|f(z)− f(w)| ≤ |f(z)− fn(z)|+ |fn(z)− fn(w)|+ |fn(w)− f(w)|≤ 3ε.


• Cauchy Criterion for sequence of functions.

Proposition. fn uniformly converges on E i given ε > 0 there exists N ∈ N such that

‖fn − fm‖E < ε.

Proof. Assume fn uniformly converges. Then use triangle inequality like before.Assume given ε < 0,∃N s.t n,m ≥ N implies ‖fn − fm‖E < ε. Fice ω ∈ E, we have fn(ω)→ f(ω). Given

ε > 0,∃N1 = N1(ω) ≥ N such that

|fN1(ω)− f(ω)| < ε.

Fo any n ≥ N we have

|fn(ω)− f(ω)| ≤ |fn(ω)− fN1(ω)|+ |fN1

(ω)− f(ω)|≤ 2ε.

which implies

‖fn − f‖E ≤ 2ε.

Corollary. fn a sequence of functions. an convergent sequenve of numbers. If ‖fn − fm‖E ≤ |an − am|then fn uniformly converges on E.

•∑fn is a innite series of complex-valued functions.

•∑fn converges if the sequence of partial sums Sn =

∑nk=1 fk converges.

•∑fn converges uniformly if the sequence of partial sums Sn converges uniformly on E.

Corollary. (Weirstraus M-test)∑fn(z) on E and

∑an convergent series of real numbers. If there exists

M > 0 such that ‖fn‖E ≤M |an| for n sucienly large, then∑n fn converges uniformly.

Proof. By Cauchy criterion

‖Sn+p − Sn+l‖E = ‖fn + · · ·+ fn+p‖E≤ ‖fn‖E + · · ·+ ‖fn+p‖E≤ M (an + · · ·+ an+p)

≤ Mε.

Power Series:

∞∑n=0

anzn(z0 := 1 by converntion

).

where an ∈ C.• Example: (geometric series)

∑∞n=0 z

n and∑nj=0 z

j = 1−zn+1

1−z .

Diverges when |z| ≥ 1 because limn→∞∣∣zj∣∣ 6= 0.

Converges when |z| < 1 , in fact∞∑n=0

zn =1

1− z.


See that ∣∣∣∣∣∣n∑j=0

zj − 1

1− z

∣∣∣∣∣∣ =

∣∣zn+1∣∣

|1− z|→ 0.

Theorem. (Abel) For every power series∑n anz

n, there exists a number R with 0 ≤ R ≤ ∞, given byHadamard's formula:

R =1

lim supn→∞n√|an|

satisfying.(1) The series converges absolute for all |x| < R. For any 0 < ρ < R, the series converges uniformly on

|z| ≤ ρ.(2) For any |z| > R, the series diverges. (In fact, anzn is unbounded).(3) On |z| < R the series is holomorphic. The derivative is obtained by term-wise derivative and the

derivaed series has the same radus of convegrence.

• Limsup?• sn a sequence of real numbers.• Mk = sup sk, sk+1, . . . = supn≥k sn. Mk is a decreaing sequence and bounde below. This meansthat limk→∞Mk exists.

• Hence dened lim supn→∞ sn := limk→∞Mk.


2/3/2015

• Today 1. Holomorphic functions 2. Series. 3. Exp and Trig functions

Theorem. Let Ω ⊂ C be a domain.f is holomorphic in Ω if and only if f ∈ C1 (Ω) and ∂f∂z = 0 in Ω.

• Let f = u+ iv where u, v are real-valued functions. Then f is holomorphic in Ω =⇒ ∆u = 0 and∆v = 0 in Ω.

• Example: Show that

u(x, y) = log

√x2 + (y + 1)

2√x2 + (y − 1)

2

is harmonic in the domain Ω =

(x, y) ∈ R2 | y > 0, x2 + (y − 1)2> 1.

Proof. Let z = x+ iy. Then

|z ± i|2 = x2 + (y ± 1)2.

So

u =1

2log|z + i|2

|z − i|2

=1

2log |z + i|2 − 1

2log |z − i|2 .

Note that

∆u = uxx + uyy = 4∂2u

∂z∂z.

Since ∂∂z = 1

2

(∂∂x − i

∂∂y

)and ∂

∂z = 12

(∂∂x + i ∂∂y

). Then

∂2

∂z∂z

(log |z + i|2

)=

∂

∂z

(1

|z + i|2· (z + i)

)Because

∂

∂z

(|z|2)

=∂

∂z(zz) = z.

=∂

∂z

(1

z + i

)= 0

Then

|z + i|2 = (z + i) (z + i) .

Hence ∆u = 0.

• Note that ∂2

∂z∂z log |z|2 = 0 for any z 6= 0.

Theorem. (Abel) Given series∑∞n=0 anz

n there exists a number

R =1

lim supk→∞k√|ak|∈ [0,+∞]

called the radius of convergence satises the following


(1) The series converges absolutely absolutely for all |z| < R. For all ρ ∈ [0, R] the series convergesuniformly on |z| ≤ ρ.

(2) For all |z| > R, then the term anzn are unbounded and the series is divergent.

(3) In |z| < R, the series is a holomorphic function, denoted by f (z). Then f ′(z) =∑∞n=1 nanz

n−1,where

∑nanz

n−1 has the same radius of convergence.

• Recall: Let sn be bounded real sequence and β = lim supn→+∞ sn. Then i) ∀ε > 0, there exists N1 ∈ N such that sn < β + ε for all n ≥ N1. ii) For all ε > 0 and for all N2 ∈ N there exists m ≥ N2 such that sm < β − ε.

Proof. Part (1) Assume wlog R > 0. Fiz z with |z| < R. Pick ρ > 0 such that |z| < ρ < R. ( 1R < 1

ρ ) Then

we want to bound∑n |an| |z|

n. Since 1

R = lim supk→∞k√|ak| < 1

ρ .Then by property i) of the limsup we

have

|an|1n <

1

ρ∀n ≥ N1.

To show uniform convergence, we pick ρ1 ∈ (ρ,R). Then by the previous argument∑|an| |z|n ≤

∑ |z|n

ρn1

≤∑ ρn

ρn1

≤ 1

1− ρρ1

< +∞,

as needed.Part (2) Leave as an exercise.Part (3)First to show

∑nanz

n−1 has radius radius of convergence R. Note z∑∞n=1 nanz

n−1 =∑∞n=1 nanz

n, thenwe claim that this implies that ∑

nanzn−1 and

∑nanz

n.

have the same radius of convergence. This is easy to see. Now we compute

lim supn→∞

n√n |an| = lim sup

n→∞n√n n√|an|

= limn→∞

n√n lim sup

n→∞

n√|an|

=1

R.

This follows since n1n = e

1n logn → e0 = 1. Also used Fact: sntn real and lim tn = T . Then lim sup sntn =

T lim sup sn.Let

g(x) =

∞∑n=1

nanzn−1.

Want to show that f ′(z) = g(z) for all |z| < R. We just need to show that

limh→0

f(z + h)− f(z)

h= g(z)


and this would have shown that this series is holomorphic in the radius of convergence. Write

f(z) =

n∑k=0

anzn +

∞∑k=n+1

anzn = fn(z) +Rn(z),

and

g(z) = · · · = gn(z) + Rn(z).

and where

gn(z) =

n∑k=1

nanzn−1.

Now ∣∣∣∣f (z + h)− f(z)

h− g(z)

∣∣∣∣ ≤ ∣∣∣∣fn (z + h)− f(z)

h− gn(z)

∣∣∣∣+

∣∣∣∣Rn(z + h)−Rn(z)

h

∣∣∣∣+ |gn(x)− g(z)| .

The rst and last terms are clearly less than some ε. So we must estimate the middle term∣∣∣Rn(z+h)−Rn(z)

h

∣∣∣.Using the formula

Ak −Bk = (A−B)(Ak−1 + · · ·+Bk−1

),

we have that if |h| < δ , δ small enough so that |z + h| < |z|+ δ then we can estimate∣∣∣∣Rn(z + h)−Rn(z)

h

∣∣∣∣ =1

|h|

∣∣∣∣∣∞∑

k=n+1

ak (z + h)k −

∞∑k=n+1

akzk

∣∣∣∣∣≤ 1

|h|

∞∑k=n+1

|ak|∣∣∣(z + h)

k − zk∣∣∣

=1

|h|

∞∑k=n+1

|ak| |h|∣∣∣(z + h)

k−1+ · · ·+ zk−1

∣∣∣=

∞∑k=n+1

|ak| (|z|+ δ)k−1

≤∞∑

k=n+1

k (|z|+ δ)k−1

Rk

< ε

where this is nite since this is a convergent geometric series. This nishes the proof.

3. Exponential functions:Dene

ez =

∞∑k=0

zk

k!


for all z ∈ C. Note that

lim supk→∞

k

√1

k!= lim sup

k→∞exp

−1

k

k∑j=1

log j

= 0

because

limk→+∞

−1

k

k∑j=1

log j

= limk→∞

(− log k) =∞,

where we used the Fact: if sn → L then lim s1+···+snn = L. By Abel's Theorem we have that

(ez)′

=

∞∑k=0

zk

k!= ez.

Also ez is holomopthic on C.

Proposition. (Addition formula)

ea+b = eaeb,∀a, b ∈ C.

Proof. Let g(z) = ezec−z where c ∈ C. I claim: g(z) ≡ g(0) = ec. Then the result follows from letting z = aand c = a+ b. It remains to prove the claim. To show a function is cosntant we dierentiate. We have

g′(z) = ezec−z

= (ez)′ec−z + ez

(ec−z

)′= ezec−z − ezec−z

= 0.

By partials with z = 0 hence g ≡consntat.

Corollary. For all z ∈ C.(1) eze−z = 1. In particular, ez 6= 0.(2) ez = ez. Need to prove this.

This implies: (|ez| = eRez and∣∣ex+iy

∣∣ = ex and∣∣eiy∣∣ = 1).

2/5/2015

• Last time we had

ez =

∞∑n=0

zn

n!, z ∈ C

and R = +∞.• We had (ez)

′= ez.

• We had the following proposition:

Proposition. (Addition formula)

ea+b = eaeb,∀a, b ∈ C.

Proof. Recall the idea was to let g(z) = ezec−z,∀z where c ∈ C . Note g′(z) = 0. By Homeowrk 1.4(a) weget g(z) ≡ g(0) = ec. Putting c = a+ b and z = a yields the result.


Corollary. For all z ∈ C.(1) eze−z = 1. In particular, ez 6= 0.

(2) ez = ez. Consequently we get (|ez| = eRez and∣∣ex+iy

∣∣ = ex and∣∣eiy∣∣ = 1).

Proof. Only need to show that ez = ez. We have that

∣∣ez − ez∣∣ ≤∣∣∣∣∣∣ez −

N∑n=0

zn

n!

∣∣∣∣∣∣+

∣∣∣∣∣∣N∑n=0

zn

n!− ez

∣∣∣∣∣∣ ,and we want to use a+ b = a+ b for a, b ∈ C. Now rst we see that the rst term becomes∣∣∣∣∣∣ez −

N∑n=0

zn

n!

∣∣∣∣∣∣ =

∣∣∣∣∣∣ez −N∑n=0

zn

n!

∣∣∣∣∣∣=

∣∣∣∣∣ez −N∑n=0

zn

n!

∣∣∣∣∣→ 0 as N →∞.

Now since ez =∑∞n=0

zn

n! then ez =∑∞n=0

zn

n! , so that the second term∣∣∣∣∣∣N∑n=0

zn

n!− ez

∣∣∣∣∣∣ =

∣∣∣∣∣N∑n=0

zn

n!− ez

∣∣∣∣∣→ 0 as N →∞.

Thus ∣∣ez − ez∣∣ = 0

as needed.Now note

|ez|2 = ezez = ezez = ez+z = e2Re(z)

so that|ez| = eRe(z)

as needed.

Trig Functions:We dened the following:

sin z =eiz − e−iz

2i, cos z =

eiz + e−iz

2. (?)

Then by the series ez =∑∞n=0

zn

n! we have the following representations

cos z =

∞∑k=0

(−1)k z2k

(2k)!, sin z =

∞∑k=0

(−1)k z2k+1

(2k + 1)!.

By (?) we have that

sin2 z + cos2 z = 1, z ∈ Ceiz = cos z + i sin z z ∈ C.

Let y ∈ R then we have Euler's formula

eiy = cos y + i sin y. (Euler) .


Then if z = x+ iy we haveex+iy = exeiy = ex (cos y + i sin y) .

Proposition. 1. If x ∈ C satises ez = 1 then z = 2kπi for some k ∈ Z.

Proof. For x, y ∈ R, write z = x+ iy and 1 = ez = exeiy then take the norm

1 =∣∣exeiy∣∣ = |ex| = ex

which implies that x = 0. But theneiy = cos y + i sin y = 1

hence we have the following two equations cos y = 1 and sin y = 0. Solving these yields y ∈ 2πZ.

Proposition. 2. Let z = x+ iy. Then

|cos z|2 = cosh2 y − sin2 x,

|sin z|2 = sinh2 y + sin2 x.

In particular, |sin z| and |cos z| are unbounded.

• Recall that the hyperbolic functions are:

cosh y =ey + e−y

2,

sinh y =ey − e−y

2

Proof. Write |cos z|2 = cos zcos z and

cos z =eiz + e−iz

2=eiyeix + eye−ix

2and

cos z =e−ye−ix + eyeix

2,

then we leave the rest of the calculation as an exercise.

3. LogarithmsTry to dene log z on C and we'll see that this is not well dened. Let w = log z to be a root of ew = z.

Write z = |z| eiθ. Let w = x+ iy for x, y ∈ R. For ew = z we have

ew = z =⇒ exeiy = |z| eiθ

=⇒ ex = |z|=⇒ x = log |z| .

Theneiy = eiθ, ei(y−θ) = 1.

By Proposition 1, y = θ + 2kπ, ∀k ∈ Z. Solog z = log |z|+ i (θ + 2kπ) , k ∈ Z

is a multi-valued function, which is not a function. If we dene log z1 = log |z1|+ iθ1 then moving along Γ(a curve about the origin and z1 in the this curve and θ 7→ θ + 2π ) we get

log z1 = log |z1|+ i2π

which is not well dened.


But if we restrict log z to a domain G not containing the origin O. (Draw a domain G not containing theorigin) dene for all z ∈ G

log z := log |z|+ iθ where − π < θ < π

then this function is well-dened. (So a loop in this domain keeps the angle unchanged because it does notcross the origin.) We call this a branch, and the origin is called a branch point. Also we could restrictθ ∈ (π, 2π) , . . . .

Denition. A point z0 is called a branch point of a function f(z), if f(z) changes its value as one startsout at a point, traces a closed path enclosing z0 and returns to the starting point.

Example. For f(z) = log z, z0 = 0 is a branch point, any z 6= 0, z ∈ C, is not a branch point.

Denition. Let G be a domain in C. If f : G → C is a continuous function such that ef(z) = z,∀z ∈ G,then f is called a branch of log z on G.

• Let f be a branch of log z on G. Let g = f + 2kπi for some k ∈ Z. Then

eg = efe2kπi = ef = z.

Proposition. 3. Let f be a branch of log z on G. Then the set of all branches of log z on G is preciselyf(z) + 2kπi; k ∈ Z.

Proof. Let g be branch of log z on G. For each z ∈ G, eg(z) = ef(z) = z. By Proposition 1, we haveg(z) = f(z) + 2πkzi for kz ∈ Z. Want to get a k independent of z ∈ G. Let

h(z) =g(z)− f(z)

2πi.

Then h ∈ C0(G), h(G) connected. Note h(G) ⊂ Z discrete, so the image must be one point, otherwise itwouldn't be connected. Hence, h(G) = k0 for some k0 ∈ Z. i.e. g(z) = f(z) + 2k0πi.

• Question: What is the largest possible domain for a well dened log z ?• Answer: C\ z = x+ yi | z ≤ 0, y = 0.• Another way to answer the question:

log z has a branch point z = 0 in C. log z also has a branch point at z = ∞ in C. To see this, let z = 1

w , then w = 9 7→ z = ∞. sothat log z = − logw which implies that w = 0, (z =∞) is a branch point.

Let G = C\ a curve joining 0 & ∞. Then there exists a branch of log z on G. From now on, we call

log |z|+ iθ, −π < θ < π

on G1 = C\ z = x+ yi | z ≤ 0, y = 0 the principal branch of log z. Sometimes we denotelog z = log |z|+ iθ.

Remark. Given G and f(z) branch of log z such that ef(z) = z and (f ∈ C0(G)). We can show that

f ′(z) =1

zon G, (??)

hence, f is analytic. That is, (log z)′

= 1z on G1.

• (??) follows from the following.


Lemma. (IVT) Let U, V be open sets in C. Suppose that f : U → C and g : V → C are both continuous,f(U) ⊂ g(V ) and

g (f (z)) = z.

If g is dierntiable and g′(z) 6= 0 for all z ∈ V , then f is dierentiable and

f ′(z) =1

g′ (f (z)).

Apply to g = ez.

• Dene for all a, b ∈ C then following

ab = eb log a.

i) if a > 0, a ∈ R then ab is signle-valued. ii) If b ∈ Z, then ab signle-valued. (e2iπkb = 1) if b ∈ Q?

2/10/2015

• A Quick way to solve problem 1 from homework 1. Let F = g + ih. Then

f = log(h2 + g2

)= log |F |2 .

Since h 6= 0 in disk then F 6= 0. Then dene the operator

∂

∂z=

1

z

(∂

∂x− i ∂

∂y

)∂

∂z=

1

z

(∂

∂x+ i

∂

∂y

).

Then

4∂2

∂z∂z=

∂2

∂x2+∂2

∂y.


It suces to show that ∂2

∂z∂z log |F |2 = 0. But then

∂

∂zlog |F |2 =

1

|F |2∂

∂z|F |2

=1

|F |2∂

∂zF F

=1

|F |2

(F∂

∂zF + F

∂

∂zF

)=

1

|F |2

(F∂F

∂z+ F

∂F

∂z

)=

1

|F |2F · ∂F

∂z, (

∂F

∂z= 0)

=1

FFF · ∂F

∂z

=1

F

∂F

∂z.

so that

∂2

∂z∂zlog |F |2 =

∂

∂z

(1

F

∂F

∂z

)=

F ∂2F∂z∂z −

∂F∂z

∂F∂z

F 2

=F · 0− ∂F

∂z · 0F 2

= 0

since ∂F∂z = ∂F

∂z = 0 = 0 and since F is holomorphic then it is complex harmonic so that ∂2F∂z∂z = 0

as needed.• Delicate Way:

In general log |F |2 6= logF + log F . where log z is a multivalued function on C. Note that F = g + ih and h 6= 0. Then

F (|z| < 1) ⊂ C\ Rez ≤ 0, Imz = 0 = G1.

On g1 take log z the principal branch of G1. By chain rule logF is holomotphic on D. Then log |F |2 = logF + log F = 2Re(logF ) is harmonic.

Last time:

• Dene ab = eb log a. So in general this is a multivalued function. But there are cases where we havesingle valued function. 1) if a > 0 a ∈ R then log a is single-valued , so is ab. 2) If b ∈ Z, then ab is single valued. This is because the log is dierered by 2kπib , sincee2kπib = 1.

3) if b ∈ Q, and b = pq reduced form. (p, q ∈ Z) then a

b = epq log a and log a = log |a|+arg a+2kπi.

So ab is q−valued function.


Example: Suppose b = 12 . Consider z

12 = e

12 log z is a two-valued function. Take the principal branch of

z12 to be

√z = |z|

12 eiθ/2 for −π < θ < π on G1. (Another branch on G1 is − |z|

12 e−iθ/2 = −

√z)

Remark: The branch points of√z are 0,∞. Pick any curve γ joint 0 and ∞. Then

√z has a branch

on C\γ. .Remark?: Riemann's idea: Idea of 5121- Complex 2.Complex Integration:

• Our goal is to show:

Theorem. 0. (Cauchy Integral formula). Let Ω be a bounded domain in C with C1 boundary ∂Ω. Then forall holomprihc f on Ω with f ∈ C1

(Ω),z

f(z) =1

2πi

ˆ∂Ω

f(ξ)

ξ − zdξ, ∀z ∈ Ω.

• A curve γ : [a, b]→ C is a continuous function. (i.e. if γ(t) = u(t) + iv(t) then u, v ∈ C0 ([a, b])).• A curve γ is simple if γ(t1) = γ(t2) implies t1 = t2.• A curve γ is closed if γ(a) = γ(b).• The direction of γ(t) is the direction along which t is increasing.• A curve is Jordan if it is simple and closed.• A curve γ is of C1 ([a, b]) , if γ′(t) = u′(t) + iv′(t) exists and continuous on [a, b].• A curve δ is of piecewise C1 if γ ∈ C0 ([a, b]) , and γ′ exists except for t0, t1, . . . , tm ∈ [a, b]. At eachtj , γ

′+(tj) = γ′(tj+), and γ′−(tj) = γ′(tj−).

• Let f(t) = u(t) + iv(t) be a continuous function on [a, b]. Deneˆ b

a

f(t)dt =

ˆ b

a

u(t)dt+ i

ˆ b

a

v(t)dt.

• This implies ˆ b

a

f(t)dt =

ˆ b

a

f(t)dt.

• This means

c

ˆ b

a

f(t)dt =

ˆ b

a

cf(t)dt,∀c ∈ C

and ∣∣∣∣∣ˆ b

a

f(t)dt

∣∣∣∣∣ ≤ˆ b

a

|f(t)| dt.

• Let γ be a C1 curve on [a, b], and f (γ (t)) be a continuous function on [a, b]. Deneˆγ

f(z)dz =

ˆ b

a

f (z(t)) z′(t)dt

where z = z(t) = γ(t).

Example2: Let γ be the boundary of the ball BR(0) in C. Find´γf(z)dz for nay continuous function f(z).

For all z ∈ ∂BR(0) thenz = Reiθ 0 ≤ θ < 2π.

So γ has the representation v(θ) = z(θ) = Reiθ and z′(θ) = Reiθi. Henceˆγ

f(z)dz = iR

ˆ 2π

0

f(Reiθ

)eiθdθ.


Theorem. 1. (Stoke's Theorem) Let Ω be a bounded domain in C and ∂Ω consists of a nite number of C1

Jordan curves. Let ω = Pdx+Qdy and P = P (x, y), Q = Q(x, y) are in C1(Ω). Thenˆ

∂Ω

ω =

ˆΩ

dω.

Remark: For ω = Pdx+Qdy. Using (dx ∧ dx = 0) and dx ∧ dy = −dy ∧ dx then

dω = dP ∧ dx+ dQ ∧ dy= (Pxdx+ Pydy) ∧ dx+ (Qxdx+Qydy) ∧ dy= Pydy ∧ dx+Qxdx ∧ dy= (Qx − Py) dx ∧ dy.

Theorem. 2. (Complex version) Let Ω be as before. Let ω = f(z)dz, whre f ∈ C1(Ω), complex -valued.

Then ˆ∂Ω

ω =

ˆΩ

dω =

ˆΩ

∂f

∂zdz ∧ dz.

Remark: dω = d (fdz) = df ∧ dz. (recall df = fxdx+ fydy = fzdz + fzdz.)

Theorem. 3(Cauchy-Green) Let Ω be as in Thereom 2. Then ∀f ∈ C1(Ω), complex-valued. we have

f(z) =1

2πi

[ˆ∂Ω

f(ξ)

ξ − zdξ +

ˆΩ

∂f∂ξdξ ∧ dξξ − z

].

• Note that Theorem 3 implies Theorem 0, since ∂f∂ξ≡ 0.

Proof of theorem3:

Proof. Use \zeta instead of \xi. oops. Let G = Ω\Bε(z). Then apply Theorem 2 to ω = f(ζ)ζ−z dζ. Since

f(ζ)ζ−z ∈ C

1(G), we have by stokes theorem that

ˆ∂G

ω =

ˆG

dω

so ˆ∂G

ω =

ˆ∂Ω

ω −ˆ∂Bε(z)

ω

=

ˆ∂Ω

f(ζ)

ζ − zdζ −

ˆ∂Bε(z)

f(ζ)

ζ − zdζ

so that ˆG

dω =

ˆΩ\Bε(z)

d

(f

ζ − z

)∧ dζ

=

ˆΩ\Bε(z)

∂f

∂ζ

dζ ∧ ζζ − z

.

We claim(1)

´∂Bε(z)

f(ζ)ζ−z dζ → 2πif(z) as ε→ 0.

(2)´∂\Bε(z)

∂f∂ζ

dζ∧ζζ−z →

´Ω∂f∂ζ

dζ∧ζζ−z . as ε→ 0.


For (1), we put ζ = z + εeiθ for ,0 ≤ θ < 2π thenˆ∂Bε(z)

f(ζ)

ζ − zdζ = (Ex)

ˆ 2π

0

f(εeiθ

)dθi

→ 2πif(z),

as ε→ 0. For (2) we get ∣∣∣∣∣(ˆ

∂\Bε(z)−ˆ

Ω

)∂f

∂ζ

dζ ∧ ζζ − z

∣∣∣∣∣ =

∣∣∣∣∣ˆBε(z)

∂f

∂ζ

dζ ∧ ζζ − z

∣∣∣∣∣≤ ‖f‖C1

ˆBε(z)

dζ ∧ ζζ − z

but dζ ∧ dζ = −2irdr ∧ dθ so that dζ = d(reiθ

)= eiθdr + ieiθdθr and ζ = z + re%iθ so thatˆ ε

0

ˆ 2π

0

drdθ → 0


2/12/2015

• Last time:

Theorem. (Cauchy-Green-Pompei formula). Let Ω be a bounded domain in C with C1 boundary ∂Ω. Forany f ∈ C1

(Ω),

f(z) = f(z) =1

2πi

[ˆ∂Ω

f(ξ)

ξ − zdξ +

ˆΩ


], ∀z ∈ Ω.

Remark. 1.We have that idζ∧dζ2 is a volume form. Indeed, write ζ = reiθ. So

dζ = eiθdr + rieiθdθ

dζ = e−iθdr − ire−iθdθdζ ∧ dζ = −2irde ∧ dθ.

Since dr ∧ dr = dθ ∧ dθ = 0, and dr ∧ dθ = −dθ ∧ dr. Andi

2dζ ∧ dζ = rdr ∧ dθ

i

2

ˆfζ ∧ dζ =

ˆ ˆf(reiθ

)rdrdθ.

Remark. 2. From Last time we proved this by applying stokes theorem to G = Ω\Bε(z) Wait

ˆΩ\Bε(z)


→ˆ

Ω


as ε→ 0. We estimate ∣∣∣∣∣ˆ

Ω\Bε· −

ˆΩ

∣∣∣∣∣ =

∣∣∣∣∣ˆBε(z)


∣∣∣∣∣=

∣∣∣∣ˆ ∂f

∂ξ2ie−iθdrdθ

∣∣∣∣≤ ‖f‖C1

ˆ ∣∣2ieiθ∣∣ drdθ≤ 2 ‖f‖C1

ˆ ε

0

dr

ˆ 2π

0

dθ

= 2 ‖f‖C1ε2π → 0

as ε→ 0. with (ζ = z + reiθ) and dζ = d(riθ).

Corollary. 1. (Cauchy Integral formula). Let Ω be a bounded domain in C with C1 boundary ∂Ω. Then forall holomorphic f on Ω with f ∈ C1

(Ω), then

f(z) =1

2πi

ˆ∂Ω

f(ξ)

ξ − zdξ, ∀z ∈ Ω.

Corollary. 2. (Cauchy Integral Theorem). Let Ω be a bounded domain in C with C1 boundary ∂Ω. Then ifF is holomorphic on Ω and F ∈ C1

(Ω), thenˆ

∂Ω

F (z)dz = 0.


Proof. Apply Corollary 1 to f(z) = zF (z) with z = 0.

Theorem. 1. Let Ω be a bounded domain in C and ∂Ωconsists of a nite number of rectiable Jordancurves. Let f be holomorphic on Ω and f ∈ C0

(Ω). Thenˆ

∂Ω

f(z)dz = 0.

(⇐⇒ ∀z ∈ Ω f(z) = 12πi

´ f(ζ)ζ−z dζ)

Corollary 2 =⇒ Corollary 1: Apply Stoke's theorem to F (ζ) = f(ζ)ζ−z with G = Ω\Bε(z).

Denition. A curve γ : [a, b] → C is rectiable if it satises the following. Let P be a partition P =a = t0 < t1 < · · · < tn = b of [a, b]. Denote

L (γ;P ) =

n∑j=1

|γ(tj)− γ(tj−1)| .

We say γ is rectiable if supP L (γ;P ) < +∞.

Example. 1. A piecewise C1 curve is rectiable: Take γ ∈ C1 . Then

L (γ;P ) =

n∑j=1

|γ(tj)− γ(tj−1)|

=

n∑j=1

∣∣∣∣∣ˆ tj

tj−1

γ′(t)dt

∣∣∣∣∣≤

n∑j=1

ˆ tj

tj−1

|γ′(t)| dt

=

ˆ b

a

|γ′(t)| dt

< ∞.• In fact, we have that

L(γ) =

ˆ b

a

|γ′(t)| dt.

Example. 2. Let γ be

γ(t) =

0 t = 0

t+ itα sin(πt

)0 < t ≤ 1

where α > 0 is a constant. It kinda looking a shrinking topologists sine curve. It can be shown that γ isNOT rectiable. When 1 < α ≤ 2, we have that γ is NOT piecewise C1.

Denition. Let γ be a rectieable curve on [a, b] and f be continuous on γ ([a, b]). Deneˆγ

f =

ˆ b

a

f(γ(t))dγ(t) (Riemman Stieljes).

That is,

RHS = lim

m∑j=1

f(γ(cj)) [γ(tj)− γ(tj−1)]


where a = t0 < · · · < tm = b is a partition cj ∈ [tj−1, tj ].

Example. 1. Want to show that this denition is consistant with our previous denition. That is ifγ ∈ C1([a, b]) then ˆ

γ

f =

ˆ b

a

f(γ(t))γ′(t)dt.

Proof. We estimate∣∣∣∣∣ˆγ

f −ˆ b

a

f (γ(t)) γ′(t)dt

∣∣∣∣∣ < 2ε+

∣∣∣∣∣∣m∑j=1

f (γ(cj)) (γ(tj)− γ(tj−1))−m∑j=1

f (γ(dj)) γ′(dj)(tj − tj−1)

∣∣∣∣∣∣< 3ε.

Pick dj such that γ′(dj)(tj − tj−1) = γ(tj)− γ(tj−1) by MVT.

Example. 2. Compute´γ

1,´γz, with γ rectiable. Note for the rst one we compute

m∑j=1

1 (γ(tj)− γ(tj−1)) = γ(b)− γ(a),

for all partitions P . Thus ˆγ

1 = γ(b)− γ(a).

Next, we compute´γz: Consider the Riemman Stieljes sum

ˆγ

z = lim

m∑j=1

γ(cj) (γ(tj)− γ(tj−1)) .

Put cj = tj and ,cj = tj−1 respectively and getˆγ

z =1

2lim

m∑j=1

(γ(tj) + γ(tj−1)) (γ(tj)− γ(tj−1))

=1

2lim

m∑j=1

(γ(tj)

2 − γ(tj−1)2)

=1

2

(γ(b)2 − γ(a)2

).

In particular,¸γ

1 =¸γzdz = 0.

• For Cauchy Integral Theorem, ∂Ω rectieable Jordan f ∈ C0 (Ω).• Case1: Ω = D = |z| < 1. Assume f ∈ C0

(Ω)

= C0(D), f holomorphic on D. We wantˆ

∂Df(z)dz = 0.

Proof. Let Dr = |z| < r. with 0 < r < 1. By corolary 2, we haveˆ∂Drf(z)dz = 0


which implies ˆ 2π

0

f(reiθ

)eiθdθ = 0

so that ∣∣∣∣ˆ∂Df(z)dz

∣∣∣∣ =

∣∣∣∣ˆ 2π

0

f(eiθ)eiθdθ

∣∣∣∣=

∣∣∣∣ˆ 2π

0

f(eiθ)eiθdθ −

ˆ 2π

0

f(reiθ

)dθ

∣∣∣∣≤

ˆ 2π

0

∣∣f (eiθ)− f (reiθ)∣∣ dθ≤ 2πε,

for all ε > 0. So that´∂D f = 0.

• Case2: (Goursat Theorem) Let ∆ be a region of traingle. Assume ∆ ⊂ Ω. f holomorphic on ∆.We want ˆ

∂∆

f = 0.

Proof. Let I =´∂∆

f . Then

I =

ˆ∂∆1

f +

ˆ∂∆2

f +

ˆ∂∆3

f +

ˆ∂∆4

f

and

|I| ≤4∑

n=1

∣∣∣∣ˆ∂∆i

f

∣∣∣∣ .There exists one ∆i such that

14 |I| ≤

∣∣∣´∂∆if∣∣∣. Denote ∆(i) = ∆i. So you have a sequence ∆,∆(1), . . . ,∆(k), · · · .

Then1

4k|I| ≤

∣∣∣∣ˆ∂∆k

f

∣∣∣∣ ,∀k.Now note that ¯∆(k) ⊂ ¯∆(k−1), with diam(∆(k))→ 0 as k →∞. By Canto's lemma,

∞⋂k=1

∆(k) = z? ⊂ Ω.

Note that f is holomoprhic on z?. We have

f(z)− f(z?)

z − z?− f ′(z?) = R(z?, z),

where R(z?, z)→ 0 as z → z?. WE have

f(z) = f(z?) + f ′(z?) (z − z?) +R(z − z?)so that ˆ

∂∆k

f =

ˆ∂∆k

f(z?)dz +

ˆ∂∆k

f ′(z?)(z − z?)dz +

ˆ∂∆k

R(z − z?)dz,

= 0 + 0 +

ˆ∂∆k

R(z − z?)dz


wher the rst two terms are zero by Example 2. Now we must show the lasdt term is also zero. We have∣∣∣∣ˆ∂∆k

f

∣∣∣∣ ≤ max |R|ˆ∂∆k

|z − z?| |dz|

≤ max |R| diam∆

2kL(∆)

2k

which implies |I| ≤ max |R| diam∆1

L(∆)1 → 0 asw z → z?.

• 3 prelims problms, like homoeworks and lecture notes.• Show a functioin is holomophic, Use z method.

Make sure you undersand the homeowrk probelms and lecture notes. Read cross material in the books.


2/19/2015

• Last time


(Ω). Thenˆ

∂Ω

f(z)dz = 0.

(⇐⇒ ∀z ∈ Ω f(z) = 12πi

´ f(ζ)ζ−z dζ)


(Ω). Then

f (k)(z) =k!

2πi

ˆ∂Ω

f(ζ)

(ζ − z)k+1dζ

for all k ≥ 0, all z ∈ Ω.

Corollary. 2. For every domain G ⊂ C, not necessarily bounded. If f ′(z) exists for all z ∈ G, then f (k)(z)exists for z ∈ G.

Proof. For all z ∈ G there exists a ball B(z, r) ⊂ G. Then apply theorem 1 to f and Ω = B(z, r).

Proof of Theorem 1:

Proof. For k = 1, we have∣∣∣∣∣f(z + h)− f(z)

h− 1

2πi

ˆ∂Ω

f(ζ)

(ζ − z)2 dζ

∣∣∣∣∣ =

∣∣∣∣∣ 1

2πi

1

h

ˆ∂Ω

(f(ζ)

(ζ − z − h)− f(ζ)

ζ − z

)dζ − 1

2πi

ˆ∂Ω

f(ζ)

(ζ − z)2 dζ

∣∣∣∣∣=

1

2π

∣∣∣∣∣ˆ∂Ω

f(ζ)

(1

(ζ − z − h) (ζ − z)− 1

(ζ − z)2

)dζ

∣∣∣∣∣=

1

2π

∣∣∣∣ˆ∂Ω

f(ζ)

(h

(ζ − z − h) (ζ − z)

)dζ

∣∣∣∣≤ 1

2πsup∂Ω|f | · |h|

∣∣∣∣ˆ∂Ω

dζ

(ζ − z − h) (ζ − z)

∣∣∣∣ (1)

Now denote d = dist (z, ∂Ω) > 0. Then for all h such that 0 < |h| < d4 . Then

|ζ − z| ≥ dso that

|ζ − (z + h)| ≥ ||ζ − z| − |h|| ≥ 3d

4.

But then ∣∣∣∣ˆ∂Ω

dζ

(ζ − z − h) (ζ − z)

∣∣∣∣ ≤ 4

3d3

∣∣∣∣ˆ∂Ω

dζ

∣∣∣∣ ≤ 4

3d3L (∂Ω) .

Plugging this into (1) we get∣∣∣∣∣f(z + h)− f(z)

h− 1

2πi

ˆ∂Ω

f(ζ)

(ζ − z)2 dζ

∣∣∣∣∣ ≤ 1

2πsup∂Ω|f | · |h| · 4

3d3L (∂Ω) =

2

3d3πsup∂Ω|f | · |h|L (∂Ω) ,

which proves as needed.


Now for k ≥ 2, we observe that

1

h

[(k − 1)!

(ζ − z − h)k− (k − 1)!

(ζ − z)k+1

]needs to be estimated. There is a little trick (like Abel's theorem). Here it is:

(?) =1

h

[(k − 1)!

(ζ − z − h)k− (k − 1)!

(ζ − z)k+1

]=

(k − 1)!

h

(ζ − z)k−1+ · · ·+ (ζ − z − h)

k−1

(ζ − z)k (ζ − z − h)k

− k!

(ζ − z)k+1,

where we use ak − bk = (a− b)(ak−1 + · · ·+ bk−1

). Now

|(?)| = (k − 1)!

∣∣∣∣∣ (ζ − z)k−1+ · · ·+ (ζ − z − h)

k−1

(ζ − z)k (ζ − z − h)k

− k

(ζ − z)k+1

∣∣∣∣∣≤ (k − 1)!

∣∣∣∣∣∣(ζ − z)

[(ζ − z)k−1

+ · · ·+ (ζ − z − h)k−1]− k (ζ − z − h)

k

(ζ − z)k (ζ − z − h)k

∣∣∣∣∣∣≤ C(k) |h| d−k−2 → 0

as h→ 0 we're we used a estimate similar to the k = 1 case.

Corollary. 3. Let Ω be a domain in C. Suppose B(z,R) ⊂ Ω. Then∣∣∣f (k)(z)∣∣∣ ≤ k!

Rksup

∂B(z,R)

|f |

≤ k!

RksupB(z,R)

|f | .

Proof. Apply Theorem 1 to f and Ω = B(z,R). Then∣∣∣f (k)(z)∣∣∣ =

∣∣∣∣ k!

2πi

ˆ∂B

f(ζ)

(ζ − z)k+1dζ

∣∣∣∣=

k!

2π

∣∣∣∣ˆ∂B

f(ζ)

(ζ − z)k+1dζ

∣∣∣∣=

k!

2πsup∂B|f |∣∣∣∣ˆ∂B

dζ

(ζ − z)k+1

∣∣∣∣≤ k!

2π

L(∂B)

Rk+1sup∂B|f |

=k!

Rksup∂B|f | .

Corollary. 4. Let Ω be a domain in C. Let K ⊂ Ω be a compact set. Then for any neighborhood V of Kin Ω,

supz∈K

∣∣∣f (k)(z)∣∣∣ ≤ C sup

V|f | ,


whre C > 0 is a constant depending only on K, dist(K, ∂V ).

Remark. A neighborhood V of K is called rel compact if V is compact. So the corol in Corollary 4. Wehave K ⊂ V ⊂ V ⊂ Ω.

Proof 1 of Corolary4:

Proof. For all z ∈ K, pick B(z,Rz) ⊂ V . So that

K ⊂N⋃i=1

B(z,Rz).

Cor3!

Theorem. (Liouville's Theorem) Let f be a holomorphic funciton on the whole plane C. If f is bounded onC, then f ≡ constant.

Proof1:

Proof. For all z ∈ C by Coro3.(Cauchy's estimate) Then

|f ′(z)| ≤ 1

Rsup

∂B(z,R)

|f | .

Note that this holds for all R > 0. So that

sup∂B|f | ≤ sup

C|f | = M.

This means that

|f ′(z)| ≤ M

R→ 0

as R→ +∞. Hence |f ′(z)| = 0 which implies f ′(z) ≡ 0 for all z ∈ C. This shows f mus the constant.

• The key is to control |∇f | by |f |C0 .

Theorem. 3. (Morera) Let f be a continuous function on a domain Ω. If´γf = 0 for any closed rectiable

curve inside Ω, then f is holomorphic on Ω.

Proof. Fix z0 ∈ Ω. For all z ∈ Ω. Pick a rectiable curve γ1 joining z0 and z1. Dene a function

F (z) =

ˆγ1

f =

ˆ z

z0

f.

Check that´ zz0f is well-dened. Then we can show

F ′(z) = f(z)

by the homework problem. And proceed by showing that f must have a primitive F . Since the primitive Fis holomorphic then f must be holomorphic.

Theorem. 4. (Series representation) Let Ω be a domain and B(z0, R) ⊂ Ω. Let f be holomorphic on Ω.Then

f(z) =

∞∑k=0

ak (z − z0)k,∀z ∈ B(z0, R)


where the series RHS converges absolutely and uniformly on B(z0, R) and

ak =f (k)(z0)

k!=

1

2πi

ˆ∂B

f(ζ)

(ζ − z0)k+1dζ

for k ≥ 0.

• Sometimes this is used as the denition of a holomorphic function.• We need a lemma to prove this theorem.

Lemma. 1. Let γ be a rectiable curve. Let Fn be a sequence of continuous functions on γ. Suppose Fnconverges uniformly to F . Then

limn→∞

ˆγ

Fn =

ˆγ

f.

Proof. By the linearity we have ∣∣∣∣ˆγ

Fn −ˆγ

f

∣∣∣∣ =

ˆγ

|Fn − F |

< ε

∣∣∣∣ˆγ

1

∣∣∣∣≤ εL(γ)→ 0

as ε→ 0, because the convergence is uniform.

Proof of Theorem 4:

Proof. For all z ∈ B, we have

f(z) =1

2πi

ˆ∂B

f(ζ)

ζ − zdζ.

Then1

ζ − z=

1

ζ − z0

1

1− z−z0ζ−z0

=1

ζ − z0

∞∑k=0

(z − z0)k

(ζ − z0)k

if |z − z0| < |ζ − z0| = R. Then use the lemma to prove the rest. Left as an exercise to the reader.


2/24/2015

• Midterm 1 Review:

We have that ∆ |f |2 = 4∣∣∣∂f∂z ∣∣∣2. This implies |f |2 is subharmonic.

Proof: ∆ = 4 ∂2

∂z∂z . Then

∆ |f |2 = 4∂

∂z

[∂

∂zf f

]= 4

∂

∂z

[f∂f

∂z

],∂

∂zf = 0

= 4∂f

∂z

∂f

∂z

= 4

∣∣∣∣∂f∂z∣∣∣∣2 .

For p > 0 we have

∆ (|f |p) = ∆(|f |2

)p/2= 4

∂2

∂z∂z

(|f |2

)p/2= 4

∂

∂z

[(p2

)(|f |2

) p2−1

f∂f

∂z

]= 4

(p2

)(|f |2

) p2−1

∣∣∣∣∂f∂z∣∣∣∣2

+4(p

2

)(p2− 1)(|f |2

) p2−2

f∂f

∂z· f · ∂f

∂z

= 4(p

2

)(|f |p−2

) ∣∣∣∣∂f∂z∣∣∣∣2

+4(p

2

)(p2− 1)(|f |p−4

)|f |2

∣∣∣∣∂f∂z∣∣∣∣2

= p2 |f |p−2

∣∣∣∣∂f∂z∣∣∣∣2

Remark: If |f |p = fp2 f

p2 then we need to worry about the branch of f

p2 , p not even.

• #2: dzz = idθ.

• #3. Two methods: Method 1. Suppose that u = log |z| has a harmonic conjugate v. Then f = u+iv is holomorphivon Ω ⊃ |z| = 1. Then use

∂u

∂r=

1

r

∂v

∂θ

∂v

∂r= −1

r

∂u

∂θ

Using these equations we get that we must have v = θ + C where C is a real constant. Sof = u+ i (θ + C) on Ω. But then ef = zeic so f − ic is a branch of log z on Ω. But log z doesnot have a branch on Ω. Contradiction!


Method2: Let f ′ = ux + ivx = ux − iuy = x−iyx2+y2 = 1

z on Ω. Then

ˆγ

f ′dz = f (γ(1))− f (γ(0)) = 0

since γ is a closed curve. On the other hand,ˆγ

f ′dz =

ˆγ

1

zdz = 2πi.

By either Cauchy's formula. Apply to f(ζ) = 1, or simply change coordinates and computeˆγ

f ′dz =

ˆ 2π

0

idθ = 2πi.

In either case we get a contradiction.

Homework Problem:

Exercise. (August 2012) Suppose the function f is analytic in a simply-connected domain Ω. Show thatthere exists an analystic function F in Ω such that F ′(z) = f(z) for all z ∈ Ω.

• Remark: We cannot dene F (z) =´ zz0f(z)dz for a general holomorphic f on Ω.

• For example if we have a hole, then this function will not be well dened.• However, if Ω is simple connected, then F is well dened. So γ1− γ2 ecnlosed a domain inside Ω. Soˆ

γ1−γ2

fdz = 0.

Important Theorems for this week:

• Rouche's Theorem• Maximum Principle• Schwarz lemma.

Zeros:

• Let f be a holomorphic function on domain Ω. A point z0 ∈ Ω is a zero of f if f(z0) = 0.• If f(z0) = f ′(z0) = · · · = f (m−1)(z0) = 0, and f (m)(z0) 6= 0. Then z0 is a zero of f withmultiplicitym. If m = 1, z0 is a simple zero of f

Lemma. 1. Let f be a holomorphic function on Ω. a point z0 is a zero of f with multiplicity m if and onlyif there exists a disk B(z0, δ) centereds at z0 of radius δ > 0, such that

f(z) = (z − z0)mh(z)

where h is holomorphic on B(z0, δ) and h 6= 0.

Proof. (⇐=)Trivial( =⇒ )By the series representation of f there exists a B(z0, R) such that

f(z) =

∞∑j=0

aj (z − z0)j


on B(z,R) with aj = f(j)(z0)j! . Then a0 = · · · = am−1 = 0, so that

f(z) = (z − z0)m∞∑j=0

am+j (z − z0)j.

Here this new series∑∞j=0 am+j (z − z0)

jand

∑∞j=0 aj (z − z0)

jconverge for the same z. So

∑am+j (z − z0)

j

dened a holomorphic function h. Now h(z0) = am 6= 0. By continuity, there exists δ ∈ (0, R) such thath(z) 6= 0, for all z ∈ B(z0, δ). Then f = (z − z0)

mh on B(z0, δ).

Theorem. 2. Let f be a holomorphic on Ω. Then the set of zero of f has no limit points in Ω, unless f isidentically zero.

Corollary. 3. Let f1 and f2 be two holomorphic on Ω. If there exists a set E ⊂ Ω such that E contains alimit point of Ω and f1 = f2 on E, then f1 = f2 on Ω.

Proof. Theorem 2 implies Coro 3. Apply Theorem 2 to f = f1 − f2.

• Applications: sin2 z + cos2 z = 1 (?). Since sin z, cos z are holomorphic, when z is real then (?)holds. Let E =real axis.

Proof of Theorem 2

Proof. Let U =z ∈ Ω | ∃B(x, δ) ⊂ Ω s.t.f ≡ 0 on B(z, δ)

. Also let V =

z ∈ Ω | ∃B(x, δ) ⊂ Ω s.t.f 6= 0 on B(z, δ)\ z

.

We claim: Ω = U⊔V .

Now clearly U ∩ V = ∅. Also U, V are open. If f 6= 0 on Ω, then Ω ⊂ V . If there eixsts z0 is a zero ofmultiplicity m, then by Lemma, z0 ∈ V . If there exists z1 such that f (k)(z1) = 0 for all k ≥ 1. By seriesrepreseation f ≡ 0 on some B(z1, ε1) so z1 ∈ U . So Ω ⊂ U

⊔V ⊂ Ω. The claim is proved. Now suppose

z? is a limit poibt of Z(f) =set of zeros of f . There exists a sequence zk of distinct points zk → z?. Thenf(z?) = limk→∞ f(zk) = 0. Suppose z? has multiplicity m then f(z) = (z − z?)mh. on B(z?, δ)., h 6= 0. Butwhen k is suciently large, we have

0 = f(zk) = (zk − z?)mh(zk).

So f (k)(z?) = 0, for all k. Hence, z? ∈ U , Hence V = ∅. Now Ω = U .

• Example: Z(f) could have a limit point on ∂Ω. Consider f(z) = sin 11−z and take zk = 1− 1

2kπ . but1 ∈ ∂D.

Theorem. 3. (Argument Principle) Let f be a holomorphic on Ω. Suppose there is a rectiable Jordancurve λ ⊂ Ω such that f |γ 6= 0 and γ encloses a domain U in Ω. Then the number of zeros of f in U is givenby

Nf =1

2πi

ˆγ

f ′

f.

Theorem. 4. (Rouche's Theorem) Let Ω be a domain and γ ⊂ Ω resdtiable Jordan curve encloses a(bounded) domain U in Ω. Suppose f and g are holo on Ω and

|g(z)| < |f(z)|

on γ. Then f and f ± g have the same number of zeros counting multiplicities.

Application:


Theorem. (Fundamental Theorem of Algebra) Any kth degree polynomial has exactly k roots with multipl-cities.

Proof. Let P (z) = akzk + · · ·+ a1z1 + a0. Pick g(z) = ak−1z

k−1 + · · ·+ a0. Then

|g(z)| < |P (z)|when |z| = R. So P (z) and P (z)− g(z) = akz

k have k roots (mult) on B(0, R).


2/26/2015

Lemma. Let f be a holomorphic function on Ω. a point z0 is a zero of f with multiplicity m if and only ifthere exists a disk B(z0, δ) centereds at z0 of radius δ > 0, such that

f(z) = (z − z0)mh(z)

where h is holomorphic on B(z0, δ) and h 6= 0.

Theorem. Let f be holomorphic on Ω. We have f ≡ 0 on Ω if one of two properties holds:(i) ∃z0 ∈ Ω such that f (k)(z0) ≡ 0 for all k ≥ 0.(ii) The zero set Z(f) has a limit point in Ω.

Corollary. (Holo continuation)

Theorem. 1 (Arg. Principle) Let f be holomorphic on Ω. Let γ ⊂ Ω be rectiable Jordan curve whichenclosed a bounded domain U ⊂ Ω. Assume f |γ 6= 0. Then the number of zeros of f in U countingmultiplicities is

1

2πi

ˆγ

f ′

f.

Proof. First, f has only nitely many zeros in U . For, otherwise Z(f) has a limit point (because everyinnite set has a limit point) , which implies f |Ω= 0, a contradiction.

Second, let z1, . . . , zp be the zero set of f in U . There exists B(zi, δi) ⊂ U where f(z) = (z − zi)mi hi(z),with i = 1, . . . , p. Denote γi = ∂B(zi, δi). Apply Cauchy integral Theorem to f ′

f on U\ ∪p1 B(zi, δi). Thus

0 =

ˆγ−

⋃pi=1 γi

f ′

f,

which implies

1

2πi

ˆγ

f ′

f=

1

2πi

p∑j=1

ˆγj

f ′

f.

For each j = 1, . . . , p we haveˆγj

f ′

f=

ˆγj

mi

z − zi+

ˆγj

h′i(z)

hi(z)

=

ˆγj

mi

z − zi+ 0, since

h′i(z)

hi(z)is holomorphic

= 2πimj ,

as neeeded.

• Remark: Note f ′

f ” = ” (log f)′. (Need to check f (γ) ⊂ C\(−∞, 0])

• so that¸γf ′

f ” = ”¸γargfdθ where log f = log |f |+ i arg f .

Theorem. 2. (Rouche') Let f, g be holomorphic on Ω. Let γ ⊂ Ω be rectiable Jordan curve. Assume

|g| < |f | on γ. , (?)

Then #Z(f) = #Z(f ± g), counting multiplicities in U .


Proof. Only show #Z(f) = #Z(f + g). Let F = f + g. By arg. principle. We have

#Z(f) =1

2πi

ˆγ

f ′

fand #Z(F ) =

1

2πi

ˆγ

F ′

f

so that

#Z(f)−#Z(F ) =1

2πi

ˆγ

(f ′

f− F ′

F

)=

1

2πi

ˆγ

f ′F − fF ′

fF

=1

2πi

ˆγ

(f ′F)f2 − fF ′

f2

Ff

=−1

2πi

ˆγ

(F/f)′

(F/f)

= (?)

We claim(F/f)

′

(F/f)=

(log

F

f

)′on γ.

We show the inequality assumption to show the claim. To see the claim we need to check that this holds(according to the remark): (

F

f

)(γ) ⊂ C\ (−∞, 0] .

In fact, ∣∣∣∣Ff∣∣∣∣ =

∣∣∣∣f + g

f

∣∣∣∣ =

∣∣∣∣ gf + 1

∣∣∣∣ ,so that ∣∣∣∣Ff − 1

∣∣∣∣ =

∣∣∣∣ gf∣∣∣∣ < 1 on γ,

by the original assumption. ThenF

f(γ) ⊂ B(1, 1) ⊂ C\ (−∞, 0] .

Take the principal branch of log,so that log Ff is holomorphic on γ. Thus

#Z(f)−#Z(F ) = (?) =−1

2πi

ˆγ

(log

F

f

)′= 0,

by the Fundamental Theorem of Line integrals. Thus we need this assumption to make sense of the log.

• You can't directly apply to homeowork problem 2. But you can nd large ball. Semi circle. Show∣∣z3 + z + 1∣∣ ∼ R3 in the circle part, which is why you only have to worry about the imaginary part.

• Remark: Rouche' Theorem can be slightly improved. Condition (?) cna be relaxed to

|g| < |f |+ |f ± g| on γ. But this is not important for us. The rst version is good enough for the prelim. In practice, you just nd a closed curve, usuallya straight line togerther with a circle.


• Here's another application of Rouche's

Theorem. (Open mapping Theorem) Let f be holo on Ω. Then for all z0 ∈ Ω, let w0 = f(z0). Assumethat z0 is a zero of f(z) − w0 with multiplicity m. Then for all small p > 0, there exists δ > 0 such that∀w ∈ B(w0, δ), we have f(z) − w, has m zeros in B(z0, ρ). In particular, B(z0, δ) ⊂ f (Ω) . That is, f isopen.

Proof. There exists B(z0, R) ⊂ Ω such that f(z)− w0 = (z − z0)mh(z) where h |

B(z0,R)6= 0. On ∂B(z0, R),

we have

|f(z)− w0| = Rm |h(z)|≥ Rm min

z∈∂B|h(z)| .

Pick δ = Rm minz∈∂B |h(z)| > 0. Then for any w1 ∈ B(w0, δ). we have

|w1 − w0| < δ < |f(z)− w0|

on ∂B(z0, R). Then by Rouche's Theorem,

m = # (f − w0) in B(z0, R)

= # (f − w0 − w1 + w0) in B(z0, R)

= # (f − w1) in B(z0, R),

which completes the proof.

Maximum Principle:

Lemma. (Mean Value Equality) Let f be holomorphic on Ω. Then for all B(z, δ) ⊂ Ω. Then

f(x) =

∂B(x,R)

f(ζ)dζ.

andffl∂B(x,R)

· = 12πR

´∂B(x,R)

.

Proof. By the Cauchy integral formula we have

f(z) =1

2πi

ˆ∂B(x,R)

f(ζ)

ζ − zdζ

let ζ = z +Reiθ. Then (since d log is well dened )

dζ

ζ − z(= d log(ζ − z)) = idθ.

so that

f(z) =1

2πi

ˆ∂B(x,R)

f(ζ)

ζ − zdζ

=1

2πR

ˆ 2π

0

f(z +Reiθ

)Rdθ

=

∂B(x,R)

fdζ.


Theorem. (Maximum Principle) Let f be holomorphic on Ω. If there exists z0 ∈ Ω such that

|f(z0)| ≥ |f(z)| ,∀z ∈ Ω.

then f ≡ const on Ω.

• One of the homework problems needs maximum principle.

Proof. Let

U = z ∈ Ω | |f(z)| = |f(z0)| .Since z0 ∈ U, U 6= ∅. Now U is closed since |f | is continuous.

Claim: U is open.Assume the claim then Ω = U . That is, ∀z ∈ Ω, we have |f(z)| ≡ const. By Homework 1, we have

f ≡ const. It remains to show the claim.Proof of Claim:For all z ∈ U , there exists B(z, δ) ⊂ Ω. Suppose there exists z1 ∈ B(z, δ) such that

|f(z1)| < |f(z)| = |f(z0)| .Pick δ1 = |z − z1|. By MVE we have

|f(z)| =

∣∣∣∣∣ 1

2πδ1

ˆ∂B(z,δ1)

f(ζ)dζ

∣∣∣∣∣≤ 1

2πδ1

ˆ∂B(z,δ1)

|f(ζ)| dζ

< |f(z0)|= |f(z).| ,

a contradiction. Hence there can't exists a z1 ∈ z1 ∈ B(z, δ) with |f(z1)| < |f(z)| = |f(z0)|, so that|f(z1)| = |f(z0)| for all z1 ∈ B(z1, δ). Thus B(z, δ) ⊂ U . Thi shows U is open.

Corollary. (IMPORTANT )Let f be holomorphic on a bounded domain Ω. Then maxΩ |f | is attained onlyon ∂Ω unless f ≡ cosnt.

Remark. By switching to the reciprocal, we can get the minimum modulus principle. It states that if f isholomorphic within a bounded domain D, continuous up to the boundary of D, and non-zero at all points,then |f(z)| takes its minimum value on the boundary of D.

Theorem. 5. (Schwarz Lemma) Let D = B(0, 1) (open). Let ,f : D→ D holomorphic with f(0) = 0. Then

|f(z)| ≤ |z| and |f ′(0)| ≤ 1.

In addition, if |f ′(0)| = 1, or |f(z)| = |z| for some z 6= 0 with z ∈ D, then f(z) = eiaz, where a ∈ R ,constant.

Proof. Let

F (z) =

f(z)z , if z 6= 0

0, if z = 0.

Then F is holomorphic on D. For by the lemma, we can write f(z) = zmh(z) on B(z, ε) for small ε, whereh |B(z,ε) 6= 0, m ≥ 1. With

f(z)

z= zm−1h(z)


holomorphic. Tricky here is that we don't know if this is continuous on boundary or not. Then note that

max|z|≤1−ε

|F (z)| ≤ max |f |1− ε

≤ 1

1− ε,∀ε > 0.

So that by the corollary

supD|F (z)| < 1

1− ε,∀ε > 0.

Letting ε→ 0 we have thatsupD|F (z)| ≤ 1

which implies|f(z)| ≤ |z| .


3/3/2015

• Last time

Theorem. (Maximum Principle)Let Ω be a bounded domain and f holo on Ω and f ∈ C0(Ω). Assume

f |Ω 6= const. If |f(z1)| = maxΩ |f(z)| then z1 ∈ ∂Ω.

Theorem. 0 (Schwarz Lemma) Let D = B(0, 1) (open). Let f : D→ D holomorphic with f(0) = 0. Then

|f(z)| ≤ |z| and |f ′(0)| ≤ 1.

In addition, if |f(z1)| = |z1| for some z 6= 0 with z1 ∈ D or |f ′(0)| = 1 then f(z) = eiaz, where a ∈ R ,constant.

Proof. Let

F (z) =

f(z)z , if z 6= 0

f ′(0), if z = 0.

Then F is holomorphic on D. Fix any z1 ∈ D\ 0. For all 0 < ε < 12 (1− |z1|), Apply Max Principle to

|z| ≤ 1− ε. Thus|F (z1)| ≤ max

|z|≤1−ε|F (z)|

≤ max|z|=1−ε

|F (z)|

=max|z|=1−ε |f(z)|

1− ε

<1

1− ε.

Letting ε→ 0, we have that |F (z1)| ≤ 1, for all z1 ∈ D. But this implies that|f(z)| ≤ |z| or |f ′(0)| ≤ 1.

Now if |f(z2)| = |z2| then this implies that |F (z2)| = 1. And if |f ′(0)| = 1 then this implies that |F (0)| = 1.

Now locally we can write f(z) = zmh(z) for some holo h so that f(z)z = zm−1h(z)→ 0.

By Strong Max principlex, we have that F (z) ≡ c with |c| = 1 and so |F (z)| ≡ 1. Thus f(z) = cz whichimplies f(z) = eiαz, with c = eiα for some α ∈ R .

Denition. Let Ω ⊂ C as domain. Dene

Aut (Ω) = f : Ω→ Ω biholo ,where biholomorphic means f−1 exists and is holomorphic.

Claim. We claim that

Aut (Ω) = f : Ω→ Ω holo,1-1,and onto

Proof. If f : Ω→ Ω is 1-1, then f ′(z) 6= 0 for all z ∈ Ω. (For if f ′(z0) = 0, then f − f(z0) = (z − z0)mh(z)

with m > 1) By IVT, f−1 is holo.

• Goal: Determine Aut (D).

Proposition. 1.ψα(z) = eiαz

α∈R forms a subgroup of Aut (D).

Proposition. 2. For any a ∈ D, let ϕa(z) = z−a1−az , for all |z| < 1. Then ϕa ∈ Aut (D).


Proof. By HW1, we have that ϕa (D) ⊂ D. Also ∂ϕa∂z = 0 which implies ϕa is holomorphic on D. Can solve

ϕ−1a = ϕ−a holomorphic.

Lemma. If f ∈ Aut (D) with f(0) = 0, then f(z) = eiαz.

Proof. By Schwarz Lemma, we have that |f ′(0)| ≤ 1. Apply Scwarz Lemma to f−1 and get that

1

|f ′(0)|=∣∣∣(f−1

)′(0)∣∣∣ ≤ 1

So that f ′(0) = 1 so that f(z) = eiαz as needed.

Theorem. 2. For all f ∈ Aut (D) we have that

f = ϕa ψα = ψβ ϕb.

Proof. Let a = f(0). Consider g(z) = (ϕa f) (z). (standard trick, use composition of Mobious transform.)Now g : D→ D and g ∈ AutD. Then g(0) = 0 which impliez g = ψα by Lemma.

• Question: What is Aut(D?) where D? is the punctured disk, that is D? = D\ 0?• Read: The Schwarz-Pick lemma. Weierstrrass Theory,

Theorem. 3. Let Ω ⊂ D be a domain and fk be a sequence holomorphic on Ω, such that fk convergesto f uniformly on every compact set of Ω. (This is W-Weisrstrass assumption ) Then f is holo on Ω, andf ′k converges to f ′ uniformly on every compact set of Ω.

• Not true in general in Real Analysis.

Proof. By Morera's Theorem for holomorpgic functions, it suces to showˆγ

f = 0

for all closed rectable curve γ ⊂ Ω. Since fn ⇒ f on γ. Thenˆγ

f = limk→+∞

ˆγ

fk = 0.

For all K ⊂ Ω compact, and for all z ∈ K we have that

f ′k(z) =1

2πi

ˆ∂B

fk(ζ)

(ζ − z)2 dζ

where B is a large open set in Ω with B ⊃ K. Now

f ′(z) =1

2πi

ˆ∂B

f(ζ)

(ζ − z)2 dζ.

Now

|f ′k(z)− f ′(z)| ≤ 1

2π

∣∣∣∣∣ˆ∂B

|fk(ζ)− f(ζ)|(ζ − z)2 dζ

∣∣∣∣∣≤ sup

z∈K|fk − f |

C

diamB

→ 0

as k → +∞.

Corollary. With (W), f ′′k ⇒cpt f′′ and f

(l)k ⇒cpt f

(l).


Lemma. 2. (Hurwitz) With assumption (W), Let γ ⊂ Ω be rectiable cruve which enclosed a boundeddomain U in Ω. Assume f |γ 6= 0. Then #ZU (f) = #ZU (fk) for all suciently large k.

Proof. By Weiestrass Theorem, f is holomorphic. Since f |γ 6= 0, pick δ = minγ |f | > 0. Now we want toapply Rouche's theorem: We want

|f − fk| <δ

2< |f | on γ

for k sucienitly large. By Roche's Theorem, we have that #ZU (f) = #ZU (fk).

Theorem. (Hurwitz) With (W ) if fk never vanishes on Ω, then either f never vanishes on Ω or , f ≡ 0 onΩ.

Proof. Assume f(z0) = 0 and f 6≡ 0 on Ω. Then f(z) = (z − z0)mh(z) onB(z0, 2δ) which implies f |∂B(z0,δ) 6=

0. Applying the previous lemma to γ = ∂B(z0, δ) yields fk(z0) = 0 for large k. This is a contradiction.

Theorem. (Univalent) With (W ) if each fk is 1-1, then f is 1-1 unless f ≡ const.

Proof. Assume f(z1) = f(z2) = w0 for some z1 6= z2 in Ω. Consider the function f(z) − w0 which nevervanishes on ∂B(z1, δ1) and ∂B(z2, δ2). (Assume f 6≡ c). Here δ1. and δ2 are small such that

∂B(z1, δ1) ∩ ∂B(z2, δ2) = ∅.Now apply Hurwitz lemma to γ1 = ∂B(z1, δ1) and γ2 = ∂B(z2, δ2). Then

fk (z′1)− w0 = 0 = fk(z′2)− w0

which implies fk (z′1) = fk(z′2) where z′i ∈ B(zi, δi). This is a contradiction of the univalent theorem.


3/5/2015

• Last time:

Theorem. (Weirstrass) If∑∞k=1 fk ⇒cpt f in Ω, then f is holo on Ω and

∑∞k=1 f

′k ⇒ f ′. (cpt means

convergence pointwise? NO)

Laurent Series

• A general series of the form∞∑

n=−∞cn (z − a)

n ≡ f(z)

is called a Laurent series centered at a ∈ C where cn ∈ C,for all n ∈ Z. We can write

f(z) =

∞∑n=0

cn (z − a)m

+

∞∑n=1

c−n (z − a)−n

= fH(z) + fP (z).

• Here fP (z) is called the principal part of f , and fH(x) is called the holomorphic part of f .• We say a Laurent series is convergent at z0 is both fP and fH converge at z = z0.• For fH , note that

1

R= lim sup

n→+∞

n√|cn|

is radius of convergence. So fH converges on B(a,R) = |z − a| < R and fH converges uniformly

on B(a, p) for all 0 < ρ < R. For fP , let w = (z − a)−1

. Then

fP =

∞∑n=1

c−nwn.

Let

r = lim supn→+∞

n√|c−n|.

Then fP converges on |z − a| > r.

Proposition. With the notation above.1) If 0 ≤ r ≤ R ≤ +∞, then the Laurent series converges on the annulus A (a; r,R) = r < |z − a| < R.2) If r ≥ R then f diverges on any domain.

Theorem. 1. Let f be a holo fn on A (a; r,R) with r < R. Then f has Laurent series expansion

f(z) =

∞∑n=−∞

cn (z − a)n ∀r < |z − a| < R

where

cn =1

2πi

ˆ|z−a|=ρ

f (ζ)

(ζ − z)n+1 dζ =

ˆ|z−a|=ρ′

, r < ρ < R.

For any r < ρ′ < R. And the expansion is unique.


Proof. For z ∈ A (a, r,R), pcik small ε > 0 such that z ∈ A (a; r + ε, R− ε). By Cauchy Integral Formula.We have that

f(z) =1

2πi

ˆ|z−a|=R−ε

f (ζ)

(ζ − z)dζ − 1

2πi

ˆ|z−a|=r+ε

f (ζ)

(ζ − z)dζ

= f1 + f2.

For f1 , write

1

ζ − z=

1

(ζ − a)− (z − a)

=1

ζ − a1

1− z−aζ−a

=1

ζ − a

∞∑k=0

(z − aζ − a

)k,

since |z − a| < R− ε = |ζ − a| for all ,ζ ∈ ∂B (a,R− ε). By Weirstrass Theorem we have that

1

2πi

ˆ∂B(a,R−ε)

f (ζ)

ζ − zdζ =

1

2πi

∞∑n=0

ˆ|ζ−a|=R−ε

(z − a)nf(ζ)

(ζ − a)n+1 dζ

=

∞∑n=0

(z − a)n 1

2πi

ˆ|ζ−a|=R−ε

f(ζ)

(ζ − a)n+1 dζ

=

∞∑n=0

cn (z − a)n,

where´|ζ−a|=R−ε =

´|ζ−a|=ρ.

For f2 , write

1

z − ζ=

1

(z − a)− (ζ − a)

=1

z − a1

1− ζ−az−a

=1

z − a

∞∑k=0

(ζ − az − a

)k,

since |z − a| > r + ε = |ζ − a| for all ,ζ ∈ ∂B (a, r + ε). By Weirstrass Theorem we have that

1

2πi

ˆ∂B(a,R−ε)

f (ζ)

ζ − zdζ =

1

2πi

∞∑n=0

ˆ|ζ−a|=r+ε

(ζ − a)nf(ζ)

(z − a)n+1 dζ

=

∞∑n=0

(z − a)−n 1

2πi

ˆ|ζ−a|=r+ε

(ζ − a)n−1

f(ζ)dζ

=

∞∑n=0

c−n (z − a)−n

,

where´|ζ−a|=r+ε =

´|z−a|=ρ. (The trick for Lauren series is to try to write into geometric series.)

Uniqueness:


For uniqueness, assume f(z) =∑∞m=−∞ dm (z − a)

m. Fix k ∈ Z. Want cl = dk. Then

f(z)

(z − a)k+1

=

∞∑m=−∞

dm (z − a)m−k−1

.

Now

ck =1

2πi

ˆf (ζ)

(ζ − a)k+1

dζ =

∞∑m=−∞

dm

ˆ|z−a|=ρ

(ζ − a)m−k−1

dζ.

Note using Polar coordinates with ζ = a+ ρeiθ and dζ = ρieiθdθ we have that

1

2πi

ˆ|z−a|=ρ

(ζ − a)m−k−1

dζ =1

2πρm−k

ˆ 2π

0

ei(m−k)θdθ

= ρm−kδm,k

= δm,k

since ˆ 2π

0

ei(m−k)θdθ =

0 m 6= k

2π ,m = k.

Isolated Singularities:

• Let a ∈ C and f be a holo on B(a;R)? = B (a;R) \ a. So by Theorem 1,

f(z) =

∞∑n=−∞

cn (z − a)n

= fH + fP .

Denition. 1) If limz→a f(z) exists and nite, then a is a removable singularity of f .2) If limz→a f(z) =∞, then a is a pole of f .3) If limz→a f(z) does not exist, then a is an essential singularity of f .

Theorem. 2. Let f be holo on B(a;R)?. TFAE,i) a is a removarble singularity of f .ii) There exists 0 < δ < R such that supB(a;δ)? |f | < +∞.

iii) The principal part fP ≡ 0. So f ≡ fH .

Proof. Now (i) =⇒ (ii) by denition.Also (ii) =⇒ (iii) For each k ≥ 1, we have that for all 0 < ρ < δ;

|c−k| =

∣∣∣∣∣ 1

2πi

ˆ|z−a|=ρ

f (ζ) (ζ − a)k−1

dζ

∣∣∣∣∣≤ sup

0<|z−a|<ρ|f | 1

2π

ˆ|z−a|=ρ

∣∣∣(ζ − a)k−1∣∣∣ dζ

= sup0<|z−a|<ρ

|f | 1

2π

ˆ|z−a|=ρ

ρk−1dζ

= supB(a;ρ)?

|f | ρk → 0


as ρ→ 0.(iii) =⇒ (i) Now f = fP + fH = fH =

∑∞n=0 cn (z − a)

nand limz→a f(z) = c0 exists and is nite.

Corollary. (Riemann extension Theorem) Let f holomorphic on B?(a; δ). If

|f(z)| ≤ C on B?(a; δ), (?)

then there exists a holomorphic F on B(a;R) such that

F (z) = f(z) ∀z ∈ B?(a;R).

Remark. 1. Note that (? ) can be replaced by |f(z)| < C log 1|z−a| , with 0 < |z − a| < δ or

limz→a

(z − a) f(z) = 0

Remark. 2. You may use Rieman extension Theorem so solve HW 4.1.So image is bounded by 1 , so D? it can be extended.

Theorem. 2. Let f be holomorphic on B?(a;R): TFAE:i) a is a pole of f .ii) There exists 0 < δ < R such that

f(z) =h(z)

(z − a)m on B?(a; δ) ,m ≥ 1.

(iii) We have that

fP =c−m

(z − a)m + · · ·+ c−1

z − a,

where c−m 6= 0. (a is a pole of order m, if m = 1 a is a simple pole. )

Proof. (i) =⇒ (ii) Note that

limz→a

1

f(z)= 0

We can assume f(z) 6= 0 on B?(a; δ) on a smaller punctured disk. Thus 1f is holomorphic on B?(a; δ) and

has removable singularity at a. Thus 1f is holomorphic on whole diskB(a; δ). Now

1

f= h1(z) (z − a)

m,

for m ≥ 1. This implies that

f(z) =h−1

1 (z)

(z − a)m =

h(z)

(z − a)m .


3/10/2015

• (Problem #1, HW3) A subtlety f, g holo on D. f, g ∈ C0

(D)then |f | = |g| on ∂D which implies that |f | ≡ |g| on D.

Natural Idea: Apply maximum principle. Let h = fg . h 6= 0 on D so h is holo on D. Assume

h ∈ C0(D). then

maxD|h| = max

∂D|h| = 1

so that |f | ≤ |g|. Switch h1 = gf and get |f | ≥ |h|

For example f(z) = (z − 1)3. Then for some g with |g(z)| = |f(z)| = |z − 1|3 . If we let

h = gf = g

(z−1)3 . Can this guy still be continuously extended to the bunary of the disk?

We have to use analyticitiy. Say, (|z| − 1)3

=(|z|2 − 1

)5

on ∂D. Then(|z|2 − 1

)3

(|z|2 − 1

)5 /∈ C0(D).

Try to x this extention argument? (He'll give you extra credit.) His hint is that, actually youdon't even need extention, just work on smaller disk. Walzona Weirstress. But he found a gapin his proof.

But this is a subtlety that he wants to mention.

• Last time:• I. Let f be holo on 0 < |z| < R Then z = 0 is an isolated singularity of f . We say that

z = 0 removable ⇐⇒ limz→0

exists and niter

⇐⇒ supB(0,δ)

|f | < +∞ for some 0 < δ < R

⇐⇒ f holo on B(0, R).

• II. z = 0 is a pole of order m.

⇐⇒ limz→0

f(z) =∞

⇐⇒ f(z) =h(z)

zmon 0 < |z| < δ

⇐⇒ fP =c−m

(z − a)m + · · ·+ c−1

z − a

on B(0; δ) where c−m 6= 0.• II. z = 0 is an essential singularity.

⇐⇒ limz→0

f(z) does not exist

⇐⇒ fP =

∞∑n=1

c−nzn

on B(0; δ),

whith innitely many c−n 6= 0.• Example: The function f(z) = e1/z has essential dinsgularity at z = 0.


To see this we have

limz=x→0+

e1/z = limz→0+

e1/x = +∞

and

limz=x→0−

e1/z = limz→0−

e1/x = 0.

Alternatively once can see that we can write

e1/z =

∞∑k=0

1

zkk!

with 1k! 6= 0.

Theorem. 1. (Casorati-Weierstrass) Let f be a holo on B (a;R)?(punctured disk). Assume that a is an

essential singularity of f . Then for any 0 < δ < R. We have

f(B (a;R)

?)= C

Proof. Suppose not. Then there exists 0 < δ < R such that f(B (a;R)

?) C. Pick w0 ∈ C\f(B (a;R)

?).

Then

|f (z)− w0| ≥ ε0 > 0, ∀0 < |z − a| < δ.

Consider g(z) = 1f(z)−w0

. Thus

|g(z)| ≤ 1

ε0∀0 < |z − a| < δ.

Thus g has removable singularity at z = a , which implies that g is holomoprihc on B(a; δ) by Riemmanextension theorem. Thus we can write

1

f(z)− w0=

∞∑n=0

cn (z − a)n

= (z − a)mh(z)

on B(a; δ), for m ≥ 0 , ( not innitely many cn's. ?) Thus

f(z) =h(z)−1

(z − a)m + w0

has a pole or removable singularity at a. This is a contradiction.

• Remark: Another way of obtaining a contradiction in the previous proof. Take

F (z) =f(z)− w0

z − a, z 6= a.

Then take

limz→a|F (z)| = lim

z→a

∣∣∣∣f(z)− w0

z − a

∣∣∣∣ ≥ ε0 limz→a

1

|z − a|=∞

which implies F has a pole. Which implies f has a pole or removable singularity at a.• Let f be a holo on 0 < R < |z| <∞. Then we say ∞ is an isolated singularity on f . Bychanging variables, w = 1

z . We have reduced to the isolated singularity at 0.


Proposition. 1.I. ∞ is removable i f(z) =

∑∞n=0

c−nzn on |z| > R.

II. ∞ is a pole of order m i f(z) =∑∞n=0

c−nzn + c1z + · · ·+ cmz

m with cm 6= 0.III. ∞ is essential i

f(z) =

∞∑n=0

c−nzn

+

∞∑n=1

cnzn

with ininitely many cn 6= 0 for n ≥ 1.

Meromorphic Function:

• Recall f is an entire function if f is holomorphic on whole space C.• For entire functions, ∞ is an isolated singularity.

I. If ∞ is removable, then f ≡ cosnt, c0. (by Prop 1 or Liouville Theorem). II. If ∞ is a pole of order m, then

f(z) = c0 + c1z + · · ·+ cmzm, cm 6= 0.

III. If ∞ is essential, then

f(z) =

∞∑n=0

cnzn

with inntely many nonzero cn. Example are ez, sin z, cos z, Bessel functions, etc.

Theorem. Aut (C) = ϕ(z) = az + b | a, b ∈ C, a 6= 0.

Proof. For all f ∈ Aut (C). We write

f(z) =

∞∑n=0

cnzn,∀z ∈ C.

There are nitely many nonzero cn's. For otherwise, f has an essential singularity at ∞. By WeistrassTheorem, this would contradict 1− 1. A contradiction. Then

f(z) = c0 + c1z + · · ·+ cmzm.

Apply the open mapping theorem to

f(z)− c0.

This implies that f(z) = c0 + c1z with c1 6= 0.

• A function f is meromorphic (on C) if f only has poles on C.• An enire function is meromorphic. .• Besides, a rational function

R(z) =Pn(z)

Qn(z)

where Pn(z) ∝ Qm(z) are reduced polynomials equations. 1z .

Proposition. For rational function R, ∞ is either removable or a pole.


Proof. Let Pn(z) = anzn + · · ·+ a0 and Qn(z) = bmz

m + · · ·+ b0. Then

R(z) ∼ anzn

bmzm→

anbm

if n = m

0 if n < m

∞ if n > m

as z →∞. Thus O(zn−m) = that stu , of order n−m.

Theorem. 2. Let f be a meromorphic function. If ∞ is either removable or a pole, then f is rational.

Proof. f is holo on |z| > R

Proposition. 1. Let f(z) = fH + fP,∞ where fP,∞ = c1z + · · ·+ cmzm on |z| > R.

On |z| ≤ R = B (0;R) Then f has only nitely many poles. For otherwise, the limit point of poles isan essential singularity of f .

Denote the poles by z1, z2, . . . , zp. Then

f(z) = fH +c−mk

(z − zk)mk + · · ·+ c−1

(z − zk)

near each k = 1, . . . , p.

Claim. We have f(z) = fP,∞ +∑pk=1 fp,k + const on C. To see this, dene

g = f − fp,∞ −p∑k=1

fp,k

for ∞, z1, . . . , zp. Can show g has removable singularity at ∞, z1, . . . , zp. This implies g is holomprihc,bounded on C. This implies g ≡ const.


3/12/2015

• 1. Analytic Continuation• 2. Residue Theorem• 1. Let f be a holomorphic on domain U . If there exists a domain G ⊃ U and a holomorphic F onG such that F (z) = f(z) for all z ∈ U then F is an analytic continuation of f .

• Remark: If F exists, then it is unique. (Recall if F1, F2 holo on G, then F1 = F2 on some E ⊂ Gwhere E contains a limit point in G then F1 ≡ F2 on G)

• Example: Power Series: Let f(z) =

∑∞n=0 anz

n which has radius R of convergence. Which implies f(z) is holo on|z| < R by Abel's Theorem. For ζ0 ∈ ∂B (0;R)

(1) If ∃B(ζ0, δ) (smaller disk centered at the boundary) and g holo on B (ζ0; δ) such thatg(z) = f(z) for all z ∈ B (ζ0, δ) ∩B (0;R) then we say ζ0 is a regular point.

(2) We say ζ0 is a singular point og f if it is NOT regular.

Proposition. i) The set of regular points of f is open in ∂B (0;R).The set of singular points of f is closed in ∂B (0;R).

ii) For ζ0 ∈ ∂B (0;R), let z0 be in−−→Oζ0 (ray from origin to ζ0). Then f(z) has Taylor series

f(z) =

∞∑n=0

an (z − z0)n

an =f (0) (z0)

n!

Let ρ be the radius of convergence.If ρ > R− |z0|, then ζ0 is a regular point.If ρ = R− |z0| , then ζ0 is a singular point.

Proposition. 2. Let∑∞n=0 anz

n be a power series with radius R of convergence. Then tehre exists onesingular point on ∂B (0, R).

Proof. (Use Heine-Borel Theorem) Pick open cover of balls on the boundary circle.

Example. 2. Show that every point on ∂B(0, 1) is a singular point of∞∑n=0

zn! ≡ f(z)

Proof. (1) f has radius of convergence R = 1. Let

ak =

1 if k = n!

0 otherwise.

Then1

R= lim sup

k→∞

k√|ak| = lim sup

ki→∞|aki |

1/ki = limn→∞

11/ni = 1.

(2) It suces to nd a subset S of singular points of f such that S is dense in ∂B(0; 1). Let

S =e2πi pq | p ∈ Z≥0, q ∈ N, p, q reduced

.

Then S is dense in ∂B(0; 1). Fix an arbitrary ζ1 ∈ S. Want to show ζ1 is singular. Write

f (rζ1) =

q−1∑n=0

rn!e2πi pn!q +

∞∑n=q

rn!.


Claim: ϕ (r) ≡∑∞n=q r

n! → +∞ as r → 1−. If true, then we'd be done. For if so, then

|f (rζ1)| ≥ ϕ (r)−

∣∣∣∣∣q−1∑n=0

rn!e2πi pn!q

∣∣∣∣∣≥ ϕ(r)− q→ +∞,

as r → 1−. Supppose the claim is NOT true. Then

limr→1−

ϕ(r) = M < +∞.

(Note ϕ(r) is increasing on [0, 1)). Then ∀N , we have

N∑n=q

rn! ≤M.

Letting r → 1−1 , then N − q ≤ M . But then this inequality would be true for any N , which is acontradiction.

2. Residue Theorem

• Let f be holo on B? (a;R). Dene

Res (f ; a) =1

2πi

ˆ|z−a|=ρ

f(z)dz

where 0 < ρ < R. Recall f has Laurent series

f(z) =

∞∑n=−∞

cnzn

where

cn =1

2πi

ˆ|z−a|=ρ

f(z)

(z − a)n+1 dz.

Proposition. (1) If a is removable, then Res (f ; a) = 0.(2) If a is a pole of order m, then

Res (f ; a) =1

(m− 1)!limz→a

dm−1

dzm−1((z − a)

mf(z)) .

In particular, if m = 1 then

Res (f ; a) = limz→a

[(z − a) f(z)] .

Proof. (2) If a is a pole of order m, then using our third characterization of a pole of order m we have that

f(z) =c−m

(z − a)m + · · ·+ c− 1

z − a+ fH(z).

Then

(z − a)mf(z) = c−m + · · · c−1 (z − a)

m−1+ fH(z) (z − a)

m

and sodm−1

dzm−1((z − a)

mf(z)) = c−1 (m− 1)! + cfH(z) (z − a) .


so that

limz→a

dm−1

dzm−1((z − a)

mf(z)) = c−1 (m− 1)!

as needed.

Theorem. (Residue Theorem) Let γ be a rectiable Jordan curve. Let U be a bounded domain enclosed byγ. Let f be a holomorphic function on U\ z1, . . . , zp and f is continuous on U\ z1, . . . , zp. Then

1

2πi

ˆγ

f(z)dz =

p∑j=1

Res (f ; zj) .

Proof. Use Cauchy Integral Theorem to f on

U\p⋃j=1

B (zi, δi)

and ll in the details to

0 =

ˆγ∪

⋃pj=1 ∂B(zi,δi)

f(z)dz =

ˆγ

f(z)dz −p∑j=1

Res (f ; zj) .

• Remark: We can dene

Res (f ;∞) = −ˆ|z|=ρ

f(z)dz = −c−1

for any holo f on R < |z| < +∞.

Corollary. Let f be a meromorphic function with poles z1, . . . , zp . Thenp∑j=1

Res (f ; zj) + Res (f ;∞) = 0.

• Example 2: Compute ˆ ∞0

xp−1

1 + xdx for 0 < p < 1.

Consider

f(z) =zp−1

1 + z.

Note that zp−1 = e(p−1) log z. We know that log has branch cuts at 0 and ∞. So we need tond a branch cut. (If we take the standard branch cut of log z then we get nothing!)

Let γ be the following:

∗ : Let R big number bigger than 1, and ε < 1 is small number. γ = CR ∪ Cε ∪ L1,2


Here we take the branch cut to be [0,+∞) of log z = log r + iθ where 0 < θ < 2π holo on U .

So f(z) = zp−1

z+1 is holo on U\ −1. Then1

2πi

ˆγ

f(z)dz = Res (f ;−1) .

Then1

2πi

ˆCR

f +1

2πi

ˆCε

f +1

2πi

ˆL1

f +1

2πi

ˆL2

f = Res (f ;−1) .

Note Res (f ;−1) ” = (−1)p−1

” =(eiπ)p−1

= eiπ(p−1).

Claim1: 12πi

´CR

f → 0 and 12πi

´Cεf → 0 asR→ +∞ and ε→ 0+. where CR =

Reiθ : δ ≤ θ ≤ 2π − δ

and Cε =

εeiθ | π2 ≤ θ ≤

3π2

, and L1 = x+ Ei | 0or δ1 ≤ x ≤ R, ε > E > 0 and L2 =

x− Ei | 0 ≤ x ≤ R, 0 < E < ε. Then ˆ

L1

f(z)dz =

ˆ R

δ1

(x+ εi)p−1

1 + x+ εidx

and so

(x+ εi)p−1 → xp−1

x+ εi =√x2 + ε2e(tan−1 ε

x )i

(x+ εi)p−1

=√x2 + ε2e(tan−1 ε

x )i(p−1) → xp−1.

Now ˆL2

f(z)dz =

ˆ 0

R

(z − εi)p−1

1 + x− εidx

where z = x− εi then

(x− εi) =√x2 + ε2e(2π−tan−1 ε

x )i

(x− εi)p−1=

√x2 + ε2e(2π−tan−1 ε

x )i(p−1) → xp−1e2π(p−1)i.

Claim 2: We have thatˆL1

f(z)dz →ˆ +∞

0

xp−1

x+ 1dx

ˆL2

f(z)dz →ˆ 0

+∞

xp−1

x+ 1dx · e2π(p−1)i.

∗ By Residue Theorem we have

1

2πi

ˆ +∞

0

xp−1

x+ 1dx[1− e2π(p−1)i

]= eπ(p−1)i

and so ˆ +∞

0

xp−1

x+ 1dx =

2πi

1− e2π(p−1)ieπ(p−1)i

=2πi

e−π(p−1)i − e2π(p−1)i

=π

sin pπ.


∗ The Claim follows from Jordan's Lemma or direct calculation.∣∣∣∣ 1

2πi

ˆCR

f(z)dz

∣∣∣∣ ≤ 1

2π

ˆ 2πδ

δ

|f(z)| dz, let z = Reiθ

≤ 1

2π

ˆ 2π

δ

Rp−1R

R− 1dθ ∼ cRp−1 → 0

as R→ +∞.

Midterm 2 will be problems similar to the homework problems.

3/26/2015

Riemann Mapping Theorem

• One of two most important theorems in complex analysin. The other one is Cauchy Integral formula.

Theorem. (A) Let Ω ( C be a simply-connected domain in C. Then Ω is biholomorphic to D = |z| < 1.(That is, there exists holomorphic f : Ω→ D such that f is onto and one-to-one)

Remark. (1) If Ω = C, then Not true (Liuoville Theorem)(2) Theorem implies Any two simply connected domains Ωi ( C are biholo to each other. This implies

Ω1 and Ω2 are (dieomorphic/homeomorphic) to each other. Here we can allow the domains to be equal toC.

There exists dieroemorphic map C→ D. Take (x, y) 7→(

x√1+x2+y2

, y√1+x2+y2

).

(?)(3) Uniformization Theorem(Math 5121) Any Riemann surface has a universal cover, which is biholoto one of the following

i) D, ii) C and iii) C.

• Our proof follows ideas of Schwarz, Poincare, etc.• First a Lemma.

Lemma. 1. Let Ω be a simply-connected domain and f holo on Ω, f 6= 0 on Ω. Then we can take ananalytic branch g(z) of log f on Ω. That is, g(z) holo on Ω satisfying

eg(z) = f(z),∀z ∈ Ω.

Proof. Note f ′

f is holomorphic on Ω. By Homework 2.5, there exists a primitive function g holomorphic on

Ω such that g′ = f ′

f on Ω.(We used simply connectedness here). Let F (z) = f(z)e−g(z) for z ∈ Ω. Then

F ′(z) = f ′(z)e−g(z) − e−g(z)g′(z)f(z) ≡ 0

which implies F (z) ≡ c. This implies eg(z) = cf(z) so that g(z)+log c is branch of f . (Here log c = log |c|+iθ0

with −π < θ0 < π)

Corollary. 2. Let Ω be simply connected and f |Ω 6= 0, holo. Then there exists a branch g of√g(z) on Ω,

g |Ω 6= 0.

Proof. We have√f(z)

def= e

1z log f(z).

• We also need the following


Theorem. (Montel) Let F be a family of holomorphic functions on Ω. If there exists a constant M > 0 suchthat |f(z)| < M, for all f ∈ F for all z ∈ Ω, then F is normal, i.e. every sequence fk has a subsequencefkl which convegres uniformly to a function f on every compact set of Ω.

• Normal does not guarantee that the limiting funciton is in F .

Theorem. (B) Let Ω ( C be a simply conneted domain. Fix a ∈ Ω. Then there exists a holomorphicf : Ω→ D such that

1) f(a) = 0, and f ′(a) > 02) f is one-to-one3) f is onto.

Proof. Uniqueness:The uniquenes is trivial. Suppose f, g : Ω→ D saties the properties. Then consider

D f−1

→ Ωg→ D

which implies g f−1 : D→ D is biholomorphic, and g f−1(0) = 0. From a lemma about the automorphicgroup of D we have that g f−1 must be a rotation. That is, g f−1 = eiαz for some α ∈ R. Thus

g(z) = eiαf(z)

so that

g′(z) = eiαf ′(z)

but

0 < g′(z) = eiαf ′(z)

which implies α = 0. Thus g = f as needed.Existence:Let F = f : Ω→ D holo f(a) = 0, f ′(a) > 0, f is one-to-one .Step1: F 6= ∅Step2: There exists g ∈ F such that g′(a) = sup f ′(a) : f ∈ F. (This is the hardest)Step3: g is onto.Proof of Step1:

Since Ω ( C, there exists a b ∈ C\Ω. By Corollary 2, there exists branch g of√z − b, z ∈ C with g |Ω 6= 0.

Claim(1) g is one-to-one. Claim (2) For all z ∈ Ω,−g(z) /∈ Ω.In fact, [g(z1)− g(z2)] [g(z1) + g(z2)] = z1 − z2 for all z1, z2 ∈ Ω. This implies (1) and (2). Let

f(z) = eiαβg(z)− g(a)

g(z) + g(a),∀z ∈ Ω.

Here α and β are real constant TBA. Note f is holo. f(a) = 0 and f is 1-1. Remains to check f ′(a) > 0 andf (Ω) ⊂ D. Now

f ′(a) =

[eiαβ

g′(z) [g(z) + g(a)]− (g(z)− g(a)) g′(z)

[g(z) + g(a)]2

]z=a

= eiαβg′(a)

2g(a)> 0

if we set

α := eiαg′(a)

g(a)=

∣∣∣∣g′(a)

g(a)

∣∣∣∣ > 0.


Now

|f(z)| = |β|[1 +

∣∣∣∣ 2g(a)

g(z) + g(a)

∣∣∣∣] .We want [g(z) + g(a)] >? . By open mapping Theorem, there exists B(g(a), δ) ⊂ f (Ω) for some small δ > 0.Then

|g(z) + g(a)| ≥ δ, ∀z ∈ Ω.

This implies |f(z)| ≤ β(

1 + 2|g(a)|δ

)< 1 for small β > 0. This completes Step1.

Proof of Step2:Denote B ≡ sup f ′(a) : f ∈ F. Since its a supremum then there exists a sequence fn ⊂ F such that

f ′n(a) → B. Since |fn(z)| ≤ 1, for all n and for all z ∈ Ω. By Montoel's Theorem, fnk(z) ⇒cpt g(z). Then

g is holo, and g′(a) = lim f ′n(a) = B > 0 which implies that 0 < B < +∞. Thus

|g(z)| ≤ 1,∀z ∈ Ω.

Remains to check g is 1-1, which follows from Hurwitz Theorem.Proof of Step3:Suppose NOT. That means there exists a w0 ∈ D\g (Ω) . Let

F (z) =

√g(z)− w0

1− w0g(z)

be a branch. We can do this because since w0 is not in the image then 1 − w0g(z) is never vanishing. Theinside is like a Mobius transformation. Now you cna check that

|F (z)| ≤∣∣∣∣ g(z)− w0

1− w0g(z)

∣∣∣∣ ≤ 1,

using the estimate |√z| = |z|

12 ≤ |z| when |z| < 1. Thus F is one-to-one. Want to modify F to be F ∈ F .

Let

F (z) = eiγF (z)− F (a)

F (z) + F (a),

where γ ∈ R TBA(he means, to be determined). Now F (a) = 0. Furthermore since√z′

= 12√zthen

F ′(a) =

[eiγ

F ′(z) [F (z) + F (a)]− [F (z)− F (a)]F ′(z)

[F (z) + F (a)]2

]z=a

= eiαF ′(a)

2F (a).

Computing we get that F (a) =√−w0 and

F ′(a) =1

2

(√−w0

)−1

(1− |w0|2

)g′(a)

[1− w0g(a)]2 .

Pluggin these back into F ′(a) we get

F ′(a) =1 + |w0|2√|w0|

g′(a) > g′(a)

which is a contradiction. (Since 1 + a2 > 2a,∀a).


Remark: Idea: Get maps

D F−1

→ Ωg→ D\ w0

by 0 7→ a 7→ 0 so that g F−1 : D→ D can be applied Schwarzz lemma so that

|g′(a)|∣∣∣F ′(a)

∣∣∣−1

< 1,

as needed.


4/31/2015

• HW4-#1 f : D? → D? we show f(z) = eαiz for α ∈ R xed. Key is to show f(0) = 0. Quizk Way: Riemann extension f(0) is dened. f holo on D. Open mapping theorem. NowB (f(0), δ) ⊂ f (D). Pick small neighborhood on boundary, inside and intersection 0 and get acontradiction.

Another way: Riemann extension to f−1(0). He said make sure you can solve it!!!!!!!!!!!!!!!!!! Maybe this is a hint that it will be on thePRELIM!!!!!!!!!!!!!!!!!!!!!!!!

• HW4-#3 Cannot dene

´ zz0gdz on 0 < |z| < 1. You want to make sure that every curbe has the same

valued. Like is´γ1gdz 6=

´γ2gdz?! So its not well dened. The mean value theorem for analuytic

functions does not hold. Method 1: Since its isolated singularity we can write

fP (z) =c−1

z+c−2

z2+ · · ·

f ′P (z) =−c−1

z2+

˜c−2

z3+ · · ·

then

f = fP + fH

f ′ = f ′P + f ′H ,

okay. Method2: Take f ′(z) = 1

z . So

0 =

ˆ∂B(0,δ)

1

zdz = 2πi 6= 0.

You want to integrate on a specic curve.• HW4 - #5-

Use Schwarz lemma. Get ∑

|f(zn)| ≤∑|zn| ≤

∑|z|n =

1

1− |z|<∞

as needed. Now

k2n

is dense in [0, 1).

Can also check that f(z) = z + f(z2). Rudin's theorem. Hadaman's theorem.

• Test2- #1 Now for |z| ≥ 1 get |f(z)| ≤ e1/|z| ≤ e and for |z| ≤ 1 get |f(z)| ≤ max|z|=1 |f(z)| ≤ C sof ≡ const = 0

• Test2 - #2 For

´∞0

log xx2+a2 dx.

Method 1: Choose branch −π < θ < π.


∗∗ Here's a better picture:

∗ Method 2: Choose Pacman. Then let branch be dened on 0 < θ < 2π. Get ˆ R

0

log z

z2 + a2dz +

ˆ 0

R

log z + 2πi

z2 + a2dz =

ˆ R

0

2πi

z2 + a2dz

which is not usefull. Thus Can use (log z)2

z2+a2 . Then

ˆ R

0

(log z)2

z2 + a2−ˆ R

0

(log x+ 2πi)2

x2 + a2= −

ˆ R

0

2πilog x

x2 + a2dx.

Method 3.

∗ Picture:∗ Use branch 0 < θ < 2π.


∗ Get Res(

log zz2+a2 ; ai

)=

log a+π2 i

2ai . Then Γ1 = −x; 0 < x < R. Thenˆ

Γ1

=

ˆ R

0

log z + πi

x2 + a2dx

and Γ2 = x+ εi | 0 < x < R so thatˆΓ2

=

ˆ R

0

log x

x2 + a2dx.

Then ∣∣∣∣ˆCε

∣∣∣∣ ≤ Kε log ε→ 0 as ε→0∣∣∣∣ˆCR

∣∣∣∣ ≤ K11

R→ 0 as R→ +∞.

Then

2

ˆ +∞

0

log x

x2 + a2dx+ πi

ˆ +∞

0

dx

x2 + a2=π

a

(log a+

π

2i).

Method 4:

∗• Similar problem:

´∞0

√x log xx2+a2 dx.

Theorem. (Riemann Mapping Theorem) For every Ω ( C that is simply connected. Fix z0 ∈ Ω. Thereexists a unique holomorphic function f : Ω→ D such that

(1) f(z0) = 0 and f ′(z0) > 0.(2) f : Ω→ D is 1-1.(3) f : Ω→ D is onto.

Proof. From last time: We have that

F = f : Ω→ D holo | f satises (1) and (2) .Lemma(1): F 6= ∅.Lemma(2): There exists g ∈ F such that g′(a) = supf∈F f

′(a).Lemma(3): g is onto.For Lemma(2) recall that we needTheorem (Montel): Let Ω ⊂ C be a domain F a family of holo on Ω. Now f there exists a constant

M > 0 such thatsupf∈F

supz∈Ω|f(z)| ≤M

then F is normal.-


Also need:Lemma (Ascoli-Arzela): Let fn be a sequence of functions on a compact set K ⊂ RN such that:(1) fn is equibounded i.e. there exists c > 0 such that

supn

supz∈K|fn(z)| ≤ C.

(2) fn is equicontinuous, i.e. for every ε > 0 ther eexists δ > 0 such that

supn|fn(x)− fn(y)| < ε,

for every |x− y| < δ. Then there existsfnk of fn such that fn ⇒ f on K.-Remark: For Lemma AA, one need K to be compact. For exan we can nf a sequneve of smooth functions

fn(x) =

1, x ≥ n0 x < n− 1

.

Then clearly |fn| ≤ 1 and |f ′n| ≤ C.Now (AA) implies Theorem (Motel)Proof: Let fn ⊂ F . Then fn is equibounded. Need to show fn is equicontinuous. Write

fn(z) =1

2πi

ˆ∂B(δ)

fn(ζ)

ζ − zdζ

Then for all z, w ∈ B(δ) we get

|fn(z)− fn(w)| =

∣∣∣∣∣ 1

2πi

ˆ∂B(δ)

fn (ζ)

(1

ζ − z− 1

ζ − w

)dζ

∣∣∣∣∣≤ 1

2π

ˆ∂B(δ)

|fn(ζ)| |z − w|(ζ − z) (ζ − w)

dζ

≤ M

2π|z − w| 1

δ2/4

≤ C |z − w|< ε.

By chososing small enough ball.Remark: Method2: Can rst show that |f ′(z)| ≤ C . Then

|fn(z)− fn(w)| =

∣∣∣∣ˆγ

f ′n(γ(t))γ′(t)dt

∣∣∣∣≤ C1 |z − w| < ε

by pick γ to be a line segment from z to w, and then realizing that its length is |z − w|.One cannot argue as follows: |fn(z)− fn(w)| 6= |f ′n(η)(z − w)| ≤ |f ′(η)| |z − w| ≤ C |z − w|.Example: Take f(z) = ez then e2πi = e0i = 1. Then e2πi − e0i = eη (2πi− 0). ContradictionRemark2: Montel's theorem can be relaxed. It turned out you don't have to nd a global bound. But

you can bound the family on a smaller compact neighbrhood. Just like the Homework Problem:This revisied Montell's theorem will have an improved version, down at the bottom.

Relevant Theorems from above:


Theorem. (2) Montel's Theorem (Improved version) Let Ω ⊂ C. Let F be holomorphic on Ω. Supposefor every compact set K ⊂ Ω, there exists constant MK > 0 such that

supf∈F

supz∈K|f(z)| ≤MK ,

then F is normal.

• Like in the Homeowork, you can bound 11−r on |z| < δ < 1 but not when we iunclude 1.

Lemma. (Ascoli-Arzela) or (AA): Let fn be a sequence of functions on a compact set K ⊂ RN suchthat:

(1) fn is equibounded i.e. there exists c > 0 such that

supn

supz∈K|fn(z)| ≤ C.

(2) fn is equicontinuous, i.e. for every ε > 0 ther eexists δ > 0 such that

supn|fn(x)− fn(y)| < ε,

for every |x− y| < δ. Then there existsfnk of fn such that fn ⇒ f on K.

Proof. There exists E ⊂ K which is countable and dense in K. (E − K ∩ QN ), with E = x1, x2, . . . .For x1, we have that fn (x1)∞n=1 has a convergent subsequence. Say fnk := fnk1

, fnk2, . . . . For x2, look

at fnk(x2) and nd its convergent subsequence. Say fnk1, fnk2

, . . . . Then this the usual diagonilization

argument. And look at the diagonal subsequence. Pick the diagonal fnn. Then fnn converges on E. Thisis because

|fnn(x)− fmm(y)| ≤ |fnn(x)− fnn (x′)|+ |fmm(y)− fmm(y′)|And then use equicontinuity and fact that E ⊂ K.


4/7/2015

Theorem. (Riemann Mapping Theorem) Let Ω ( C simply-connected domain, there exists a biholomorphicf : Ω→ D.

• Can we extend f continuously to ∂Ω→ ∂D such that f : Ω→ D.• A neccessary condition is Ω has to be bounded.

For, Ω = f−1(D)compact. Assume Ω is bounded. The answer is Yes, if ∂Ω satises certain

condition.• Example1: ∂Ω is very nice. (i.e., ∂Ω is smooth) Let Ω = D and ∂Ω = ∂D = S1.

Every function f : D→ D biholomrphic is of the form

f(z) = eiαz − b1− bz

for α ∈ R, b ∈ D and |b| < 1. For |z| = 1, we have that

|f(z)| =

∣∣∣∣ z − 1

1− bz

∣∣∣∣=|z| |z − b|∣∣1− bz∣∣

=|1− bz|∣∣1− bz∣∣ = 1.

In fact , f ∈ Aut (∂D). f is holo since ∂

∂z f = 0 . So f : D→ D biholo.

• Example2: Let Ω = D+ = D ∩ H. Want D+ → D. Then L1 = z+1z−1 and R = eiπz and f = z2 and

g = z+iz−i . Then L1 bring top hald disk to third quadrant, then eiπ rbing to rst qudrant then bring

to the top half plane. Get

F = g f R L1

=

(z+1z−1

)2

+ i(z+1z−1

)2

− i

=(z + 1)

2+ i (z − 1)

2

(z + 1)2

+ i (z − u)2 .

F can extend continuously to D+. But F /∈ C1(D+

), since F ′ (±1) = 0.

(F−1 =

√z + 1

).

• Example3: LetΩ = D\ x+ yi | y = 0, 0 ≤ x < 1 .

Find G : Ω → D biholo. Then G = F √z . G cannot extend continuous to Γ = ∂Ω. Say, pick

14 ∈ Γ. Bring Ω→ D+ (top have disk)→ D. Then

√z =√reiθ/2 for 0 < θ < 2π.

Then zn = 14ei/n → 1

4 and wn = 14ei(2π− 1

n ) → 14 .

But√zn → 1

2 and√wn → − 1

2 . With F(

12

)6= F

(− 1

2

). And G (zn) → F

(12

)6= F

(− 1

2

)←

G (wn) .


Theorem. 1. Let Ω be a bounded simply-connected domain in D. If ∂Ω is a single/simple? Jordan curve,then any biholomorphic f : Ω→ D can extend to a homeomorphism Ω onto D.

Proof. Step1: f can be dened on ∂Ω.

Step2: f ∈ C0(Ω)

Step3: f is 1-1.

Then we are done. For D ⊂ f(Ω)is compact and f

(Ω)⊂ D. This implies that f

(Ω)

= D. Then f−1

is continuous. Any continuous function form a compact space to a Hausdor space implies that f−1 iscontinuous.

Lemma1: For every ζ ∈ ∂Ω, the limit

limz→ζ, z∈Ω

f(z) exists.

Consequently, we can dene

f (ζ) = limz→ζ, ζ∈Ω

f (z) ,

Proof: Oberseve the fact ∀ zn ⊂ Ω with zn → ζ ∈ ∂Ω. If f (zn)→ f (ζ) ∈ D, then |f (ζ)| = 1.Proof of fact: Suppose |f (ζ)| < 1. Let g = f−1. Then ζ = g (f (ζ)) = limn→∞ g (f (zn)) = limn→∞ zn =

ζ. Since g : D → Ω, and ζ = g (f (ζ)) ∈ Ω. This contradicts ζ ∈ ∂Ω. Suppose limz→ζ,z∈Ω f(z) does NOTexist. Then there exists znζ ; wn ⊂ Ω such that limn→∞ f(zn) = a and limn→∞ f(wn) = b where a, b ∈ Cand a 6= b. Then (f (Ω) ⊂ D) Furthermore, |a| = |b| = 1, by previous Fact.

Proposition: For all a, b ∈ ∂D, there exists a Mobius transform ϕ such that ϕ(a) = 1 and ϕ(b) = −1.Proof: Have that with

ϕ =z − α1− αz

ϕ(a) = 1 and ϕ(b) = −1 then implies

α =z + b− ab+ a

.

QED.Now the proposition implies that there exists ϕ such that F = ϕ f : Ω→ D . Then

limn→∞

F (zn) = eiπ/4 and limn→∞

F (wn) = ei5π/4.

Now let ζ0 = F−1(0) and d ≡ |ζ0 − ζ| > 0. Now for all 0 < δ < d2 . Then for zn, wn ∈ B(ζ; δ) ∩ Ω for all

n ≥ N . Then let γ be a continuous curve in B (ζ; δ) ∩ Ω such that γ(0) = zn and γ(1) = wn. Now F (γ(t))must intersect the real imaginary axis. Then

P = last F (γ(t)) ∩ y = 0Q = 1stF (γ(y)) ∩ x = 0 .

Then the curve not linear−−→PQ ⊂3rd quadrant. By reectivity

−−→PQ w.r.t axis to get a closed curve γ1 encloses

a domain U ⊂ D. Note that 0 ∈ U . Dene

h(w) = (G(w)− ζ)(G(w)− ζ

)(G(−w)− ζ)

(G(−w)− ζ

),∀w ∈ D.

Note h is holomorphic on D.Claim: supU |h(w)| ≤ δM3 for all 0 < δ < d

2 .

Assume the claim is true. Letting δ → 0+ we get that h(w) ≡ 0. Note |h(0)| ≤ δM3 → 0 which impliez

h(0) = 0. But h(0) = |ζ0 − ζ|4 = d4 > 0. A contradiction.


Proof of CLaim: To see the claim sup∂U |h(w)| ≤ δM3. On−−→PQ , We have

|h(w)| ≤ |G(w)− ζ|M3

= |γ(θ)− ζ|M3

< δM3.

• Done with prelim material• Do prelim problems from now on.• Last homework some of Damin's favorite problems (special attention to the ones you haven't seenbefore, hint hint?) and old prelim problems.


4/9/2015

• Last time:

Theorem. Let Ω be a bounded simply-connected domain in D. If ∂Ω is a single/simple? Jordan curve, thenany biholomorphic f : Ω→ D can extend to a homeomorphism Ω onto D.

Lemma. 1. For all ζ ∈ ∂Ω, we have limz→ζ,z∈Ω f(z) exists and nite.

Lemma. 2. For any ζ ∈ ∂Ω, dene f(ζ) = limz→ζ,z∈Ω f(z). Then f ∈ C0(Ω).

Proof. For all ε > 0, there exists δ > 0 such that |f(z)− f(ζ)| < ε, for all z ∈ B (z, δ)∩Ω. For all ζ1 ∈ B(ζ; δ2 )pick zn → ζ1 with zn ∈ Ω. Then

|f(ζ1)− f(ζ)| ≤ |f(ζ1)− f(zn)|+ |f(zn)− f(ζ)|< ε+ ε = 2ε,

for large n, with zn ∈ B(ζ1; δ4

)⊂ B (ζ; δ). This shows that

|f(w)− f(ζ)| < 2ε, ∀w ∈ B(ζ; δ) ∩ Ω.

Lemma. 3. Let f ∈ C0(Ω)be given in Lemma. Then f is one-to-one.

Proof. Incorrect Proof:For all ζ1 6= ζ2 ∈ ∂Ω, note that

f (B(ζ1; δ) ∩ Ω) ∩ f (B (ζ2; δ) ∩ Ω) = ∅

with 0 < δ DOES NOT IMPLY that f (ζ1) 6= f (ζ2).

Theorem. 2 (Lindelof) Let Γ(t) be a curve in D such that |Γ(t)| < 1, if 0 ≤ t < 1 and Γ(1) = 1. Let g be abounded holomorphic function on D such that

limt→1

g (Γ(t)) = β ∈ C.

Then limt→1− g(t) = β. (limr→1− g(r) = β with r ∈ D ∩ Imz = 0).

Proof of Lemma 3.

Proof. Suppose ζ1 6= ζ2 ∈ ∂Ω such that f (ζ1) = f (ζ2) = eiα. Replace f by e−iαf , if necessary, we canassume f (ζ1) = f (ζ2) = 1. Can pcik a curve γ1, γ2 such that γj ([0, 1)) ⊂ Ω, γj(1) = ζj with j = 1, 2. Let

Γj(t) = f (γj(t)) is curve in D satisfying Theorem 2. Let g = f−1 on D, g is holo, bounded and

limt→1

g (Γ1 (t)) = limt→1−

γ1(t) = ζ1

and by Theorem 2 it implies

limr→1−

g(r) = ζ1.

But similarly

limr→1−

g(r) = ζ2.

But ζ1 6= ζ2 which gives us a contradiction.

Proof of Theorem2:


Proof. We need to show for all ε > 0, there exists δ > 0, such that

|g(r)− β| < ε, ∀r, δ < r < 1.

Can assume β = 0 (replace g by g−β) and |g(z)| < 1 on D. (g by g(z)supD|g|

). Since limt→1− g (Γ(t)) = 0. There

exists δ > 0 with |g (Γ(t))| < ε, for all 1− 2δ < t < 1. Fix r ∈ (1− δ, 1) . Let

t1 = max t | ReΓ(t) = r

with Γ ([t, 1]) ⊂ D ∩ x ≥ r. Reect E1 = Γ ([t1, 1]) with respect to real axis to get E2. Reect E1 ∪ E2

with respect to x = r to get E3. Then E = E1 ∪E2 ∪E3. Let U be the bounded open sets enclosed by E.Dene

h(z) = [g(z)][g (z)

] [g (2r − z) g (2r − z)

]= [E1] [E2] [E3] .

Note that h is holomorphic on U , with |h| < 1. (Note ∂∂z g(z) = ∂

∂z g (z) = ∂∂z g = 0).

Claim: supU |h(z)| < ε. If so, r ∈ U and h(r) = |g(r)|4 which implies |g(r)| < ε1/4 then for all1− δ < r < 1. Remains to show the claim. Note

supz∈E1\1

|g(z)| < ε =⇒ supE1\1

|h(z)| < ε.

Then E1 = Γ ([t1, 1]). Similarly, supE\1,2r−1 |h(z)| < ε. Let hC(z) = h(z) (1− z)c (2r − 1− z)c where c > 0

is a constant TBA, so hC is holo on U , and hC ∈ C0(U). Now

supz∈U|hC | < εM c.

Letting c→ 0+, we have that supz∈U |h(z)| < ε. This proves the claim.

Schwarz Reection Principle

Lemma. (Symmetry Principle)Let Ω1 and Ω2 be two domains such that Ω1 ∩ Ω2 = ∅ and ∂Ω1 ∩ ∂Ω2 = γ be a rectable curve. Let fj be

holo on Ωj , and fj ∈ C0 (Ωj ∪ γ) , j = 1, 2. If f1 = f2 on γ, then

f(z) =

f1(z) z ∈ Ω1

f1(z) = f2(z) z ∈ γf2(z) z ∈ Ω2

is

holo on Ω ≡ Ω1 ∪ γ ∪ Ω2.

Proof. Remains to show f is holo on γ. Let T be a triangle domain, T ⊂ Ω. Need to show thatˆ∂T

fdz = 0

then we can apply Morera's theorem. For T ⊂ Ω1 ∪ γ by Cauchy-Gorsart we have thatˆ∂T

fdz = 0.

If T ∩ Ω1 6= ∅ and T ∩ Ω2 6= 0. Let Γ0 = T ∩ γ and Γj = T ∩ Ωj . Thenˆ∂T

f =

ˆΓ1+Γ0

f +

ˆΓ2−Γ0

f = 0.


Theorem. 3. Let Ω be a domain subset of H. Let I = ∂Ω ∩ Imz = 0 . Let f be holo on Ω, and f ∈C0 (Ω ∪ I) . If f (I) ⊂ R, then there exists F on Ω+ ∪ I ∪ Ω−, holo, such that F = f on Ω+.

Proof. Let

F (z) =

f(z) z ∈ Ω+ ∪ γf (z) z ∈ Ω−.

Then F is holo on Ω−(∂∂z f (z) = ∂

∂z f (z) = 0 then or because ∂∂z f (z) = 0) On I, for all x ∈ I ⊂ R, with

f(x) = f (x).

Question 3: Let f ∈ C0(D)f is holo on D. If there exists an open arc γ ⊂ ∂D such that f |γ≡ c ∈ C,

then f ≡ c on D.Proof1:

Proof. Can assume c = 0 (f 7→ f − c) Let L1 : H→ D wtih Γ 7→ γ and

z 7→ z − iz + i

.

So f1 = f L1 : H→ C is holo.Pick an x0 and pick a small ball such that the boundary still contains Γ. Now look f1 on that semicircle

curve (H∪B(x0, δ)) on H, and applying Scwartzreection preinciple, we can reect, Call it F1. Now F1 |Γ≡ 0so that F1 ≡ 0 Thus this implies f ≡ 0.

• Q1 is his favorite problem,


4/14/2015

• #2) from the homework Method1: Just show that ∞ is NOT an essential singularity. Suppose ∞ is an essential singu-larity. Then we know that f(z) can be written as

f(z) =

∞∑m=0

cmzm

with innitenly many nonzero terms. Use an inequality to get a contradiction.

For all k ∈ N f(z)zk

has essential singularity at ∞. By Casaroti-Weirstraa theorem: there exists

zl∞l=1 s.t.

f(zl)

zkl→∞ as l→∞.

Write

Pn(z) +Pn−1(z)

f(z)+ · · ·+ P1(z)

f(z)n−1= 0

Write

Pn(z) = an,NzN + an,N−1z

N−1 + · · ·+ an,0.

Then

1 +an,N−1

a+ · · ·+ an,0

zN+

n−1∑j=1

Pj(z)

an,NzNf(z)= 0. (2)

Pick k = max degPj(z); 1 ≤ j ≤ n− 1 − N . Let k = max d, 0. Apply (2) to zl. Lettingl→ +∞ we get that

1 + 0 +∑ zkl

f (zl)= 0

which implies 1 = 0, a contradiction. Remark: Avoid ∞±∞, use inequality. Method2: (Tim Smits, etc)

∗ Assune, without loss of generality, that there exists R0 > 1 such that

max|z|≤R

|f(z)| > 1.

Choose a suciently large R > R0 such that

1 ≤ δj |z|dj

≤ |Pj(z)|≤ Cj |z|dj , for all |z| = R

Indeed

|Pj(z)| =∣∣aj,mzm + aj,m−1z

m−1 + · · ·+ aj,0∣∣

= |aj,mzm|∣∣∣1 +

aj,m−1

z+ · · ·+ aj,0

zm

∣∣∣≥ |aj,m|Rm (1− ε) .

Here Cj , δj depdends only onthe leading terms of Pj , dj = degPj .


∗ Let |z0| ≤ R such that

|f(z0)| = max|z|≤R

|f(z)| > 1

By max principle, |z0| = R. So

δjRdj ≤ |Pj(z0)| ≤ cjRdj , 1 ≤ j ≤ n− 1.

Now (1) =⇒

|Pn(z0)f(z0)| =|Pn−1(z0)||f(z0)|

+|Pn−2(z0)||f(z0)|2

+ · · ·+ |P1(z)||f(z0)|n−1

≤ |Pn−1(z0)|+ · · ·+ |P1(z0)| .so

δnRdn |f(z0)| ≤

n−1∑j−1

cjRdj

which implies

|f(z0)| ≤n−1∑j=1

cjδnRdj−dn

which implies

max|z|≤R

|f(z)| ≤ CRd

where d = max d | 1 ≤ j ≤ n − dn ≥ 1 .∗ By Cauchy Integral formula∣∣∣f (k)(0)

∣∣∣ =

∣∣∣∣∣ k!

2πi

ˆ|ζ|=R

f (ζ)

ζk+1dζ

∣∣∣∣∣≤ C

max|ζ|=RRk+1 |f(ζ)|

≤ CRd−k−1,

for k ≥ d letting R→∞ we get that f (k)(0) = 0. Hence

f(z) =

d∑k=0

ckzk.

is a polynomial.• #3 from the homeowor: Given

|f ′(z)|(

1− |z|2)

+ |f(0)| ≤ 1. (3)

Show that F is normal.

Proof. From (3) we have that

|f ′(z)| ≤ 1− |f(0)|1− |z|2

≤ 1

1− |z|2.

For all compact set K ⊂ D ,

supK|f(z)| ≤?


there exist r ∈ (0, 1) such that K ⊂ Br. On Br, we have that

|f ′(z)| ≤ 1

1− r2, for all |z| ≤ r.

Then

|f(z)| =

∣∣∣∣∣ˆ

[0,z]

f ′(z)dz

∣∣∣∣∣≤ r

1− r2≤ 1

1− r2,

Hence

supK|f(z)| ≤ sup

Br

|f(z)| ≤ 1

1− r2.

Rest of the Course:

• 1. Picard's Little Theorem• 2. Picard's Big Theorem• 3? Runge's approximation Theorem

Theorem. (PLT) If f is an entire function which omits two values, then

f ≡ constant.

• Remark f(z) = ez only omits 0, and its not a constant.• Our proof uses dieretial geometry and maximum principle. Due to Alphors (1938).• Let us rst consider the Scwarz lemma.• The new idea for proving Schawrz lemma. The key observation is D admits a nice metric

idz ∧ dz(1− |z|2

)2 ≡ ωP , Poincare

• Notation: hdz ⊗ dz ∼ hidz∧dzz ∼ |h| |dz|2.

• The term1(

1− |z|2)2 = h

satises the proerty

∂2

∂z∂zlog h = 2h (?)

or using Laplace operator

∆ log h = 8h.

• Proof of (?) from log h = −2 log(

1− |z|2)get

∂

∂zlog h =

−z1− |z|2

.


Note |z| = zz then

∂

∂z

∂

∂zlog h =

−1

1− |z|2+

−zz(1− |z|2

)2 =−1(

1− |z|2)2 .

• Exercise: Compute ∂2

∂z∂z log 11+|z|2 .

• Can't do the rest. Refer to Bobby's notes.


4/16/2015

Last time:

• Schwarz Lemma: If f : D→ D holo, then

|f ′(z)|2(1− |f(z)|2

)2 ≤1(

1− |z|2)2

i.e. f?ωP ≤ ωP .

Proof. (Ahlfors. 1938) Let F = f?ωPωP

=(1−|z|2)

2|f ′(z)|2(1−|f(z)|2)

2 . Then F ≥ 0, F ∈ C∞ (D). For all z ∈ D s.t.

F (z) > 0 (⇐⇒ |f ′(z)| > 0).Then

∂z∂z logF = ∂z∂z log(

1− |z|2)2

+ ∂z∂z log |f ′(z)|2

−∂z∂z log(

1− |f(z)|2)2

= ∂z∂z log(

1− |z|2)2

+ 0− ∂z∂z log(

1− |f(z)|2)2

=2(

1− |z|2)2 − 2∂z

[−f ′(z)f(z)

1− |f(z)|2

]

=−2(

1− |z|2)2 +

2 |f ′(z)|2(1− |f(z)|2

)2

=2(

1− |z|2)2 [F (z)− 1] .

If F attains supD F at z0 ∈ D then so is logF . Also ∆ logF ≤ 0. (∆ = 4∂z∂z) which implies F (z0) ≤ 1.That is, supD F ≤ 1.The subtle point is supD F need NOT be attained in D. Two ways to overcome this.

1 (Ahlfors): Let F =(r2−|z|2)|f ′(z)|2r2(1−|f(z)|2)

2 . That is, use metric

4r2(r2 − |z|2

)2

idz ∧ dz2

on |z| < r .

To show|f ′(z)|2(

1− |f(z)|2)2 ≤

r2(r2 − |z|2

)2 .

Letting r → 1−, yields the proof. The advatange of using F is, there exists z2 ∈ B(0; r) such that F (z2) =

sup|z|<r F . Then14∆ log F = r2

(r2−|z|2)2

[F − 1

]. At z2, we have that

∆ log F ≤ 0 =⇒ F (z) ≤ F (z2) ≤ 1.

2. Yau (1978)


If (where before ∆ logF ≥(

11−|z|2 (F − 1)

)) ∆P logF ≥ F − 1 where ∆P is the Laplacian of a compelte

metric space then supF < 1 then

∆ = ∂2xx + ∂2

yy = ∂z∂z.

• Schaws Lemma: If f : B(0;R)→ U be holo function, U domain with a metric ω satisfying K(ω) ≤−κ < 0 where κ ≡ constant > 0, then

f?ω ≤ 1κ ωP on B(0;R)

where ωP = 4R2

(R2−|z|2)2−dz∧dz

2 .

Remark: If ω = g idz∧dz2 , dene K (ω) = 2∂z∂z(− log g)g . (This is the Gauss curvature). Recall for

D, ωP = 4

(1−|z|2)2idz∧dz

2 , where K (ωP ) ≡ −1.

Proof. Let F = f?ωωP

=(R2−|z|2)

2g(f(z))|f ′(z)|2R2 . Then ∆ logF = R2

(R2−|z|2)2 [κF − 1] which implies (by

Ahlfors/You) that F ≤ 1κ .

Corollary. 2. If f : C→ U is holo, and U admits a metric ω with K (ω) ≤ −κ < 0, then f ≡ const.

Proof. For any R > 0,

g (f (z)) |f ′(z)| ≤ 1

κ

R2(R2 − |z|2

)2 .

Let R→ +∞ which goes to zero, which implies |f ′(z)| ≡ 0.

• Remark1: Coro2 implies Liuville's Theorem. If f : C → B(0;L) with L2

(L2−|z|2)2idz∧dz

2 , which

means it has negative curvature. (2) C does not have metric with negative curvature. K(ω) But 4dz∧dz

(1+|z|2)2 is a metric on C with K > 0.

∗ Negative curvature is a space that looks the gabriels horn.Triangle (angles)add up to lessthan π. Positive curvature is a space that looks like a sphere, triangle angles add up tomore than π.· Negagtive curvature looks like its bloated in, while posive curvature is like bloatedout.· zero curvature is a at space.

(3) C? = C\ 0 No metric with negative curvature C→ C? with ez.

Theorem. (Picard's Little Theorem) Any holo f : C→ C\ a, b is a constant.

Proof. We need to construct a metric ω on C\ a, b s.t. K(ω) ≤ −κ < 0.Idea: Find a metric ω on D? = D\ 0 with negative curvature, that is K(ω) ≤ −κ < 0. Get a covering

map eiz.With ωP? = 4dz∧dz

|z|2 log2|z|2 on 0 < |z| < 1.

We claim K (ωP?) ≡ −1.


Proof:

∂z∂z log1

|z|2(

log |z|2)2 = −∂z∂z log |z|2 (goes to 0)− 2∂z∂z log

(log |z|2

)

= −2∂z

[1

log |z|2z

|z|2

]

= −2∂z

[1

z log |z|2

]

=2

|z|2 log2 |z|2.

Claim:


4/21/2015

• 3 questions from last two problem set. One or two from his favorite questions.• Question2: Very easy question.

Hint: We know√z is not a nice function. But

√z2 always has an analytic branch on ANY

domain. The reason, because√z2 goes back to the absliute value of the number.

Indeed√z2 = e

12 log z2

if z = reiθ then z2 = r2e2iθ then log z2 = log r2 + 2iθ so that√z2 =

elog r+iθ = reiθ. If you take θ 7→ θ + 2kπ does no change since eiθ = ei(θ+2kπ) so

√z2 = reiθthen we can take .

∗ the baosutely value thing, take a branch cut.

∗ If equal to zero then dene it to be zero.√z2 = 0 if |z| = 0. You can also use limit to

show holo at 0, or use info that we arelady know since its holo around it. Also

√h(z) always has a branch in B(0; δ). If h(0) 6= 0, Indeed, choose δ > 0 s.t. h |B(0;δ) 6= 0.

A smaller disk B(0; δ) simply connected, then by Riemman mapping theorem then√h has a

branch. Check your notes. In general, h(0) = 0 use Taylor expansion.

• Picard's Little Theorem: Any entire function f : C→ C\ a, b is constant.

Proof. Reduces to construct a metric ω1 on C\ 0, 1 (Replace f by f(z)−ab−a ). A metric such that K (ω1) <

−δ < 0 with (δ > 0) , so negative curvature.Let the metric be in the following form:

ω1 =

(1 + |z|2)2

|z|2[

log|z|2

1 + |z|2

]2

·(

1 + |z − 1|2)2

|z − 1|2[

log|z − 1|2

1 + |z − 1|2

]2−1

· idz ∧ dz2

≡ g1 · g2 ·idz ∧ dz

2.

where

g−11 =

(1 + |z|2

)2

|z|2[

log|z|2

1 + |z|2

]2

,

g−12 =

(1 + |z − 1|2

)2

|z − 1|2[

log|z − 1|2

1 + |z − 1|2

]2

.

So

K (ω1) ≡ − 2

g1g2∂z∂z log (g1g2)

= − 2

g1g2(∂z∂z log (g1) + ∂z∂z log (g2))

=2

g2

(− 1

g1∂z∂z log (g1)

)− 2

g1

(− 1

g2∂z∂z log (g2)

).


Only need to compute the Gauss curvature of − 2g1∂z∂z log (g1) by letting z 7→ z−1. Note that ∂z∂z log |z|2 =

0 for all z 6= 0. And so

∂z∂z log(

1 + |z|2)2

=2(

1 + |z|2)2 .

Then

∂z∂z log

(log

|z|2

1 + |z|2

)2

=1(

1 + |z|2)2

|z|2[log |z|2

1+|z|2

]2by noting that |z|2

1+|z|2 = 1− 11+|z|2 . Then

(1) K (ω1) < 0 on C\ 0, 1(2) Also limz→0K (ω1) = −8 [log 2]

2= limz−1K(w2)

(3) limz→∞K(ω1) = −4. Which implies that ∞ < −δ2 ≤ K(ω1) ≤ −δ1.This nishes the proof.

• Homework Problem3: At most one connected component. Use Picard's Little Theorem Because it is vey precise. The entire function can only omit possibly one point.

Theorem. 2 (Picard's Big Theorem) Let f be holo on B?(a; δ). If f has an essential singularity at a, then∀0 < δ1 < δ we have that

f (B? (a; δ1)) = C or C\ one point .• This is an extensino of the Casorati-Weisrtrass Theorem.• Remark: PBT =⇒ CW and PBT =⇒ PLT.• Fact 1: Let f be a meromorphic function on Ω ⊂ C. Then f cna be viewed as a holomorphic

function with values in C = C ∪ ∞, the Riemann sphere. Indeed, let z = a be a pole of f . Thenwe note that locally f can be written as the following form

f(z) =h(z)

(z − a)m in B? (a; δ) ,

where h is holo on B (a; δ). Then f : B (a; δ)→ C . Let I(z) = 1z for all z 6= 0. Then (I f) is holo

on B(a; δ).

• Fact 2: There is a metric ωs on C such that ωs |C= idz∧dz2(1+|z|2)

2 and K (ωs) ≡ +1 on C.

With C = C\ ∞ Near ∞ let w = 1z then [w = 0! z =∞] and z = 1

w and z = 1w then lets

compute the metric: We have dz = − 1w2 dw and dz = − 1

w2 dw and |z|2 = 1|w|2 Then

idz ∧ dz

2(

1 + |z|2)2 =

idw ∧ dw

2(

1 + |w|2)2 .

This is invariant. That means this is always of the form

ωs ∼idz ∧ dz

2(

1 + |z|2)2

2dz ⊗ dz(1 + |z|2

)2

for every z ∈ C. Then take the square root of this guy:

dS (0, z) =

ˆ |z|0

dr

1 + r2= 2 tan−1 |z| ≤ π.


Denition. Let F be a family of meromorphic functions on a domain Ω ⊂ C. We say F is normal, if forall fn ⊂ F , has convergent subsequence fnk with respect to dS such that

dS (dnk , g)→ 0 as k → +∞(or replace with convergence uniformly on every compact subset) where g is a meromorphic function on Ω.

Theorem. (Marty's criterion)Let F be a family of meromorphic functions on Ω. Then F is normal ⇐⇒ for every compact K ⊂ Ω,

we have thatf?ωS

is uniformly bounded for every f ∈ F .This is equivalent to: For every compact K ⊂ Ω ,

∃MK > 0,|f ′(z)|

1 + |f(z)|2≤MK , ∀f ∈ F .

Proof. Lets rst prove (⇐=): By A−A lemma, for all fn ⊂ F , we have that fn is equibounded, anddS (fn, 0) ≤ π, ∀n.

To check fn is equi-continuous on K. Then for all z1 ∈ K, pick B(z1; δ) ⊂ Ω, for all z2 ∈ B(z1, δ) we havethat

dS (fn (z1) , fn (z2)) ≤ˆf(γ)

ωS =

ˆγ

f?ωS

≤ MK |z1 − z2| ,which shows this family is equicontinuous. Then apply Arzella-Ascoli's Theorem.


4/23/2015

• C - The Riemann sphere.

Proposition. Aut(C)

= SL (2,C) \ ±I2. Here

SL (2,C) =

(a bc d

)| a, b, c, d ∈ C, ad− bc = 1

.

Proof. Let ϕ ∈ Aut(C). If ϕ (∞) =∞, then ϕ ∈ Aut (C) which implies ϕ(z) = az + bz for a, b ∈ C.

Now if ϕ (∞) 6= ∞, let ψ(z) = 1ϕ(z)−ϕ(∞) then ψ (∞) = ∞ and ψ ∈ Aut (C) . Which implies that

ψ(x) = cz + d. But then

ϕ(z) =1

cz + d+ ϕ(∞)

=az + b

cz + d,

by letting a = cϕ(∞) and b = dϕ(∞) + 1. Let F : SL (2;C)→Aut(C)by(

a bc d

)7→ az + b

cz + d,

and notice that kerF = ±I2. Since its surjective and a homomorphism then First isomorphicm theroem

says that Aut(C)∼= SL(2;C)/ ±I2.

• Remark: We have that Aut(C)∼= GL (2;C) / λI2;λ ∈ C\ 0.

Proposition. 2. We have dS on C with

ωS =4idz ∧ dz/2(

1 + |z|2)2

=4dz ⊗ dz(1 + |z|2

)2 ,

For every z, w ∈ C we have that

dS (z, w) = infγ

ˆγ

2 |dz|√1 + |z|2

,

and

dS (z, w) =

ˆγ

2 |dz|√1 + |z|2

, γ=great circle passing thoughz, w

= 2 tan−1

∣∣∣∣ z − w1 + wz

∣∣∣∣ (1).


• One way to see (1). Assume w = 0 and get that

dS (z, 0) =

ˆ |z|0

2dr

1 + r2= 2 tan−1 |z| .

• If w 6= 0 let ϕ(z) = z−w1+wz ∈ Aut

(C). Note that

ϕ?ωS =4 |ϕ′(z)|2 dz ⊗ dz(

1 + |ϕ(z)|2)2

=4dz ⊗ dz(1 + |z|2

)2 .

• Then

dS (z, w) = dS (ϕ(z), ϕ(w))

= dS (ϕ(z), 0)

= 2 tan−1 |ϕ(z)|

= 2 tan−1

∣∣∣∣ z − w1 + wz

∣∣∣∣ .Theorem. (Marty's criterion) Let F be a family of meromorphic functions on Ω ⊂ C. Then F is normal⇐⇒ for every compact K ⊂ Ω , there exists MK > 0 such that

supf∈F

supz∈K

|f ′(z)|1 + |f(z)|2

≤MK .

Proof. (⇐=) Done last class.

( =⇒ ) Suppose NOT. Then there exists a K such that and fk ⊂ F such that|f ′k(z)|

1+|fk(z)|2 is unbounded

on K. Since F is normal. Then fkdS⇒ g on K. If supK |g| <∞, by Weirstrass, g is holo on K, and fk ⇒ g.

Now|f ′k(z)|

1 + |fk(z)|2→ |g′(z)|

1 + |g(z)|2

which the thing on the right is bounded on K. If g (z0) =∞, for some z0 ∈ K. Pick B (z0; δ) ⊂ Ω, such that1g is bounded on B (z0; δ), so is 1

fnkfor large k. By Weiretrass, we have that 1

g is holomorphic and 1fnk⇒ 1

g

. But ∣∣∣∣( 1g

)′(z)

∣∣∣∣1 +

∣∣∣ 1g ∣∣∣2 ←∣∣∣∣( 1fnk

)′(z)

∣∣∣∣∣∣∣∣1 +(

1fnk

)2

(z)

∣∣∣∣ =

∣∣f ′nk(z)∣∣

1 + |fnk(z)|2bounded on.

Lemma. 2. Let Fbe a family of meromorphic functions on Ω. If for every f ∈ F then

f (Ω) ⊂ C\ P,Q,R

where P,Q,R are 3 distinct points, then F is normal.


Proof. Can assume P,Q,R = 0, 1,∞ otherwise, ϕ(z) = (z, P,Q,R) = z−Pz−R

Q−RQ−P . Now ϕ : C → C,

replace f by ϕ f . so f (Ω) ⊂ C\ 0, 1. Note it suces to show supf∈F f?ωS is bounded on each B (z0;R)

and B (z0, R) ⊂ Ω. Recall in the proof of PLT, we construct a metric ω on C\ 0, 1. Now

ω =

(1 + |z|2)2

|z|2[

log|z|2

1 + |z|2

]2

·(

1 + |z − 1|2)2

|z − 1|2[

log|z − 1|2

1 + |z − 1|2

]2−1

· idz ∧ dz2

with K (ω) ≤ −δ < 0. Note ωS = 4idz∧dz/2(1+|z|2)

2 . We have

ωSω

=

(1 + |z|2)2

|z|2[

log|z|2

1 + |z|2

]2

·(

1 + |z − 1|2)2

|z − 1|2[

log|z − 1|2

1 + |z − 1|2

]2

≤ C on C\ 0, 1 .

where

log|z|2

1 + |z|2= log

[1− 1

1 + |z|2

]∼ − 1

1 + |z|2as |z| → ∞

using log (1− x) ≤ x. And similarly

log|z − 1|2

1 + |z − 1|2∼ −1

1 + |z − 1|2,

as z →∞. This implies that ωS ≤ Cω, which implies that

f?ωS ≤ Cf?ω

≤ C

δ

4R2(R2 − |z|2

)2

2dz ∧ dz2

.

for every B(0; r) ⊂ B (0;R). So

supf∈F

f?ωS ≤ CR2

(R2 − r2)2 on B(0;R),

which nishes the proof.

Theorem. (Picard's big theorem) If f has an essential singularity at z = a and f is holomorphic on B? (a; δ)then for every δ1 satisfying 0 < δ1 < δ , we have

f (B? (a; δ1)) = C or C\ one point .


Proof. Suppose NOT. Then there exists a δ1 < δ such that

f (B? (a; δ1)) ⊂ C\ b, c .

Can assume b, c = 0, 1 by considering the map f 7→ f(z)−bc−b . Can ssume a = 0 by considerig the map

z =7→ z − a. So we havef (B? (a; δ1)) ⊂ C\ 0, 1 = C\ 0, 1,∞ .

Let for every n ∈ N be fn(z) = f(zn

)for 0 < |z| < δ2. The F ≡ fn omits 0, 1,∞. By Lemma 2, F is

normal. This means that there exists subsequence fnk of fn such that

fnkdS→ g

uniformly on every compact K ⊂ B? (0; δ1). Let K =|z| = δ1

2

. If supz∈K |g(z)| < ∞, then g is holo on

K. We have fnk ⇒ g on K. Now fnk(z) = f(znk

)is bounded for large k, which implies

|f(z)| ≤M,∀ |z| = δ12nk

.

By maximum principle,|f(z)| ≤M, ∀0 < |z| < δ1/2

which implies z = 0 is a removable singularity, which is a contradiction. If g (z0) =∞ for some z0 ∈ K, pick

B (z0; δ0) ⊂ Ω such that 1g is holo. We also have 1

fnk→ 1

g on B (z0, δ). A contradiction.


4/28/2015

• Homeowor Problems Set 6• #1, If g : Ω→ D and Ω ⊂ D then take g−1.• #2: Consider

√f has a branch on B (0, δ)if f(0) 6= 0 . If f(0) = 0 then f(z) = zmh(z) on B(0; δ) .

Now f(z2) = z2mh(z2) because√f(z2) =

√z2mh.

Now zm is a branch of√z2m. Take zmh.

• #3, Should be C\ f(z) | f(z) < M Problem 7, 2010 Aug. and this is |w| ≥M \ one point. The statement is Not true! C\ z | |f(z)| ≤M is NOT true.

Example: f(z) = sin z and |f |2 = sinh2 y + sin2 x and then |f(z)| ≤ 2 contains a strip whichexlcudes the real axis. So C\ z | |f(z)| < M has two components.

Remark1: The statemeent is true if C is replaced by C. The point: C\ z | |f(z)| < M eachconpoene tof is unbounded.

Remark2: The original statement is true if f is a polynomial.∗ More generally, we say a function f is proper if f−1 (compact set in C) is compact in C.∗ Prop: If f is proper, then C\ z | |f(z)| < M has at most one component.

Remark3: An entire function is proper if and only if f is a polynomial.∗ (Weisrtrass) ⇐⇒ For every unbounded zn , f (zn) is unbounded.∗ Idea: If C\ z | |f(z)| < M has two components E1

⊔E2. Then one of them say E1, has

to be bounded. Then apply max princinple to E1 and get a contradiction.

• #4, Apply the downward box to e−z2

and get e−a2

4b2√π.

Original integral´∞−∞ eiax−bx

2

dx.• #5, Use

F (z) =

f(z) z ∈ D1

f( 1z )

z /∈ D.

On ∂D we have |f(z)| = 1 and using 1z = z get f(z) = 1

f(z)and so its well dened on ∂D.

Now must verify this is holomorphic:

∂

∂z

(1

f(

1z

)) =1(f)2 (− ∂

∂zf

)

=1(f)2(− ∂

∂zf

(1

z

))

=1(f)2 (−0) = 0,

if f(

1z

)6= 0. N ow this happens if and only if f(z) 6= 0 in z ∈ D. and 1

z and z are symmetricwith respect to ∂D.

Now for allz ∈ ∂D|f(z)| = 1 implies f has nitely many zeros unless f ≡ eia. Then appply symmetry princinples on a smaller circle inside D that contains all the zeros andapply it there.

Now f(a1(= 0 implies F(

1a1

)= 1

f(a1)=∞ which implies F has poles at a1, . . . , aN .

• #5b) With F (z) = eiα∏Nj=1

z−aj1−ajz . Let h(z) =

∏Nj=1

F (z)(1−ajz)z−aj . Then h(z) has removable singu-

larity at a1, . . . , am. Now consider 1a1, . . . , 1

am.


Then h is holo on C. Since |h|∂D ≡ 1 and h is nowhere zero in D. which implies h ≡ c = eia inD on C.

• #6, By Montel's Theorem, the sequence fk we have that fkl⇒cpt g on Ω. Now E = z | lim fk = f ⊂z | g = f has a limit point. Thus fkl ⇒cpt f . We need to show the whole sequence convergesuniformly to f , not just the subsequence. Method1(Prove by contradiction): We have that for every compact K ⊂ Ω and for every ε > 0there exists N ∈ N such that

supz∈K|fk(z)− f(z)| < ε, ∀k ≥ N.

There exists a compact K ⊂ Ω compact, ∃ε,∀N ∈ N,∃k ≥ N such that

supz∈K|fk(z)− f(z)| ≥ ε

which implies there exists zl∞l=1 such that |fkl (zl)− f (zl)| ≥ ε. Since K is compact zlm →z? ∈ K by Bolzona Weistrass, Thus there exists

fklm⇒ f

by Montel's again, but this is a contradiction of the epsilon.

Method2: It sufccies to show E = Ω. For every K ⊂ Ω compact let K =⋃Nj=1B (aj ; δ) for

nitely many. For 1 ≤ j ≤ N we get

sup1≤j≤N

|fk (aj)− fl (aj)| < ε k, l ≥ L

for every z ∈ K, we have that z ∈ B(aj ; δ) and choosing δ small enough we get

|fk(z)− fl(z)| ≤ |fk(z)− fk(aj)|+ |fk(aj)− fl(aj)|+ |fl(aj)− fl(z)|< 3ε.

Now fk Cauchy sequence on K and fk ⇒ f . To show E = Ω, suppose z0 ∈ Ω\E. Montel's,

there exists fnk (z0)→ A 6= f(z0). Nowfnkl

⇒K g and g = f on Ω. Then

A = lim fnkl = g(z0) = f(z0)

which is a contradiction.• #7: If

´γ1f =

´γ2f If yes, then there exists F =

´ zz0f such that F ′ = f .

Now ˆγ1−γ2

f =

˛Γ

f =

p∑i=1

Res (f, ai) = 0

by the second problme is not equal to zero.• Final Exam:

3 problems, haven't yet made it 1. problem from the last two homework sets. HW5/6 1. problem from his favorite questions.

∗ Q1 punturee disk∗ Q2 rst problem in HW6∗ Q3 If I give you a holomorhic funciton on the disk, if there exist an arc on the circle γ(arc only on a part of the circle) with f |γ≡ c then shjow that f ≡ c on D.· Use Swarzat reection princinple. Consider H extend, thus f has to identically zero.simialr to problem, its like problem 5, just use symmetry princinple.


∗ Q+ Residue(no an original favorite problem) 1 from a previous prelim, one haven't done yet on HW, like one where you can use Swarz lemma.

• 5121, Cpx2, 5411 PDE• 5030 dierential geometry.

math 5120 - complex analysis- lecture notesmariano/notes/math 5120 - complex analysis...math 5120 -...

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