Selected Solutions to Ahlfors

Selected Solutions to  Complex Analysis  by Lars Ahlfors Matt Rosenzweig

1

Contents Chapter 4 - Complex Integration

Cauchy’s Integral Formula . . 4.2.2 Exercise 1 . . . . . 4.2.3 Exercise 1 . . . . . 4.2.3 Exercise 2 . . . . . Loca ocal Proper perties ies of Analytica ical 4.3.2 Exercise 2 . . . . . 4.3.2 Exercise 4 . . . . . Calculus of Residues . . . . . 4.5.2 Exercise 1 . . . . . 4.5.3 Exercise 1 . . . . . 4.5.3 Exercise 3 . . . . . Harmonic Functions . . . . . 4.6.2 Exercise 1 . . . . . 4.6.2 Exercise 2 . . . . . 4.6.4 Exercise 1 . . . . . 4.6.4 Exercise 5 . . . . . 4.6.4 Exercise 6 . . . . . 4.6.5 Exercise 1 . . . . . 4.6.5 Exercise 3 . . . . .

4

. . . . . . . . . . . . . . . . . . . . . . . . Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Chapter 5 - Series and Pro duct Developments

Power Series Expansions . . . . . . 5.1.1 Exercise 2 . . . . . . . . 5.1.2 Exercise 2 . . . . . . . . 5.1.2 Exercise 3 . . . . . . . . Partial Fraction ions and Factorization 5.2.1 Exercise 1 . . . . . . . . 5.2.1 Exercise 2 . . . . . . . . 5.2.2 Exercise 2 . . . . . . . . 5.2.3 Exercise 3 . . . . . . . . 5.2.3 Exercise 4 . . . . . . . . 5.2.4 Exercise 2 . . . . . . . . 5.2.4 Exercise 3 . . . . . . . . 5.2.5 Exercise 2 . . . . . . . . 5.2.5 Exercise 3 . . . . . . . . Entire Functions . . . . . . . . . . 5.3.2 Exercise 1 . . . . . . . . 5.3.2 Exercise 2 . . . . . . . . 5.5.5 Exercise 1 . . . . . . . . Normal Families . . . . . . . . . . 5.5.5 Exercise 3 . . . . . . . . 5.5.5 Exercise 4 . . . . . . . .

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4 4 4 5 6 6 6 7 7 7 8 10 10 10 11 12 13 13 13 14

2

14 14 15 16 16 16 17 18 18 19 19 20 20 21 22 22 23 23 24 24 25

Conformal Mapping, Dirichlet’s Problem

27

The Riemann Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Elliptic Functions

Weierstrass Theory . 7.3.2 Exercise 2 7.3.3 Exercise 1 7.3.3 Exercise 2 7.3.3 Exercise 3 7.3.3 Exercise 4 7.3.3 Exercise 5 7.3.3 Exercise 7 7.3.5 Exercise 1

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Chapter 4 - Complex Integration Cauchy’s Integral Formula 4.2.2 Exercise 1 Applying the Cauchy integral formula to  f (  f (z ) =  e z ,

 

 

1 f (z ) 1 =  f (0)  f (0) = dz 2πi |z|=1 z

 ⇐⇒  ⇐ ⇒ 2πi  2 πi =  =

|z|=1

ez dz z

Section 4.2.2 Exercise 2

Using partial fractions, fractions, we may express express the integrand integrand as z2

1 i = +1 2(z 2(z  + i)

i  − 2(z 2(z − i)

Applying the Cauchy integral formula to the constant function  f (  f (z ) = 1,

 

1 1 i dz = dz  = 2 2πi |z |=2 z + 1 2

    1 2πi

|z|=2

1 dz z + i

−

i 2

    1 2πi

1

|z|=2 z − i

dz = dz  = 0

4.2.3 Exercise 1 1. Applying Cauchy’s Cauchy’s differentiation formula formula to f  to  f ((z ) =  e z , (n − 1)! 1 =  f  − (0) = (n 1)

2πi

 

|z|=1

ez dz zn

 

2πi ez = dz (n 1)! |z|=1 z n

 ⇐⇒  ⇐ ⇒ −

2. We consider the following following cases: cases: (a) If  n  n  0,  0 , m  0, then it is obvious from the analyticity of  z of  z n (1 the integral is 0.

 ≥

(b) If  n  n

− z)

 ≥

m

and Cauchy’s theorem that

 ≥ 0,  0 , m <  0, then by the Cauchy differentiation formula,

 

z n (1 z )m dz = dz  = ( 1)m

−

|z|=2

−

 

|z|=2

zn dz = dz  = (z 1)|m|

−



0 ( 1)m 2πi ( m 1)! (n

− | |−

n! m +1)!

−| |

= (−1)|m| 2πi

(c) If  n  n <  0,  0 , m

 

zn

|z|=2

 ≥ 0, then by a completely analogous argument, 0 (1 − z ) | |− (1−z ) dz = dz  = dz = dz  =

 

m

|z|=2

m

z |n|



n

( 1) (n

−

1

2πi

| |−1)!

  n m 1

| |−

n< m n m

m< n

 

m! |n|−1 2πi m −|n|+1)! = (−1) |n|−1

(m

 | | − 1 m ≥ n

(d) If  n  n <  0,  0 , m <  0, then sincen sincen( z  = 2, 0) =  n(  n ( z  = 2, 1) = 1, we have by the residue formula that

||

 

|z |=2

(1

− z)

||

m n

z = 2πires( πires(f  f ;; 0) + 2πi 2πires( res(f  f ;; 1) =

 

|z|

4

= 12

(1

 | | − 1  ≥ | |

− z)

m n

z dz + dz +

 

= 12

|z−1|

(1

− z)

m n

z dz

Using Cauchy’s differentiation formula, we obtain

 

(1

|z |=2

=

− z)

2πi ( n 1)!

 

m n

z dz = dz  =

(1

− z)−|

|

m

z |n|

dz + dz  +

 

z −|n| dz (1 z )|m|



|z −1|= 12 −  ( |m| + |n| − 2)!  ( −1)|m| 2πi  ( −1)|m|−1 (|n| + |m| − 2)! |z|= 12

| | − 1)! + (|m| − 1)! · (|n| − 1)! |m| + |n| − 2 − |m| + |n| − 2  = 0 = 2πi 2 πi |n| − 1 |n| − 1 −4 3. If  ρ  ρ  = 0, then it is trivial that | |= |z − a| |dz|  = 0, so assume otherwise. If  a If  a =  = 0, then | | −  ·

(m

    z







ρ

|z|=ρ

|z|−4 |dz| =

 

1

2πi ρ3

ρ−4 2πiρdt = πiρdt =

0

Now, assume that a that  a = 0. Observe that

 

1

|z − a|

  | − |    − z

|z|=ρ

4 a − |dz |  =

1

=

i

(ρe2πit

0

 

|z |=ρ (z −

1 2 a) (z

4

1

= (z

− a)2(z − a)2 1

− a)2 |dz| =

 −  

ρ2πie 4πit i dt = dt  = a)2 (ρ ae2πit )2 ρ |z|=ρ (ρ

−

−

  0

(ρe2πit

z

−

a 2 ρ z ) (z

− a)

−

1 2 a) (ρe−2πit

dz = dz  = 2

  −

 −iρ 2

a

2πie 4πit ρ dt a)2 ie4πit z ρ2

|z|=ρ (z − a  ) 2 (z − a)2

We consider two cases. First, suppose a  > ρ. Then z Then  z (z a)−2 is holomorphic on and inside 2 and ρa   lies inside z  = ρ  =  ρ . By Cauchy’s differentiation formula,

 {| |

 

|z|=ρ

 | |

}

|z − a|−4 |dz| = 2πi −aiρ2



(z

−

− a)−2 − 2z(z − a)−3



2

z = ρa

=



2πρ 2

a2 ( ρa

2 2  −2πρ( πρ (ρ2 + |a| ) 2πρ( πρ (ρ2 + |a| ) = 2 2 (ρ2 − |a| )3 (|a| − ρ2 )3   lies outside |z |  = ρ  =  ρ,, so the function  z (z −

dz

 {|z| = ρ  =  ρ }

ρ2

− a)2 1 − 2 a( − a) ρ2 a



=

2

Now, suppose a  < ρ. Then ρa inside z  = ρ  =  ρ . By Cauchy’s differentiation formula,

 | |  {| | }

 

|z |=ρ



2 2 |z − a|−4 |dz| = 2πi −iρ2 (z −  ρ  ) −2 − 2z(z −  ρ  ) −3

a

−2πρ (a + = 2 (|a| − ρ2 )2 a −

a

ρ2 a  ) ρ2 a

a



= z =a

ρ2  ) a

−2 is holomorphic on and

2πρ a2 (a

−

ρ2 2 a  )



1

a

−2 − (a

ρ2 a  )



2 2  − 2πρ( πρ(|a| + ρ2 ) 2πρ( πρ(|a| + ρ2 ) = = 2 2 (|a| − ρ2 )3 (ρ2 − |a| )3

4.2.3 Exercise 2 C  be a holomorphic Let f  : C holomorphic function function satisfying the following following condition: condition: there exists R >  0 and  n N n such that f (z )  < z z  R.  R . For every every  r  R,  R , we have by the Cauchy differentiation formula that for all m all  m > n, n, n m! z m! (m) f  f  (a) dz 2π |z|=r z m+1 rm−n

 →  | |  | | ∀ | | ≥

Noting that m that  m may write

 ∈

 ≥

  ≤ ≤  

| |  | | ≤ ||

− n ≥ 1 and letting  r  → ∞, we have that f  that  f (

m)

(a) = 0. Since f  Since  f  is  is entire, for every  a

 ∈ C , we

 f (n) (a)  f ( f (z ) =  f (  f (a) + f  (a)(z )(z − a) + · · · + (z − a)n + f n+1 (z )(z )(z − a)n+1 ∀z ∈ C n!

where f  where  f n+1   is entire. entire. Since f  Since f n+1 (a) =  f (n+1) (a) = 0 and a Hence, f  Hence,  f  is  is a polynomial of degree at most  n.  n . 5

 ∈ C  was arbitary, we have that f  +1 ≡  0 on C. n

Local Properties of Analytical Functions 4.3.2 Exercise 2



Let f  Let  f  : C . Consider the function g function  g((z ) =  f  z1 C be an entire function with a nonessential singularity at n n−1 at z  = 0. 0. Let Let n N  be minimal such that limz→0 z g (z ) = 0. Then Then the functi function on z g (z ) has an analytic continuation  h(  h (z) defined on all of  C. By Taylor’s Taylor’s theorem, theorem, we may express express  h(  h (z ) as

→

 ∞

 ∈

 h  (0)  h  (0) 2  h (n−1) (0) n−1 − z g (z ) =  h(  h (z ) = h =  h(0) (0) + z  + z + ··· + z + hn (z )z n ∀z  =0 n 1

1!

cn−1

where h where  h n  :

C

2!

          − −

(n

cn−2

c0

→ C is holomorphic. Hence,

cn−1 cn−2 + n −2 + n 1 − z z

lim g (z )

z

→0

And

1)!

  −    

· · · + c0

 = lim zh n (z ) = 0 z

→0

cn−1 cn−2 + n−2 + + c0  = lim f ( f (z ) =  f (0)  f (0) n − 1 z →∞ z →0 z z cn−1 2 since f   f   is entir entire. e. Note Note that we also obtain obtain that c0 = f (0). (0). Hence, Hence, g (z ) + zcnn− + c0 (we −2 + z n−1 are abusing notation to denote the continuation to all of  C) is a bounded entire function and is therefore identically zero by Liouville’s theorem. Hence, lim g (z )

∀z = 0, f ( f (z ) =  c

···

n 1z

−

 −



···



n 1

− + cn−2 z n−2 + · · · + c0

Since f  Since  f (0) (0) = c =  c 0 , we obtain that f  that  f  is  is a polynomial.

4.3.2 Exercise 4  C Let f  Let  f  :  C  be a meromorphic meromorphic function function in the extended extended complex plane. plane. First, I claim that f   f   has finitely many many poles. Since the poles of  f  are f  are isolated points, they form an at most countable subset ∞ ˜  pk k=1 of  C. By definition, the function f ( f (z ) =  f  z1  has either a removable singularity or a pole at  z  = 0. ∞ In either case, there exists r >  0 such that f  is f ˜ is holomorphic on D (0; r). Hence, Hence,  pk k=1  D(0;  D (0; r). Since Since ∞ this set is bounded,  pk k=1  has a limit point p point  p.. By continuity, continuity,  f (  f ( p)  p) =  and therefore  p  is a pole. Since p Since  p is an isolated point, there must exist N  exist  N  N  such that k  N,  N , pk  =  p.  p . Our reasoning in the preceding Exercise 2 shows that for any pole pk =  of order mk , we can write in a neighborhood neighborhood of  p  p k

∪ {∞ {∞}} → ∪{∞}



{ }

 { }

 ∈  ∈

  

 ∀  ≥

cmk cmk −1 f (z ) = + + (z  pk )mk (z  pk )mk −1

−

−

···

  

c1 + + c0 +gk (z ) z  pk

−



f k (z )

where g where  g k  is holomorphic in a neighborhood of  p  p k . If  p =  p  =

  

 { }  ⊂

 ∞    ∞

 ∞ is a pole, then analogously,  c 1 ∞ −1 · · · + + + c0 +˜ g∞ (z ) ∞ −1 z

cm cm f ( f ˜(z ) = m∞ + z ∞ zm

  



˜∞ (z ) f 

where g˜∞  is holomorphic holomorphic in a neighborhood neighborhood of 0. For clarification, clarification, the coefficients coefficients cn   depend on the pole, 1 ˜ but we omit the dependence for convenience. Set  f ∞ (z ) = f ∞ z  and

  − n

h(z ) =  f (  f (z ) − f ∞ (z )

f k (z )

k =1

I claim that h that  h  is (or rather, rather, extends extends to) an entire, entire, bounded function. function. Indeed, Indeed, in a neighborhood of each  z k , n h  can be written as  h(  h (z ) =  gk (z )  f  ( z ) and in a neighborhood of  z  z ∞ as h as  h((z ) =  g ∞ (z ) i k=1 f k (z ), =k k ˜ (z ) =  h 1  is evidently bounded in a neighborhood of 0 since which are sums of holomorphic functions. h z the f  the  f k z1  are polynomials and f  z1 f ∞ z1  = g˜∞ (z), which is holomorphic holomorphic in a neighborhood neighborhood of 0. By Liouville’s theorem, h theorem,  h  is a constant. It is immediate from the definition of  h of  h  that f   that  f  is  is a rational function.

    −  

−



6

−



Calculus of Residues 4.5.2 Exercise 1 Set f ( f (z ) = 6z 3 and g (z ) = z 7  2z  2 z 5  z + 1. Clearl Clearly y, f, g   are entire, entire, f ( f (z ) > g (z ) z   = 1, and 7 5 3 f ( f (z ) + g +  g((z ) = z 2z + 6z 6z z  + 1. By Rouch´e’s e’s theorem, f  and f  + g  +  g  have the same number of zeros, which is 3 (counted with order), in the disk z   g (z ) z  = 2. By Rouch´e’s e’s theorem, z 4 6z +3 has 4 roots (counted (counted with order) in the open disk z   g (z ) z  = 1. By Rouch´e’s e’s theorem, z 4 6z  + 3 = 0 has 1 root in the in the open disk z  > 0, >>  0,

−1 2

z √  √  √  )(z + √ 2i) − 3i)(z )(z + 3i)(z )(z − 2i)(z

γ 1  : [ R, R]

 → C, γ 1(t) =  t;  t ; γ 2  : [0, [0, π] → C, γ 2 (t) =  Re

−

it

and let γ  let  γ  be  be the positively oriented closed curve formed by  γ 1 , γ 2 . By the residue formula and applying iz the Cauchy integral formula to ze+ai  to compute res(f  res( f ;; ai), ai),

 

√ 

√ 

f ( f (z )dz = dz  = 2πires( πi res(f  f ;; 3i) + 2πi 2 πires( res(f  f ;; 2i)

γ 

It is immediate from Cauchy’s integral formula that

√ 

√  √  √  √  √  √  · 3π − − √ 3|= √  √  √  z 2 (z  + i 2)−1 (z 2 + 3)−1 (i 2)2 √  √  √  ·  − dz = dz  = 2 πi = 2π (z − i 2) ((i ((i 2)2 + 3)(2i 3)(2i 2) | − √ 2|=

  |  

2πires( πi res(f  f ;; 3i) =

z i

√ 

2πires( πires(f  f ;; 2i) =

z

i



(i 3)2 z 2 (z + i 3)−1 (z 2 + 2)−1 dz = dz  = 2πi = (z i 3) ((i ((i 3)2 + 2)(2i 2)(2i 3)



Using the reverse triangle inequality, we obtain the estimate

   

γ 2

Hence, 2

 

∞

0

f (z )dz

 ≤ |

R2

πR 3 3 R2

 0 , R → ∞ − | | − 2| →  0,

∞ x2 x2 = dx = dx  = ( 3 x4 + 5x 5x2 + 6 5x2 + 6 −∞ x4 + 5x

√  √  − 2)π 2)π ⇒

8

 

∞

0

√  − √ 

x2 dx ( 3 2)π 2)π = 4 2 x + 5x 5x + 6 2

(e) We may write cos(x cos(x) eix = Re 2 x2 + a2 ( x + a2 ) eiz z 2 +a2 ,

So set f  set  f ((z ) =

which has simple poles at

 ±ai. ai. First, suppose that a that  a   = 0. For R For  R >> 0, >>  0, define [0, π] → C, γ 2 (t) =  Re γ 1  : [−R, R] → C, γ 1 (t) =  t;  t ; γ 2  : [0, it

and let γ  let  γ  be  be the positively oriented closed curve formed by  γ 1 , γ 2 . By the residue formula,

        

i(ai)

f ( f (z )dz = dz  = 2πires( πi res(f  f ; ai) ai) = 2πi

γ 

π

f ( f (z )dz =

0

γ 2

π

≤

since e since  e −R sin(t)



eiR[cos(t)+i sin(t)] it Re dt = R2 e2it + a2 Re−R sin(t) R2

0

≤ 1 on [0, [0, π]. Hence,

−

a2

dt

−a

πe ·  e(2ai = (2ai)) a

 

π



eiR cos(t) e−R sin(t) it Re dt R2 e2it + a2

0

 ≤ R2πR  0 , R → ∞ − a2 →  0,

 

 

∞ cos(x cos(x)

1 ∞ eix πe−a = Re dx = dx  = x2 + a2 2 −∞ x2 + a2 2a 0 If  a  a  = 0, then the integral does not converge. log(z ) (h) Define f  Define  f ((z ) =   (1+  [ π2 , 32π ). For R For  R >> 0, >>  0, z 2 ) , where we take the branch of the logarithm with arg( z ) define 1 1 −it  1 it γ 1  : [ R, ]  t ; γ 2  : [, π] e ; γ 3  : [ , R]  t ; γ 4  : [0, [0, π] C, γ 1 (t) =  t; C, γ 2 (t) = C, γ 3 (t) =  t; C, γ 4 (t) =  Re R R R and let γ  let  γ  be  be the positively oriented closed curve formed by the  γ i .

 ∈ −

−  −  →

 −

 →

   

 ≤    | |    ≤   | | −| | − log R

π

f (z )dz

0

γ 2

 →

3π

R(log |R| + 2 )  ≤  π  0 , R → ∞ 1 |R2 − 1| → 0, R 1 log R + it R(log |R| + π) → 0, Rdt ≤ π  0 , R → ∞ 2 R 1 R2 − 1 dt

R2

π

f ( f (z )dz

0

γ 4

−1 + 3π 2 1

 →

By the residue formula and applying the Cauchy integral formula to  f (  f (z )/(z + i) to compute res(f  res( f ;; i),

 

π

f ( f (z )dz = dz  = 2πires( πi res(f  f ;; i) = 2πi

γ 

2

log(z ) π 2 ·  (log(z | = =  = 2πi · z  + i) 2i 2 z i

Hence, π2 = 2

− 1 log(te log(teiπ ) f ( f (z )dz+ dz + f ( f (z )dz = dz  = dt+ dt+ γ 1 γ 3 −R 1 + t 2

 

 

    R

=2

1

R

 

 

R

log(t log(t) +π 1 + t2

 

R

1

R

R

1

R

 

1

 

− R log( t ) log(t log(t) dt = dt  = dt+ dt+ 1 + t2 −R 1 + t2

1 dt = dt  = 2 1 + t2

 

R

1

R

||

R

1

R

 

log(t log(t)  π 2 + 1 + t2 2

∞ where we’ve used 0 1+1t2 dt = dt  = limR→∞ arctan(R arctan(R) − arctan(0) = π2 . Hence, R ∞ log(t log(t log(t) log(t)

  1

R

1 + t2

   ⇒

dt = dt  = 0

0

Lemma Lemma 1. Let  U, V 

1 + t2

dt = dt  = 0

⊂ C  be open sets, F  : U  → V   V   a holomorphic function, and  u : V  → C  a harmonic   → C is harmonic.  function. Then  u  u ◦ F  :  U  → Proof.   Since u Since u ◦ F  is F  is harmonic on U  on  U  if and only if it is harmonic on any open disk contained contained in U  in  U  about  about every point, we may assume without loss of generality that V   V   is an open disk. Then there there exists a holomorphic holomorphic function G  :  V  → C  such that  u  = Re(G Re(G). Hence, Hence, G ◦ F  :  U  → C  is holomo holomorph rphic ic and Re (G ◦ F ) F ) =  u ◦ F , F , which shows that u that  u ◦ F   F   is harmonic. In what follows, a conformal map  f  : Ω → C  is a bijective holomorphic map. 9

1

−R 1 log(t log(t) dt+ dt + π dt 1 + t2 −R 1 + t 2

Harmonic Functions 4.6.2 Exercise 1 Let u  :  D  (0; ρ) bounded. I am going to cheat a bit and assume Schw Schwarz’s arz’s theorem theorem R  be harmonic and bounded. for the Poisson Poisson integral integral formula, formula, even though Ahlfors discusses it in a subsequen subsequentt section. section. Let

 →

1 P u (z ) = 2π

 

2π

Re

0

reiθ z u(reiθ )dθ iθ re + z

−

denote the Poisson integral for u  on some circle of fixed radius r < ρ. Sinc Sincee u   is continuous, P u (z ) is a harmonic function in the open disk  D(0;  D (0; r) and is continuous on the boundary z  = r  =  r . We want to show that u that  u and  and P   P u  agree on the annulus, so that we can define a harmonic extension of  u of   u by  by setting u setting  u(0) (0) = P  =  P u (0). Define g(z ) =  u(  u (z ) P u (z )

 {| |

}

−

and for  for   >  0 define g (z ) =  g(  g (z ) +  log

| | ∀ z r

0 < z

 | | ≤ r

Then g   is harmonic in D (0; r) and continu continuous ous on the boundary boundary.. Further urthermor more, e, since since u   is bounded by hypothesis and P u  is bounded by construction on D(0; r), we have that g  is bounded on D(0; r). g (z ) is harmonic in D in  D  (0; r) and continuous on the boundary since both its terms are. Since log r−1 z ,z 0, we have that limsup g (z )    0 such that 0 < z δ  g (z )   0. 0. Sinc Sincee g  is harmonic on the closed annulus δ  z  r , we can apply the maximum principle. Hence, g Hence,  g   assumes its maximum in z  = δ   =  δ  z  = r  =  r . But, g But, g (z )  0 z  = δ   =  δ , by our choice of  δ   δ , and since u, since u, P u  agree on z  = r  =  r , we have that g that  g (z ) = 0 z  = r  =  r.. Hence, g (z )  0 0 < z  r

 | | ≤  ⇒  ⇒

{  ≤  ≤ | | ≤ }  ≤ ∀ | |

 ≤

{| |

 {| |

}

}∪{| } ∪{| | } ∀| |

 ≤  ∀  | | ≤ Letting  Letting    → 0, we conclude that  g (z ) ≤ 0  ∀0  <  |z | ≤ r,  r , which shows that u that  u  ≤ P   on the annulus. annulus. Applying Applying the same argument to h  = P   =  P   − u, we conclude that  u  =  P  on 0  <  |z | ≤ r.  r . Setting u(0) = P  =  P  (0) defines a u

u

u

u

harmonic extension of  u of  u  on the closed disk.

4.6.2 Exercise 2 If  f   f  : Ω = r1  < z  < r2 identically zero, then there is nothing to prove. prove. Assume Assume otherwise. otherwise. Since C  is identically the annulus is bounded, f  bounded,  f  has  has finitely many zeroes in the region. Hence, for  λ R, the function

 {

 | |

}→

 ∈

g(z ) =  λ log z + log f ( f (z )

||

|

|

is harmonic in Ω a1 ,  , an , where a1 ,  , an  are the zeroes of  f . f . Applying Applying the maximum maximum principle to g (z ), we see that g (z )  takes its maximum in  ∂ Ω.  ∂ Ω. Hence,

\ { ··· } ···  | | λ log |z | + log |f ( f (z )|  = g  =  g((z ) ≤ max {λ log(r log(r1 ) + log(M  log(M (r1 )), )), λ log(r log(r2 ) + log(M  log(M (r2 ))} ∀z ∈ Ω \ {a1 , · · ·  , a }  |z| = r Thus, if  |  =  r,, then we have the inequality λ log(r log(r) + log log (M (r)) ≤ max {λ log(r log(r1 ) + log(M  log(M (r1 )), )), λ log(r log(r2 ) + log(M  log(M ((r2 ))} We now find λ find  λ  ∈ R  such that the two inputs in the maximum function are equal. r1 M (r2 ) λ log(r log(r1 ) + log(M  log(M (r1 )) =  λ log(r log(r2 ) + log(M  log(M (r2 )) ⇒ λ log  = log r2 M (r1 ) n

  

               ≤ 

Hence, λ Hence,  λ =  = log

M (r)

M (r2 ) M (r1 )

log

r1 r2

−1

. Exponentiatin Exponentiatingg both sides of the obtained obtained inequality inequality,

 exp log(M  log(M ((r2 )) + log

M (r2 ) M (r1 )

log

r2 r

log

r1 r2

10

 = exp log(M  log(M (r2 ) + log

  M (r1 ) M (r2 )

α

=  M (  M (r1 )α M (r2 )1−α where α where  α  = log

     r2 r

−1

r2 r1

log

. I claim that equality holds if and only if  f  if  f ((z ) =  az λ , where a where  a

 ∈ C, λ ∈ R.

It is obvious that equality holds if  f (  f (z ) is of this form. Suppose quality holds. Then by Weierstrass’s extreme value theorem, for some z0  = r  =  r,, we have

 | |

|f ( f (z0 )|  = M   =  M ((r ) =

 r1 r

λ

M (r1 )

 ⇒

 

z0λ f ( f (z0 )  = r  =  r 1λ M (r1 )

But since the bound on the RHS holds for all r1 < z  < r2 , the Maximum Modulus Principle tells us that z λ f ( f (z ) =  a C r1 < z  < r2 . Hence, Hence, f (z ) =  az −λ . But λ  is an arbitrary real parameter, from which the claim follows.

 ∈  ∀

 | |

 | |

4.6.4 Exercise 1 We seek a conformal mapping of the upper-half plane

H+

φ(z ) =  i is a conformal map of  D  onto inverse is given by

H+

onto the unit disk

D.

lemma The map φ map  φ  given by

1+z 1 z

−

and is a bijective continuous map of  ∂  of  ∂ D  onto φ−1 (w) =

R

∪{∞}, where 1 → ∞. Its

w i w + i

−

Proof.  The statements about conformality and continuity follow from a general theorem about the group of  linear fractional transformations of the Riemann sphere (Ahlfors p. 76), so we just need to verify the images. For z or  z D, 2 1 + z  1 z 1 z Im(φ Im(φ(z )) = Im i  = 2 >  0 1 z 1 z 1 z

 ∈





− −| |  · − − | − |  ⇐⇒ ⇒ z ∈ ∂ D. In particular, φ since |z |   0.  >  0. Furthermore, observe that φ−1 (w)  = 1 ⇐  Im( w) = 0. ˜ =  U  ◦  ◦ φ  :  ∂ D → C  is a piecewise continuous function since  U  is U   U  is bounded and we therefore can ignore the fact that  φ(1)  φ (1) = ∞. By Poisson’s formula, the function

 

 

1 P U  ˜ (z ) = 2π is a harmonic function in the open disk

D.

 

2π

eiθ + z ˜ iθ Re iθ U ( U (e )dθ e z

−

0

By Lemma 1, the function

 

2π

1 P U  =  P ˜ ◦ φ−1 (z ) = U (z ) = P  U 

2π

Re

0

eiθ + φ−1 (z ) ˜ iθ U ( U (e )dθ eiθ φ−1 (z )

−

is harmonic in H+ . Fix w Fix  w 0 D  and let x let  x 0  + iy0  =  z0  =  φ −1 (w0 ). Let P  Let  P w0 (θ) denote the Poisson kernel. We − 1 iθ apply the change of variable  t  = ϕ  =  ϕ (e ) to obtain

 ∈

1 dθ 1 P w0 (θ) = 2π dt 2π

=

2 π (z0

   − 1

− −

z0 i z0 +i

|  −

z0 i z0 +i

−

  −    · −

−

2

t i t+i

1

2

y0 i)(t )(t + i) (t

t i t+i

− 2 t−i

2

=

t+i

2

− − i)(z )(z0  + i)|

1 2π

=

11



2 |z0 + i|2 − |z0 − i|2  ·  ((t  (( t + i) − (t − i)) 2 − 2 |t + i|2 (z0 − i) − − ( z  + i ) 0 + t i t i



2 y0 1 2 = π 2 z0 t π (x0

 · |  − |

 ·  − yt)02 + y2 0

Hence,

 

1 ∞ P U  ( x, y ) = U  π −∞ (x

−

y U ( U (t)dt t)2 + y2

˜ (z) at the points is a harmonic function in H+ . Furthermore, since the value of  P  of  P U  U  ˜ (z ) for z  = 1 is given by U ( of continuity and since φ since  φ −1 (∂ D) = R , we conclude that

 | |

∪{∞} ˜ ◦  ◦ φ−1(x, 0) =  U ( P  (x, 0) =  P ˜ ◦ φ−1 (x, 0) = U   U (x, 0) at the points of continuity  x  ∈ R. U  U 

U 

4.6.4 Exercise 5 I couldn’t figure out how to show that log f ( f (z )  satisfies the mean-value property for  z 0  = 0, r  = 1 without π first computing the value of  0   log log sin(θ sin(θ )dθ. dθ.

 

|

Since sin(θ sin(θ )  θ θ  [0,  [0 ,  π2 ], 1 origin. Hence, for δ for  δ > 0,

 ≤  ∀  ∈

    π

2

0

|

θ

 π

 ≥ sin( )  is continuous on [0, [0, 2 ], where we’ve removed the singularity at the θ

        |            −

θ log dθ = dθ  = lim δ →0 sin(θ sin(θ )

π

2

δ

θ log dθ = dθ  = sin(θ sin(θ) π

By symmetry, it follows that the improper integral exists. Again by symmetry, π

 

2

0

1 log sin( sin(θ)dθ = dθ  = 2

π

2

0

π

2

π

  log(sin(θ log(sin(θ ))dθ ))dθ =  =

2

0

π

2

π

log θ dθ

| | − lim →0

0

δ

2

log sin(θ sin(θ ) dθ

|

δ

|

log sin(θ sin(θ) dθ  exists and therefore

|

  log cos( cos(θ)dθ, dθ, hence π

2



1 1  sin(2θ  sin(2θ) dθ = dθ  = 2 2

log

0

π

0

where we make the change of variable ϑ  = 2θ  to obtain the last equality. Since we conclude that π log sin( sin(θ )dθ = dθ  = π log(2)

 −

0

 

π

0

π

 

2

sin(2θ sin(2θ)dθ

0

log sin(θ sin(θ ) dθ

|

|

−  π4  log(2)

 π  log(2) 4

sin(ϑ sin(ϑ)dϑ

 

 

2

0

1 log sin( sin(θ) cos( cos(θθ)dθ = dθ  = 2 1 = 4

π

 

π

sin(θ)dθ = dθ  = 2 0   log sin(

π

 

  log sin( sin(θ )dθ, dθ,

2

0

We now show that for f ( f (z ) = 1 + z, +  z, log f ( f (z )   satisfies the mean-value property for z0 = 0, r  = 1. Observ Observee that

|

|

1 1 1 log 1 + eiθ  =  log (1 (1 + cos(θ cos(θ ))2 + sin2 (θ)  =  log 1 + 2 cos( cos(θ) + cos2 (θ) + sin2 (θ)  =  log 2 + 2 cos( cos(θ ) 2 2 2

 







           2π

2π

log 1 + e

0

   

 1 1 + cos(θ cos(θ) = log 2 +  log 2 2

||

Substitutin Substitutingg and making the change change of variable variable 2ϑ 2 ϑ  = θ  =  θ,, iθ





dθ = dθ  =

0

     

 1 log log 2 +  log cos cos2 (θ ) 2

π

dθ = dθ  = 2π log log 2+

0

     |

|

|

 

1 + cos(2ϑ cos(2ϑ) log dϑ = dϑ = 2π log log 2+ 2

π

0

By symmetry symmetry,, integrat integrating ing log cos2 (θ) over [0, [0, π] is the same as integrating log sin sin2 (θ)  over [0, [0, π]. Hence, 2π

π

log sin2 (ϑ) dϑ = dϑ  = 2π log log 2 + 2

log 1 + eiθ dθ = dθ  = 2π log log 2 +

0

0

12

π

log sin(ϑ sin(ϑ) dϑ = dϑ = 0

0

|

log log cos cos2 (ϑ)dϑ

4.6.4 Exercise 6 C  be an entire holomorphic function, and suppose that  z −1 Re(f  Let f  Let  f  : C Re(f ((z )) formula (Ahlfors (66) p. 168), we may write

 → 0,  0 , z → ∞. By Schwarz’s

→

 

1 ζ  +  + z dζ  f ( f (z ) = Re(f  Re(f ((ζ )) )) 2πi |ζ |=R ζ  z ζ 

|f ( f (z ) Fix z Fix  z

 ∈ C  and let

R

2

∀ |z| < R

  ∀  ≥    ∀ ≤ | |       ||  −   ≤ |≤ −| | { | |}      | |    −  ≤ ≤  ·   − | |

Let  Let  >  0 be given and R and  R0  > 0  >  0 such that R By Schwarz’s formula,

 −  −

R

 R 0 ,

Re(f (z )) z

 < . Let R Let R be  be sufficiently large that R that  R >

R

2

> R 0.

z  < R,

2

R 2π

2π

Reiθ + z dθ Reiθ z

0

2π

R 2π

0

R + z R + R dθ = dθ  = R  R = 4R R z R R2

· −

>  max R0 , z . By Cauchy’s differentiation formula, 1 f  (z )  = 2π

R

|w|= 2

f (w) dw (w z )2 2π

1 R 4R 2π 2

1

R

0

2

1 2π

   −

2π R 2

iθ f ( R 2e )

R iθ e

0

z

2

2  dθ

R2 dθ  = 8 2 dθ = (R 2 z )2 z

− ||

Letting R Letting  R , we conclude that f  (z )  8  8 . Since z Since  z C was arbitrary, we conclude that f  (z ) 8 z Since  Since   >  0 was arbitrary, we conclude that  f  (z) = 0, which shows that f  that  f  is  is constant.

 → ∞

 |

|≤

 ∈

|  ∀  ∈ C.

 |

4.6.5 Exercise 1 Let f  : C  C  be an entire holomorphic function satisfying  f (  f (R)  R  and f (i R) f ( f (z ) f ( f (z ) vanishes on the real axis. By the limit-point uniqueness theorem that

−

 →

·  ⊂ i · R. Since f  Since  f ((R) ⊂  R ,

 ⊂

f ( f (z ) = f  =  f ((z ) z

Since f  Since  f ((iR)

 ∀  ∈ C

 ⊂ iR,  f (  f (z ) + f ( f (−z ) vanishes anishes on the imaginary imaginary axis. By the limit-point limit-point uniqueness uniqueness theorem that f ( f (z ) =  −f ( f (−z ) ∀z ∈ C

Combining these two results, we have f ( f (z ) =

 −f (−z) =  −f ( f (−z ) =  −f ( f (−z ) ∀z ∈ C

4.6.5 Exercise 3 Let f  : D f (z )  = 1 z  = 1. 1. Let Let φ  : C  be holomorphic and satisfy f ( fractional transformation z i φ(z ) = z + i

 →

 |

|

 ∀ | | −

C

∪ {∞ {∞}} → C ∪{∞}  be the linear

◦  ◦ ◦ + →  | |  ∀|  ∀ | |

C. By the maximum Consider the function g  =  φ −1 f  φ  : H maximum modulus principle, principle, f ( f (z )  1 z  1. + + − 1 ˜ Hence, g Hence,  g  : H f (z )  = 1 z  = 1, φ 1,  φ (f (z )) R z  = 1. Hence, f ( f (R) R. By the Schwarz H . Since f ( C  satisfying g Reflection Principle, g Principle,  g  extends to an entire function  g  : C  satisfying  g (z ) =  g(  g (z ). Define f ˜ =  φ g φ−1 : C C

→

 |  ⊂

 ∈  ∀|  ∀ | | → ◦ ◦ → Then f   f ˜  is meromorphic in C   since φ  has a pole at z =  − i and φ−1 has a pole at z   = 1.

| ≤  ∀ | | ≤

In partic particula ular, r, ˜ f   f   has finitely many poles. We proved proved in Problem Problem Set 1 (Ahlfors (Ahlfors Section 4.3.2 Exercise Exercise 4) that a function meromorphic in the extended complex plane is a rational function, so we need to verify that f  doesn’t f ˜ doesn’t have an essential singularity at . But in a neighborhood of 0,

 ∞



  ◦

1+ 1 f ˜  = φ  =  φ g i z 1

1

z

−1 z

  ◦

 = φ  =  φ g i

z + 1 z 1

−

which which is evidently evidently a meromorp meromorphic hic function. Alternativ Alternatively ely,, we note that

 ∀ |z| ≥ 1,

−

−  onto  H . So the image of  f ˜ in  in a suitable neighborhood of   is not dense in theorem then tells us that f  cannot f ˜ cannot have an essential singularity at .

 ∞  ∞

H

13

 ∞

C.

  ≥ f ( f ˜(z )

 1 since g   maps

The Casorati-Weierstrass

Chapter 5 - Series and Product Developments Power Series Expansions 5.1.1 Exercise 2 We know that in the region Ω = z  : Re(z Re(z )  > 1  >  1 , ζ (z ) exists since

    

 {

}

1 1 1 1 = Re(z) Im(z)i = Re(z) log(n)Im(z )i = Re(z) z n n n n n e

 

∞





and therefore n=1 n1z  is a convergent harmonic series; absolute convergence implies convergence by comN  1 pleteness. Define ζ  Define  ζ N  Clearly,  ζ N  N (z ) = N  is the sum of holomorphic functions on the region Ω. I claim n=1 nz . Clearly, ζ  that (ζ  (ζ N  ζ  on any compact subset K    Ω. Since Since K  is K  is compact and z  Re(z  Re( z ) N )N ∈N  converge uniformly to ζ  on is contin continuous, uous, by Weierstrass eierstrass’s ’s Extreme Extreme Value Theorem Theorem z0 K   K   such that Re(z Re( z0 ) = inf z∈K  Re(z  Re( z ). In 1 1 1 particular, Re(z Re(z0 )  > 1  >  1 since z since  z 0  Ω. Hence, z . So by the Triangle Inequality, n nRe(z) nRe(z0 )

 ∈

  N 

∀z ∈

1  Ω,  Ω , nz n=1

  ≤  ≤ ∃      ≤ ≤ N 

n=1

By Weierstrass’s M-test, we attain that ζ n holomorphic in Ω and

N 

1 nz

n=1

∈

 ⊂  ⊂

1

nRe(z0 )

<

 →

∞



n=1

 →  ζ    ζ   uniformly on K .

1 nRe(z0 )

<

 ∞

Therefore Therefore by Weierstrass’s eierstrass’s theorem, theorem, ζ  is

N  N  ∞ − log(n − log(n log(n) log(n)   − log(n)z − ζ  (z ) = lim lim ζ N (z ) = lim lim log(n log(n)e = li m = N →∞ N →∞ N →∞ nz nz n=1 n=1 n=1







Section 5.1.1 Exercise 3 Lemma 2. Set  an = ( 1)n+1 . If 

−

∀z ∈ C  with Re (z) ≥ Re (z0).

     



∞

an n=1 nz   converges

for some  z0 . Then  Then 



∞

an n=1 nz   converges

Proof. If  n∞=1 nazn0  conveges, there exists an  M >  0 which bounds the partial sums. Let m Let  m summation summation by parts, parts, we may write N 

N 

an an 1 1 = = z −z0 z z z − z 0 0 n n n N  n=m n=m

Hence,

N 

an nz n=m

Observe that



1 (n + 1)z −z0

1

−n−

 ≤ |  

z z0

 M 

1 N z−z0

  −   | |  −

m 1

− a n



n=1

nz0

−

n

n=m

k=1

N  1

1

|  + M  n −

= e− log(n+1)(z−z0 )

N  1

z z0

 + M 

−

n=m

e− log(n)(z−z0 )

14

ak kz0



1 (n + 1)z −z0

1 (n + 1)z−z0

1

−n−

z z0

 −   −

1  = z z0

 ≤ N  ∈  ∈ N. Using 1

−n−

z z0



  log(n+1)

log(n)

uniformly on 

  

e−t(z−z0 ) dt

1

 

  log(n+1)

≤ |z − z0|



 |Re(z Re(z ) − Re(z Re(z0 )| − log(n+1)(Re(z)−Re(z0 )) − e− log(n)(Re(z)−Re(z0)) e−t(Re(z)−Re(z0 )) dt = dt = e |z − z0|

log(n)

≤ e− log( )(Re( )−Re( )) − e− log( +1)(Re( )−Re( n

z

z0

n

z

z0 ))

1

=

nRe(z) Re(z0 )

−

 − (n + 1)Re(1 )−Re( z

z0 )



Since this last expression is telescoping as the summation ranges over  n,  n , we have that

   N 

an nz n=m

 ≤

M  N Re(z)−Re(z0 )

2M 

≤ mRe( )−Re( z

Corollary 3. If 

of  Re (z )



mRe(z)−Re(z0 )



N  an n=1 nz−z0  are

Hence, the partial sums of 

∞

mRe(z)−Re(z0 )

M 

+

z0 )

+

M 

an n=1 nz  converges

  m N 



+ M 

1 N Re(z)−Re(z0 )

 − ≤

Re(z ) Re(z0 )

−

1

1

 − mRe( )−Re( z

4M  mRe(z)−Re(z0 )

z0 )



 →  0,  0 , m → ∞

Cauchy and therefore converge by completeness.

for some  z some  z  = z  =  z0 , then 



∞

an n=1 nz  conveges

uniformly on compact subsets 

 ≥ Re (z0)}.  ⊂ {Re(z Proof. Let K  Let  K  ⊂ Re(z ) ≥ Re(z  Re( z0 )}  be compact. Since z Since  z  →  Re(z  Re(z ) is continuous, there exists z exists  z 1 ∈ K  such  K  such that ∞ Re(z Re(z ) ≥ Re(z Re(z1 ) ∀z  ∈ K . Sinc Sincee Re( Re(z1 ) ≥ Re(z Re(z0 ), conver con verges. ges. The proof of the preceding lemma lemma =1  { {



shows that we have a uniform bound

   N 

an nz n=m

 ≤

an nz1

n

4M  mRe(z)−Re(z0

4M   ≤ ) mRe( )−Re( z1

z0 )

where M  where  M  depends  depends only on z on  z 0 . The claim follows immediately from the M  the  M -test -test and completeness.

 

Since the series f ( f (z ) = n∞=1 annz  converges if we take z R>0 (the well-known alternating series), we have by the lemma that n∞=1 annz   converges Re(z Re(z )  > 0.  >  0. We now show that this series is holomorphic holomorphic on the region Re(z Re(z )  > 0  >  0 . Define a sequence of functions (f  ( f N  N )N ∈N by

 {

}

 ∈

 ∀

N 

( 1)n+1 f N  N  = nz n=1

−

C : Re(z It is clear that f  that  f N  Re(z )  > 0  >  0 N  is holomorphic, being the finite sum of holomorphic functions. Set Ω = z and let K  let  K   Ω be compact. Since the  f N   are just the partial sums of the series, we have by the corollary to N  the lemma that f  that  f N   f    f   uniformly on K  on  K .. By Weierstrass’s theorem,  f  is  f  is holomorphic in Ω. N  ∞ an 1−z To see that (1 2 )ζ (z ) = n=1 nz on Re(z Re(z )  > 1  >  1 , observe that

 {  ∈

 ⊂  ⊂

−

 →

   −  − N 

1 z

(1 − 2 − )

 {

N 

}

N 

( 1)n+1 1 = z n nz n=1 n=1

1 nz n=1

}



N 

1 2 (2n (2n)z n=1

  −

N 

( 1)n+1 1 =2 z n nz N  0 given.

5.1.2 Exercise 2

−

Differentiating 1

2αz + αz  + z 2



− 12

with respect to z to  z , we obtain  p1 (α) =

2α 2z 3 2(1 2αz + αz  + z 2 ) 2

−

−

| =0  =  α z

−

To compute higher order Legendre polynomials, we differentiate 1 obtain the equality α (1

−z

− 2αz + αz  +

3 z2) 2

=

∞



n=1

n 1

nP n (α)z −

α z = (1 1 2αz + αz + z 2

⇒ √  − − 15

2αz + αz  + z 2

− 12

 2

− 2αz + αz + z )

and its Taylor series to

∞



n=1

nP n (α)z n−1

Hence,

∞



αP n (α)z

n

n=0

∞

 −

P n (α)z

n+1

∞



=

n=0

n 1

nP n (α)z −

n=0

∞

 −

n

2αnP n (α)z +

n=0

∞



nP n (α)z n+1

n=0

Invoking elementary limit properties and using the fact that a function is zero if and only if all its Taylor coefficients are zero, we may equate terms to obtain the recurrence αP n+1 (α)

− P  (α) = (n + 2)P  2)P  +2 (α) − 2α(n + 1)P  1)P  +1 (α) + nP  (α) n

n

n

n

⇒ P  +2(α) = n +1 2  [(2n  [(2 n + 3)αP  3)αP  +1 (α) − (n + 1)P  1)P  (α)] n

n

So,

 −  − −    − − − P 2 (α) =

1 3

P 3 (α) = P 4 (α) =

1 4

1 3α2 2

n

1 5α (3α (3α2 2

1)

3α)

1 3 (3α (3α2 2

1 7α (5α (5α3 2

1

1 5α3 2

2α  =

1)  =



− 3α

1 (35α (35α4 8

− 30 30α α2 + 3)

5.1.2 Exercise 3 Observe that

∞

∞

sin(z sin(z ) 1 ( 1)n 2n+1 ( 1)n 2n = z = z z z n=0 (2n (2n + 1)! (2n (2 n + 1)! n=0



−

−



      −  −   −   −     − −  −

So,   sin(z ) = 0 in some open disk about z about  z  = 0. Hence, the function z function  z  log sin(z )  is holomorphic in an open disk about z about  z  = 0, where we take the principal branch of the logarithm. Substituting, z



log

sin(z sin(z )  = log 1 z

1

∞

=

 sin(z  sin(z ) z

1 2 3! z

 =

1 2 3! z

( 1)n 2n n=0 (2n+1)! z

∞

1

∞

1 4 5! z

1 6 + 7! z m

−

m

m

m=1

m=1

Set P  Set  P ((z ) =

z

 →

[z 8 ]

m

− 5!1 z4. Then log

   −   −  − sin(z sin(z )  = z

=

z2 3!

z 6  P (  P (z ) + [z [ z 8 ]  P (  P (z )2 + [z [z 8 ]  P (  P (z )3 + [z [z 8 ] + + + 7! 1 2 3

 z 4  z 6  1 + + 5! 7! 2



z4 (3!)2

 −

2z 6 (3!)(5!)







z6 + + [z [z 8 ] 3(3!)3

1 4 1 6  −16 z2 − 180 [z 8 ] z − z + [z 2835

=

Partial Fractions and Factorization 5.2.1 Exercise 1 From Ahlfors p. 189, we obtain for z   1.  >  1. So it suffices to consider the case  ˜hc  = 2hc 1. Then

 ≥

 → ∞

 → ∞  −

∞ >

 | |

∞



n=1

1

|b | n

˜ c +1 h

=

∞

1



2hc 1+1

−

|b | n

n=1

=

∞

1



|a | n

n=1

hc

But this shows that the genus of the canonical product associated to ( an ) is at most h most  hc 1, which is obviously a contradiction. Taking f  Taking  f  to  to be a polynomial shows that this bound is sharp. I claim that the genus of  f  is f ˜ is bounded from above by 2 h +1. Indeed, 2h 2h + 1  2deg(g  2deg(g (z )) = deg(g deg(g (z 2 )), and we showed above that  ˜hc  2h  2 hc + 1  2h  2 h + 1. This bound is also sharp since we can take

−

 ≤

 ≥

 ≤

f ( f (z ) =

∞

  −  ⇒ z n2

1

n=1

 f (  f (z 2 ) =

 −  1

n=0 =0



z z en n

f ( f (z ) is clearly an entire function of genus 0, and the genus of the canonical product associated to ( n)n∈Z is 1, from which we conclude the genus of  f  of  f ((z 2 ) is 1 .

5.2.4 Exercise 2 Using Legendre’s Legendre’s duplication duplication formula formula for the gamma function (Ahlfors (Ahlfors p. 200), Γ



1  = 6

 

√ πΓ 2 ·  1 6

Γ

1 6

6

π πz )  (Ahlfors

Applying the formula Γ(z Γ(z )Γ(1



 

− z) = sin( √  1  = π2 Γ 2 3

2

Γ

3

3

p. 199), we obtain

 · 3

 

−1 √  2 1 1  1 1 2 − · 6 2 Γ  + = π2 3 Γ Γ 1 2

1  sin π 3 π 19

1 3

1 = 2− 3

1 2

     3 π

Γ

1 3

2

2hc +1



5.2.4 Exercise 3 It is clear from the definition of the Gamma function that for each  k

 

 ∈ Z≤0,

1 + kz Γ(z Γ(z ) k = 0 z Γ(z Γ(z ) k  = 0

f ( f (z ) =

 

extends to a holomorphic function in an open neighborhood of  k of  k.. We abuse notation and denote the extension z >0 also by 1 + k Γ(z Γ(z ) and z and  z Γ(z Γ(z ). lemma For any  k Z ,

 

 ∈

Γ(z Γ(z ) =

Γ(z Γ(z  + k )



k  +  j j =1 (z  + j

 − 1)

∀z ∈/ Z

Proof.  Recall that Γ(z Γ(z ) has the property that the Γ( z + 1) =  zΓ(  z Γ(zz ). We proceed by induction. The base case Γ(z +k) is trivial, so assume that Γ(z Γ( z ) = k (z+j −1)  for some k some  k N. Then

 ∈

j =1

Γ(z Γ(z + (k ( k  + 1))



k +1  +  j j =1 (z  + j

 − 1)

=

Γ((z Γ((z + k) + 1)



=

k+1  +  j j =1 (z  + j

 − 1)

 ∈ Z≤0,

(z  + k )Γ(z )Γ(z + k)



=

k +1  +  j j =1 (z  + j

 − 1)

Γ(z Γ(z  + k )



k  +  j j =1 (z  + j

 − 1)!

= Γ(z Γ(z )

Corollary 4.   For any  k

lim (z

z

→k

−

( 1)k k )Γ(z )Γ(z ) = k!

− ||

 ∈ Z≤0. Immediate from the preceding lemma is that Γ(z Γ(z + |k | + 1)   Γ(1) (−1) lim (z  + |k |)Γ(z )Γ(z ) = lim (z + |k |) +1 = = |k|! →−| | →−| | (−1) (−2) · · · (− |k |)  j | − 1) =1 (z + | j

Proof. Fix k Fix  k

k

z

Let k Let  k

k

z

 ∈ Z≤0. Then



k

 

k j

 

 

(1 kz )Γ(z )Γ(z ) 1 1 1 (z k) Γ(z Γ(z ) res(Γ; k ) = Γ(z Γ(z )dz = dz  = dz = dz  = dz z 2πi |z−k|= 12 2πi |z−k|= 12 1 k 2πi |z−k|= 12 z k

 

Since the function 1 + integral formula,

z k

−

−

−

−

Γ(z Γ(z ) extends to a holomorphic function in a neighborhood of  k , by Cauchy’s

 

1 (z k ) Γ(z Γ(z ) dz = dz  = (z 2πi |z−k|= 12 z k

−

−

−

( 1)k k) Γ(z Γ(z ) z=k  = k!

|

− ||

where use the preceding lemma to obtain the last equality. Thus, res (Γ; k ) =

( 1)k k!

− ∀k ∈ Z≤0 ||

5.2.5 Exercise 2 Lemma 5.

   ∞

log

0

1

−

1 e−2πx



dx = dx  =

π 12

Proof.   Let 1 >> 1  >> δ > 0. >  0. Consider the function   log(1z−z) , which which has the power power series representation log(1 z

− z) = − 1 z

∞ 1



n n=1

∞ 1 z = z n−1 n



n n=1

20

∀ |z| 0

 ∈ R

∞

2π

log(1

0

1 e−2πx

1

0

For x or  x

2πe −2πx dx = dx  =

e−2πx

0

∞

e−2πx )

∞

dx = dx  =

− e−2

πx

e−2πx dx = dx  =

log 1

0

)dx

π 12

, Stirling’s formula (Ahlfors p. 203-4) for Γ(z Γ( z ) tells us that Γ(x Γ(x) =

where

1 J (x) = π

The preceding lemma tells us that 1 1 J (x) = x π

 ·

 

∞

0

where we’ve used 0 < 0  <

x2  log x2 + η 2 x2 x2 +η 2



1

 

∞

0

1 e−2πη

−

 ≤  1 ∀η. Set



1

2πx x− 2 e−x eJ (x)

x  log η 2 + x2

1 1 x π

dη

 ≤  ·



1

−

1 e−2πη

   ∞

log

0

1



dη

1 e−2πη

−



dη = dη  =

1 1 π 1 = x π 12 12 12x x

 ·  ·

θ(x) = 12xJ  12xJ (x)

It is obvious that θ (x) >  0 and θ(x) <  1 since inequality is strict. We thus conclude that Γ(x Γ(x) =

√ 

√ 

x2 x2 +η 2

<  1 almost everywhere, and therefore the preceding θ(x)

1

2πxx− 2 e−x e 12x 0 < θ (x)  > 0, >>  0, define γ 1  : [0, [0, R]

 π  → C, γ 1(t) =  t;  t ; γ 2  : [0, [0, ] → C, γ 2 (t) =  Re 4

it

iπ 4

; γ 3  : [0, [0, R]

 → C, γ 3(t) = (R − t)e

and let γ  let  γ  be  be the positively oriented closed curve defined by the  γ i .

 

γ 2



f ( f (z )dz =

  

   ≤

π

π

4

0

 ≥ 2t  2 t  (this is immediate from

π

 

4

0

e−R cos(2t) Rdt

0

Since cos(2t cos(2t) is nonnegative and cos(2t cos(2 t)  π for t for  t  [0,  [0 , 4 ], we have

 ∈

4

e−R cos(2t)−iR sin(2t) Rieit dt

π

e−R cos(2t) Rdt ≤

 

4

e−2Rt Rdt = Rdt  =

0

21

 −12



π

e−R 2

d  cos(2t  cos(2t) dt



=

π

 −2 sin(2 sin(2t) ≥ −2 on [0, [0, 4 ])

− 1  → 0,  0 , R → ∞

Since f  Since  f  is  is an entire function, by Cauchy’s theorem, 0=

 

f (z )dz = dz  =

γ 

and letting R letting  R

 

 

f (z )dz + dz +

γ 1

 

f ( f (z )dz + dz +

γ 2

 

f ( f (z )dz

γ 3

 → ∞,

∞

 

R

2 e−x dx = lim lim ei 4 R→∞

iπ

 

R

2 2 e−(R−t) e 2 dt = lim lim ei 4 e−i(R−t) dt = dt  = e  e i 4 R →∞ 0 0 0 √  2 ∞ e−x dx = where we make the substitution  y  = R  =  R − t. Substituting dx  = 2−1 π, π

π

π

 

 

∞

2

e−iy dy

0

0

 

∞

2

cos(x cos(x )dx

0

  −

∞

i

 

∞

2

sin(x sin(x )dx = dx  =

0

√ 

2 π e−ix dx = dx  = e  e −i 4 = π

2

0

√ π √ π √  − i 2√ 2 2 2

Equating real and imaginary parts, we obtain the Fresnel integrals

√ π cos(x cos(x )dx = dx  = √  2 2 0 √ π ∞ 2 sin(x sin(x )dx = dx  = √  2 2

   

∞

2

0

Entire Functions 5.3.2 Exercise 1 We will show that the following two definitions of the   genus  of an entire function  f  are  f  are equivalent: 1. If  m g(z )

f ( f (z ) =  z e

∞

 −  1

n=1

z an

h 1 j =1 j



e

( azn )

j

where h is the genus of the canonical product associated to ( an ), then the genus of f  of  f  is   is max max (deg(g (deg(g(z )), )), h). If no such representation exists, then  f  is  f  is said to be of infinite genus. 2. The genus genus of  f  is  f  is the minimal h minimal  h

 ∈ Z≥0 such that m g(z )

f ( f (z ) =  z e

∞

 −  1

n=1

z an

h 1 j =1 j



e

( azn )j

where deg(g deg(g(z ))

 ≤ h.  h. If no such  h  exists, then f  then  f  is  is said to be of infinite genus.

Proof.   Suppose f  Suppose  f  has  has finite genus  h 1  with respect to definition definition (1). If  h  h 1  =  h,  h , then deg(g deg(g(z ))  h 1 . Hence, f  is f  is of a finite genus h genus  h 2  with respect to definition (2), and  h 2  h 1 . Assume otherwise. By definition of the genus genus of the canonical product, the expression expression

 ≤

 ≤

       ∞

h1

n=1 j =h+1

1  j

z an

h1

j

=

j =h+1

1  j

 ∞ 1

j n=1 an

zj

defines a polynomial of degree  h 1 . Hence, we may write f ( f (z ) =  z e

h 1 j h+1 j

−∞=1  =1

m g(z )

n

( azn )j

∞

 − 

n=1

1

z an

h1 1 j=1 j



e

( azn )j =  z m eg˜(z)

∞

 − 

n=1

1

z an

h1 1 j =1 j



e

( azn )j

where we ˜g(z ) is a polynomial of degree h degree  h 1 . Hence, f  Hence,  f  is  is of finite genus h genus  h 2  with respect to definition (2) and h2  h1 . Now suppose that f  has f  has finite genus h2  with respect to definition (2). Reversing Reversing the steps of the previous argument argument,, we attain that f   f   has finite genus h1   with respect respect to definit definition ion (1), and h1 h2 . It fol follo lows ws immediately that definitions (1) and (2) are equivalent if  f   f   has finite genus with respect to either (1) and (2), and by proving the contrapositives, we see that (1) and (2) are equivalent for all entire functions  f .  f .

 ≤

≤

22

5.3.2 Exercise 2 lemma Let a Let  a

 ∈ C  and  r > 0. Then inf  |z − |a||  =  |r − |a||   and and | |= z

sup sup z

r

|z |=r

| − |a|| = r  =  r +  + |a|

Proof.  By the triangle inequality and reverse inequality, we have the double inequality

|r − |a|| = ||z| − |a|| ≤ |z − |a|| ≤ |z| + |a| = r  =  r +  + |a| Hence, Hence, inf |z − |a|| ≥ |r − |a||  and sup |z − |a|| ≤ r +  r  + |a|. But these values are attained at  z  = r  =  r and z and  z  =  −r,

respectively.

By Weierstrass’s extreme value theorem, f  and g   attain both their maximum and minimum on the circle z  = r  =  r at z at  z M,f  , zM,g and zm,f , zm,g , respectivel respectively y. The preceding lemma shows shows that zM,g = r and zm,g  =  r.  r . Consider the expression

 {| |

 | |

}

   −   −    −

 −

   | − |   | − | ||      − | | − | | | |  || −| | ||  || || ≤ | ≤| |   |  − |  − | | ≥ | − | || ∀  ∈      |  − | ≥  |  − | | − | || |  − | 

We have

 | |

zm



f (z ) = g (z ) zm

∞

n=1

∞

n=1

1

1

z an

=

z

∞

n=1

|a | n

∞

z z

an an

∞

reiθM,f  an rei(θM,f −arg(an )) f ( f (zM,f  ) f (zM,f  ) = = = g (zM,g ) g( r) r + an r + an n=1 n=1

∞ sup |z|=r z

r + an

n=1

Hence, f ( f (zM,f )

 |

=

∞ r + a n

n=1

r + an

an  = rei(θm,f −arg(an ))

an

r

∞

f ( f (zm,f ) f ( f (zm,f ) zm,f  an = = g (zm,g ) g (r ) r an n=1

Hence, f ( f (zm,f )

 |

=1

g(zM,g ) . Since zm,f 

we have that

an

an

| ≥ |g(z

m,g )

an

n

∞ z m,f 

n=1

N

an =1 an

zm,f 

|.

5.5.5 Exercise 1 Let Ω be a fixed region and  be  be the family of holomorphic functions  f  : Ω I claim that  is   is normal. Consider the family of functions

 → C  with Re(f  Re(f ((z )) > ))  > 0  0 ∀z ∈ Ω.

 F 

 F 

G  = Since Re(f  Re(f ((z )) > ))  > 0  0 f 

 ∀  ∈  ∈ F , we have



 → C : g =  g  = e  e−

g  : Ω

  

f 

for some f  some  f 

  ≤

 ∈  ∈ F 

 

e−f (z)  = e−Re(f (z))−iIm(f (z))  = e−Re(f (z))

 1

Hence,   is uniformly bounded bounded on compact compact subsets subsets of Ω and is therefore therefore a normal normal family. family. Fix a sequence sequence − f n (f n )n∈N , and consider the sequence g sequence  gn  = e  =  e . (gn ) has a convergent subsequence (g ( gnk ) which converges converges to a holomorphic holomorphic function function  g   on compact sets (Weierstrass’s (Weierstrass’s theorem). Since gnk  is nonvanishing for each k each  k,, g  is either either identicall identically y zero or nowhere zero by Hurwitz theorem. theorem. If  g  g  is identically zero, then it is immediate D that f nk   tends to   uniform uniformly ly on compact sets. Now, Now, suppose that g  is nowhere zero. g(K ) 0 is compact by continuity continuity.. By the Open Mapping Theorem, Theorem, for each each z K , there exists r >   0 such that D(g (z ); r)  g(Ω)  g (Ω) D 0 . The disks D disks  D((g(z ); 4−1 r) form an open cover of  g of  g((K ), ), so by compactness,

 G  ⊂ F   ⊂

 ∞  ⊂ \ { } n

  ⊂

g(K )

i=1

 ∈

n

D(g(zi ); 4−1 ri ) ⊂

 ⊂ \ { }



i=1

n

D(g (zi ); 2−1 ri ) ⊂ 23



i=1

D(g (zi ); ri )

 ⊂ D \ {0}

On each D each  D((g(zi ); ri ), we can choose a branch of the logarithm such that log( z ) is holomorphic on D on  D((g (zi ); ri ), − 1 and in particular uniformly continuous on  D(  D (g (zi ); 2 ri ). For each i each  i,, choose δ  choose  δ i  > 0  >  0 such that w, w

 ∈ D(  D (g (z ); r ) |w − w |  < δ   ⇒ |log(w log(w) − log(w log(w )|  <   ∀z ∈ K . Set δ  Set  δ  =  = min1≤ ≤ δ  , choose k choose  k 0 ∈ N  such that k that  k  ≥ k 0 ⇒ |g (z ) − g (z )|  < δ  ∀  K . Then for 1  ≤ i  ≤ n,  n, ∀k ≥ k0 log log e− ( ) − log(g log(g(z ))  <  ∀z ∈ g −1 D(g(z ); 2−1 r ) It is not a not  a priori  true priori  true that log(e log(e− ( ) ) =  −f  (z ); the imaginary parts differ by an integer multiple of 2 πi. πi . But the function given by 21 log(e log(e− ( ) ) + f  (z )   is continuous and integer-valued on any open disk about each z each  z  in Ω, and therefore must be a constant  m  ∈ Z  in that disk as a consequence of connectedness. Taking a new covering of  g(  g (K ), ), if necessary, such that D that  D((g (z ); r ) is contained in the image under g under  g of  of such a disk (which we can do by the Open Mapping Theorem), we may assume that for each  z  ∈ g −1 (D(g (z ); r )), i n

i

i

i

i

nk

  

f nk z

f nk z

nk

f nk z

πi

i





nk





i

i

i



i

i

      | − − 

i

2πmi   = lim log e−f nk (z) + f nk (z )  = log(g log(g (z )) + lim f nk (z ) k

→∞

Taking k Taking  k 0

k

→∞

 ∈ N  larger if necessary, we conclude that ∀k ≥ k0 log e− ( ) − log(g log(g (z ))  = f  (z ) [ log(g log(g (z )) + 2πm 2πm ]|  <  ∀z ∈ g −1 D(g (z ); 2−1 r )  ⊂ =1 g−1 D(g(z ); 2−1r ) , we conclude from the uniqueness of limits that f  (z) converges to Since K  ⊂

     f nk z

n i

nk

i

limk→∞ f nk (z) uniformly on K  on  K ..



i

i

i

i

nk



Suppose in addition that Re(f  Re(f ) :  f   is uniformly bounded on compact sets. I claim that  is   is then locally bounded. Let K  Let  K   Ω be compact, and let  L > 0 be such that Re(f  Re( f )( )(zz )  L z  K  f  . Then

 ⊂  ⊂

 {

 ∈  ∈ F}

 

ef (z)  = e  =  e Re(f (z))





 F   ≤  ∀  ∈  ∀  ∀  ∈  ∈ F 

L

≤ e ∀z ∈ K  ∀  ∀f  ∈  ∈ F 

Hence, g  = e  =  e f  :  f   is a locally bounded family family, and therefore therefore its derivativ derivatives es are locally bounded. Since Re(f  Re(f )) > 0  >  0 f  , we have that f  (z ) f  f  (z )ef (z)  = g  (z )

 ∈  ∈ F   ∀  ∈  ∈ F 

  | | ≤   | |  ∈  ∈ F}   ⊂  ⊂  ⊂    ∀  ∈   ⇒ | | ≤

which shows that f   :  f   is a locally bounded family. family. Since K  is K  is compact, there exist z1 , n  r i and r and  r 1 ,  , rn  >  0 such that K  that  K  and  D((zi ; ri )  Ω. By Cauchy’s theorem, i=1 D (zi ; 2 ) and D

···

 {

 r i f  (z )dz z  D zi ;

f ( f (z ) =

2

[zi ,z ]

f ( f (z )

 M i ri z

 ∀  ∈ D

n

  zi ;

 r i 2

 ∈  ∈ F}  on  D(  D (z ; 2−1 r ).

where [z, [z, zi ] denotes the straight line segment, and  M i  is a uniform bound for f   :  f  Setting M  Setting  M  =  = max1≤i≤n M i and r and  r  = max1≤i≤n ri , we conclude that

 {

· · ·  , z  ∈  K 

i

i

|f (  ∀f  ∈  ∈ F  f (z )| ≤ M r ∀z ∈ K  ∀ Normal Families 5.5.5 Exercise 3 Let f  Let  f  : C

→ C  be an entire holomorphic function. Define a family of entire functions  F  by F = {g : C → C : g(  g (z ) =  f (  f (kz) kz ), k ∈ C} Fix 0  ≤ r1 < r2 ≤ ∞. I clai claim m that that F   is normal normal (in the sense sense of Definit Definition ion 3 p. 225 225)) in the annulu annuluss r1  <  |z |  < r2  if and only if  f  is  f  is a polynomial. Suppose f  Suppose  f  =  a 0  + a  +  a1 z  + · · · + a z is a polynomial, where  a   = 0. By Ahlfors Ahlfors Theore Theorem m 17 (p. 226 226), ), it suffices to show that the expression 2 |g  (z )| ρ(g ) = 2 g ∈ F  1 + | g (z )| n

n

n

24

is locally bounded. bounded. Since g(z ) = f ( f (kz) kz ) for some k The function F  function  F ((z ) given by

  ∈ C, it suffices to show that 1+2|| (( ))||   is bounded on C.

   

2 f  f  z −1

f  z f  z

2 a1 z 2n + 2a 2a2 z 2n−1 +

2



· · · + na z +1 F ( F (z ) = = |z|2 + |a0z + a1z −1 + · · · + a |2 1 + |f ( f (z −1 )|2 is continuous in a neighborhood of 0 with F (0) F (0)  = +∞  since a   = 0. Hence, Hence, | F ( F (z )| ≤ M 1  ∀ |z | ≤ δ , which shows that 2 |f  (z )| 1 2  ≤  M 1 ∀ |z | ≥ δ  1 + |f ( f (z )| 2|  ( )| is contin continuou uouss on the compact compact set D(0; 1 ) and therefor thereforee bounded bounded by some M 2 . Takin akingg M  = 1+| ( )| max {M 1 , M 2 }, we obtain the desired result. Now suppose that F  is   is normal in r in  r 1  <  |z |  < r2 . If  f  is  f  is bounded, then we’re done by Liouville’s theorem. Assume otherwise. Let (f  ( f  ) ∈  ⊂ F  be   be a sequence given by f  (z ) =  f (  f (κ z ) for some  κ  ∈ C where κ → ∞, n  → ∞. Sinc Sincee  F   is normal, (f  (f  ) has a subsequence (f  (f  ) ∈   which either tends to  ∞ ,  {r1  <  |z| < r2}, or converges to some limit function g  in likewise fashion. uniformly on compact subsets of  {  ≤ |z| ≤ r2 − δ }. If  f   → g, then I claim that f  Fix δ >  0 small and consider the compact subset  {r1  + δ  ≤ is bounded on C, which gives us a contradictio contradiction. n. Indeed, Indeed, fix z0  ∈ C. Sinc Sincee (f  (f  ) converges uniformly on {r1 + δ  ≤  ≤ |z| ≤ r2 − δ }, (f  ) is uniformly bounded by some  M > 0 on this set. Let |κ (r1 + δ )| ≥ |z0|. By the Maximum Modulus Principle,  |f ( f (z )|  is bounded on the disk  D(0;  D (0; |κ (r1  + δ )|) by some |f (w)|  for some n

n

n

n

n

n

n

f  z f  z

2

δ

n n

N

n

n

n

n

n

nk k

N

nk

nk

nk

nk

nk

w  on the boundary. Hence,

|f ( f (z0 )| ≤ |f ( f (w)|  = |f 

nk (z )

| ≤ M    M   for some z some  z  ∈ {|z |  = r  =  r 1  + δ }

Since z Since  z 0  was arbitrary, we conclude that  f  is  f  is bounded. I now claim that f  that  f  has  has finitely many zeroes. Suppose not. Let (a ( an )n∈N  be the sequence of zeroes of  f  of  f  ordered  ordered − 1 by increasing modulus, and consider the sequence of functions f n (z ) = f n (r an z ), where r1  < r < r2 is fixed. Our preceding preceding work shows shows that (f  ( f n ) has a subsequence (f  ( f nk ) which tends to  on the compact set z  = r  =  r . But this is a contradiction since f  since  f nk (r) = 0 k N. If we can show that f   f   has a pole at , then we’re done by Ahlfors Section 4.3.2 Exercise 2 (Problem Set 1). 1). Let Let f n (z ) =  f ( (f nk )k∈N  be a subsequence which tends to   on compact compact sets. sets. Let M >  0  f (nz), nz), and let (f  N  such that for be given. given. Fix r1   < r < r 2 . Then Then f nk  uniformly on z  = r  =  r , so there exists k0 k  k 0 , f nk (z )  > M  z  =  r.  r . Taking k0  larger if necessary, we may assume that f (z )  >  0 z  rn  r nk0 . Let z Let  z C, z  rn  r nk0 , and choose k choose  k  so that  n k r > z . By the Minimum Modulus Modulus Principle, Principle, f   assumes its minimum on the boundary of the annulus nk0 r w  rn k . But

{| |

}

 ≥  | |  ∈ | | ≥

 ∀  ∈

 ∞

→∞

 ∀  ∀ | |

 {| |

 | |  ≤ | | ≤

 {

min



 ∞

inf 

|w|=n

k0 r

|f ( f (w)| , |

inf 

}

w =nk r

|

}

 ∞

|

|f (w)

 ∈  | |  ∀ | | ≥  | |

 > M 

and therefore,

|f ( |f (w)| > M  f (z )| ≥ inf  ≤| |≤  ∀ |z| ≥  r n . Since Since z   was arbitrary, we conclude that  |f ( f (z )| > M  ∀ Since M >  0 was arbitrary, we conclude that f  that  f  has  has a pole at  ∞. nk0 r

w

nk r

k0

5.5.5 Exercise 4 Let   be a family family of meromo meromorph rphic ic function functionss in a given given region region Ω, which which is not normal normal in Ω. By Ahlfors Ahlfors Theorem 17 (p. 226), there must exist a compact set  K   Ω such that the expression

 F 

 ⊂  ⊂

ρ(f )( f )(zz ) =

2 f  (z )

|

| f  ∈ 2  ∈ F  1 + |f ( f (z )| 25

is not locally bounded on K  on  K .. Hence, we can choose a sequence of functions (f  ( f n ) such that 2 f n (zn ) ,n 2 1 + f n (zn )

 ⊂ F  and   and of points (z (z ) ⊂ K  n

|

|   ∞  → ∞ | | Suppose for every z every  z  ∈ Ω, there exists exists an open disk D disk D((z ; r ) ⊂ Ω on which F  is   is normal, equivalently ρ equivalently ρ((f ) f ) is lo− − 1 1 cally bounded. Let M  Let  M  > 0 bound ρ bound  ρ((f ) f ) on the closed disk D disk  D((z ; 2 r ). The collection D(z ; 2 r ) :  z  ∈ K  forms an open cover of  K . By compactness, compactness, there exist finitely finitely many disks D(z1 ; 2−1 r1 ), · · ·  , D(z ; 2−1 r ) z

z

such that



z

z

n



n

n

K 

 ⊂  ⊂



i=1

D(zi ; 2−1 ri ) and i  = 1,

 ∀

· · ·  , n |ρ(f )(  ∈ F  f )(zz )| ≤ M   ∀z ∈ D(  D (z ; 2−1 r) ∀f  ∈ i

i

Setting M  Setting  M  =  = max1≤i≤n M i , we conclude that

|ρ(f )(z  ∀z ∈ K  ∀  ∀f  ∈  ∈ F  )(z )| ≤ M  ∀ This is obviously a contradiction since lim →∞ ρ(f  )(z )(z ) = +∞. We conclude that there must exist  z 0 ∈  Ω such that F  is   is not normal in any neighborhood of  z of  z 0 . n

n

n

26

Conformal Mapping, Dirichlet’s Problem The Riemann Mapping Theorem 6.1.1 Exercise 1 Lemma Lemma 6. Let  f  : Ω

 function  g  : Ω

 → C  be a holomorphic function on a symmetric region  Ω  (i.e.

Ω = Ω). Then Then the  the 

 → C, g(z) =  f (  f (z )  is holomorphic.

Proof.   Writing z Writing  z  = x  =  x + iy, iy , if  f (  f (z ) =  u(  u(x, y) + iv( iv(x, y), where u, where  u, v  are real, then g then  g((z ) =  u(  u(x, y) iv( iv (x, y) = 1 u ¯(x, y) + i +  i¯ v¯(x, y). It is then evident evident that that g   is continuous and u, v  have C  partials. partials. We verify verify the CauchyCauchyRiemann equations. ∂ ¯ ∂ u ¯ ∂u ∂ ¯ ∂ u ¯ ∂u (x, y) = (x, y ); (x, y) = (x, y) ∂x ∂x ∂y ∂y

− −

−

∂ v¯ (x, y) = ∂x

 −

∂v  − ∂x (x, −y );

−

−

∂ v¯ ∂v (x, y) = (x, y ) ∂y ∂y

−

The claim follows immediately from the fact that  u, v  satisfy the Cauchy-Riemann equations. D be the unique conformal Let Ω C be simply connected symmetric region,  z 0  Ω be real, and f  and  f  : Ω  map satisfying f  satisfying  f ((z0 ) = 0, f  (z0 )  > 0  >  0 (as guaranteed by the Riemann Mapping Theorem). Define  g(  g (z ) =  f (  f (z ). D  is holomorphic by the lemma and bijective, being the composition of bijections; hence,  g is Then g Then  g  : Ω conformal. Furthermore, g urthermore,  g((z0 ) = 0 since z since  z 0 , f ( f (z0 ) R. Since

 ⊂  →

 ∈

 →

 ∈

∂u ∂u 0  < f  (z0 ) = (z0 ) = (z0 ) =  g  (z0 ) ∂x ∂x we conclude by uniqueness that f  that  f  =  g.  g . Equivalently Equivalently,,  f (  f (z ) = f  =  f ((z ) z

 ∀  ∈ Ω.

6.1.1 Exercise 2 Suppose now that Ω is symmetric with respect to  z 0  (i.e. z

 ∈ Ω ⇐  ⇐⇒ ⇒ 2z  2 z0 − z ∈ Ω). I claim that  f    f   satisfies f ( f (z ) = 2f ( f (z0 ) − f (2 f (2zz0 − z ) =  −f (2z (2z0 − z ) Define g Define  g  : Ω → D  by  g (z ) =  −f (2z (2z0 − z ). Clearly, Clearly,  g  is conformal, being the composition of conformal maps, and g (z0 ) = 0. Further urthermo more, re, by the chain chain rule, g  (z0 ) = f  (z0 ) >   0. We conclu conclude de from the unique uniquenes nesss statement of the Riemann Mapping Theorem that  g(  g (z ) =  f (  f (z ) ∀z ∈ Ω.

27

Elliptic Functions Weierstrass Theory 7.3.2 Exercise 1

Let f  Let  f  be  be an even elliptic function periods  ω 1 , ω2 . If  f  is  f  is constant then there is nothing to prove, so assume otherw otherwise ise.. First, First, suppose suppose that 0 is neither neither a zero zero nor a pole of  f . f . Observe Observe that that since f   f   is even, its zeroes and poles poles occur in pairs. pairs. Since Since f  is f  is elliptic, f  has f  has the same number number of poles as zeroes. zeroes. So, let a1 ,  , an , and b1 ,  , bn  denote the incongruent zeroes and poles of  f   f   in some fundamental parallelogram P a , where ai aj mod M mod  M,, bi bj mod M  mod  M  i, j  and where we repeat for multiplicity. Define a function  g by

 ≡ −

···

···

 ≡ −

 ∀

 n

g(z ) =  f (  f (z )

℘(z ) ℘(z )

k =1

− ℘(a ) − ℘(b ) k

k



−1

and where ℘  is the Weierstrass p-function with respect to the lattice generated by ω1 , ω2 . I claim claim that that g is a holomorphi holomorphicc elliptic function. function. Since ℘(z ) ℘(ak ) and ℘(z ) ℘(bk ) have double poles at each z M  for all k , g  has a removable singularity at each z M . M . For each each k , ℘(z ) ℘(bk ) has the same poles as ℘ and is therefore therefore an elliptic elliptic function of order 2. Since b Since  b k = 0 and and ℘  is even, it follows that ℘(z ) ℘(bk ) has zeroes of order 1 at z = bk . From our conventio convention n for repeating repeating zeroes and poles, we conclude conclude that g has a removable singularity at bk . The argum argumen entt that g  has removable singularity at each ak  is completely analogous. Clearly, Clearly, g(z + ω1 ) =  g(  g (z  + ω2 ) =  g(  g (z ) for z for  z /  a i  + M  bi + M  M 

−

−

 ∈

−

 

 ±  ±

 ∈

∈

 ∪  ∪

−

 ∪  ∪

so by continuity, we conclude that g  is a holomorphic elliptic function with periods ω1 , ω2  and is therefore equal to a constant  C .  C . Hence, n ℘(z ) ℘(ak ) f ( f (z ) =  C  ℘(z ) ℘(bk )

− −



k=1

Since f  Since  f  is even, its Laurent series series about the origin only has nonzero nonzero terms with even powers. powers. So if  f  if  f    vanishes or has a pole at the origin, the order is 2m, 2 m, m N. Suppose Suppose that that f  vanishes f  vanishes with order 2m 2m. The functi function on given by f ( f ˜(z ) =  f (  f (z ) ℘(z )m is elliptic with periods ω periods  ω 1 , ω2 . f  has f ˜ has a removable singularity at  z  = 0, since ℘ since  ℘((z )k has a pole of order 2k 2 k at z  = 0. Hence, Hence, we are reduced to the previous case of elliptic elliptic function, so applying applying the preceding preceding argument, argument, we conclude that n n ℘(z ) ℘(ak ) C  ℘(z ) ℘(ak ) ˜ f ( f (z ) =  C   f (  f (z ) = m ℘(z ) ℘(bk ) ℘(z ) k ℘(z ) ℘(bk ) k

 ∈

·



=1

− −



 ⇒

=1

− −

If  f  has 2 m  at the origin, then the function given by  f  has a pole of order 2m f ( f (z ) f ( f ˜(z ) = ℘(z )m

is elliptic with periods ω1 , ω2  and has a removable removable singularity singularity at the origin. origin. From the same argument, argument, we conclude that n ℘(z ) ℘(ak ) f ( f (z ) =  C℘  C ℘(z )m ℘(z ) ℘(bk )



k =1

28

− −

7.3.2 Exercise 2 Let f  be f  be an elliptic function with periods ω1 , ω2 . By Ahlfors Ahlfors Theor Theorem em 5 (p. 271 271), ), f  has f  has the same number of zeroes and poles counted counted with multiplicit multiplicity y. Let a1 ,  , an , b1 ,  , bn  denote the incongruent zeroes and n poles of  f ,  f , respectively, where we repeat for multiplicity. By Ahlfors Theorem p. 271, k=1 bk ak  M ,  M , so n n  replacing a replacing  a 1 by a by  a 1  =  a 1  + k=1 bk ak , we may assume without loss of generality that ak = 0. k=1 bk Define a function function g  g by

···



···



 −

 n

g(z ) =  f (  f (z )

k =1

σ (z σ (z

−a ) −b ) k

k



−1

−  ∈  −

where σ where  σ  is the entire function (Ahlfors p. 274) given by σ (z ) =  z

 − 

z 1 z 2 z ew+2(w) ω

1

ω =0 =0



g  has removable singularities at a at  a i + M,  +  M, bi  + M   +  M    for 1 Recall (Ahlfors p. 274) that  σ  satisfies σ (z  + ω1 ) = where η where  η 2 ω1

 −σ(z)e−

η1 (z +

ω1

2

 ≤ i ≤ n.  n . I claim that  g  is elliptic with periods ω periods  ω 1 , ω2 .

) and σ and  σ((z + ω2 ) =

 −σ(z)e−

η2 (z +

ω2

2

) z

∀  ∈ C

 − η1ω2  = 2πi  ≡ b  + M, a  + M , πi   (Legendre’s relation). Hence, for z for  z   M , i

 n

g(z  + ω1 ) =  f (  f (z + ω1 )

k =1

σ (z σ (z

 

−1

− a  + ω1) − b  + ω1) k

k

=  e

n k=1  a k



−b f ( f (z ) k

 − −  − n

=  f (  f (z )

σ (z

k=1

σ (z σ (z

−

ak ) bk )

ω1

2

)

ω1 k+ 2

)

η 1 ( z ak +

− η1 (z −b

− a )e σ (z − b )e

k =1

n

η1

i

k k

−1



−1

=  g(  g (z )

By continuity, we conclude that  g(  g (z  + ω1 ) =  g(  g (z ) z

 ∀  ∈ C. Analogously, for z ≡ b  + M, a  + M , for  z  

 n

g(z  + ω2 ) =  f (  f (z + ω2 )

k =1

σ (z σ (z

i

 

−1

− a  + ω2) − b  + ω2) k

k

=  e

n k=1  a k



−b f ( f (z ) k

n

=  f (  f (z )

σ (z

k=1

σ (z σ (z

−

ak ) bk )

η 2 ( z ak +

− η2 (z −b

− a )e σ (z − b )e

k =1

n

η2

 − −  −

i

k k

ω2

2

ω2 k+ 2

) −1 )



−1

=  g(  g (z )

By continuity, we conclude that g that  g((z + ω2 ) =  g(  g (z ) z C. Since g Since  g  is an entire elliptic function, it is constant by Ahlfors Theorem 3 (p. 270). We conclude that for some  C  C,

 ∀  ∈ n

f ( f (z ) =  C 



k=1

 ∈  ∈

σ (z σ (z

−a ) −b ) k

k

7.3.3 Exercise 1 Fix a rank-2 lattice M  lattice  M 

 ⊂  ⊂ C  and  u ∈/  M .  M . Then σ(z − u)σ(z + u) ℘(z ) − ℘(u) =  − σ (z )2 σ (u)2

Proof.   I first claim that the RHS is periodic with respect to M . M . Let ω1 , ω2   be generators of  M   M   and let η1 ω2 η2 ω1  = 2πi. πi . For z For  z /  M ,  M ,

 −

−

∈

σ (z  + ω1 u)σ (z  + ω1  + u) = σ(z + ω1 )2 σ (u)2

 −

 −

σ (z

η1 (z u+

− u)e

ω1

σ (z  + u)eη1 (z+u+ ω1 σ (z )2 e2η1 (z+ 2 ) σ (u)2

−

2

)

u)  −σ(zσ−(zu)2)σσ((uz + =  f (  f (z ) 2 )

=

29

ω1

2

)

=

 −

ω1

σ (z u)σ(z  + u)e2η1 (z+ 2 ω1 σ (z )2 σ (u)2 e2η1 (z + 2 )

−

)

The argument for ω for  ω 2  is completely analogous. The RHS has zeroes at the same reasoning used above, we see that ℘(z )

 ±u and a double pole at 0. Hence, by

u) − ℘(u) =  −C σ(zσ−(zu)2)σσ((uz +  ∈ C for some C  some  C  ∈ 2 )

To find conclude that C  = C  = 1, we first note that ℘(z )  ℘(  ℘(u) has a coefficient of 1 for the z −2 term in its Laurent Laurent expansion. expansion. If we show that the Laurent expansion expansion of the f  the  f ((z ) also has a coefficient of 1 for the  z −2 , then it follow follow from the uniquenu uniquenuess ess of Laurent Laurent expansions expansions that C  that  C  =  = 1.

 −

−

−u + 1 ( z−u )2

 −   −    −         − −  −   −              − −  ·   −        (z 2

− u2 )  −

σ (z u)σ(z  + u) = σ(z )2 σ(u)2

−

ω =0 =0



=

1

z

z

 1

ω =0 =0

−u e ω

z

z 2 σ (u)2

−u e ω

z



2

σ(u)2

ω





z ω

1

ω =0 =0

ω =0 =0

ω =0 =0

eω+2(ω)

z +u ω

1

z

e

z

z +u 1 ω +2

2

e

z +u 1 ω +2

( z+ωu )2

2

( z+ωu )

2

2

2

eω+2(ω) z

1

z

z ω

1

z +u ω

 1

ω

1

ω =0 =0

−u + 1 ( z−u )2 ω

2

ω

g1 (z )

2  1 u + 2 z

z

 1

ω =0 =0

−u e ω

z

−u + 1 ( z−u )2 ω

σ (u)2

2

1

ω=0 =0



z +u ω

 1

ω

ω =0 =0

z ω

eω+2(ω) z

1

z

2

e

z +u 1 ω +2

( z+ωu )2

2

g2 (z )

Observe that both g1 (z ) and g2 (z ) are holomorphic in a neighborhood of 0 since we have eliminated the double double pole at 0. Hence, Hence, the coefficient of the z −2 in the Laurent expansion of  f  of  f ((z ) is given by g2 (0). (0). But since σ since  σ  is an odd function, it is immediate that  g 2 (0) = 1.

7.3.3 Exercise 2 With the hypotheses hypotheses of the preceding problem, problem, ℘ (z ) = ζ (  ζ (z ℘(z ) ℘(u)

−

− u) + ζ (z + u) − 2ζ (z)

Proof. For z or z =  u + M , M , we can choose a branch of the logarithm holomorphic in a neighborhood of  ℘ of  ℘((z ) ℘(u). Taking the derivative of the log of both sides and using the chain rule,

 

−

℘ (z ) ∂  = log( σ (z ℘(z ) ℘(u) ∂z



−

− u) +  σ  (z + u) −  2σ  2 σ  (z )  ζ (z − u) + ζ (z + u) − 2ζ (z ) − u) σ(z + u) σ(z) = ζ ( ( )  ζ (w) ∀w ∈ C  (Ahlfors p. 274). ( ) =  ζ ( =

where we’ve used

σ (z σ (z



− − u)) + log(σ log(σ(z − u)) − log(σ log(σ (u)2 σ(z )2 )

σ w σ w

7.3.3 Exercise 3 With the same hypotheses as above, for  z =

   −u + M ,

ζ (z  + u) =  ζ (  ζ (z ) + ζ (u) +

 1 ℘ (z ) 2 ℘(z )

− ℘ (u) − ℘(u)

Proof.  Since the last term has a removable singularity at  z  = u  =  u + M , by continuity, we may also assume that z =  u + M . First, First, observe observe that by replacing replacing switching switching u  u  and z  and  z  in the argument for the last identity, we have that ℘ (u) = [ζ (u z ) + ζ (z + u) 2ζ (u)] =  ζ (  ζ (z u) ζ (z  + u) + 2ζ  2 ζ (u) ℘(z ) ℘(u)

 

−

 −

−

−

30

− −

where we’ve used the fact that  σ (z ) is odd and therefore  ζ (  ζ (z ) = ℘ (z ) ℘ u = (ζ (z ℘(z ) ℘(u)

− −

σ  (z )   is σ (z )

also odd. Hence,

− u) + ζ (z + u) − 2ζ (z)) − (ζ (z − u) − ζ (z + u) + 2ζ  2 ζ (u)) = 2ζ  2ζ (z + u) − 2ζ (z ) − 2ζ (u)

The stated identity follows immediately.

7.3.3 Exercise 4 By Ahlfors Section 7.3.3 Exercise 3,





− ℘ (u) − ℘(u) Differentia Differentiating ting both sides with respect to z to  z  and using −ζ  (w) =  ℘(  ℘(w) ∀w ∈ C \ M , M , we obtain   ( ℘ (z ) − ℘ (u))℘ ))℘ (z ) −℘(z + u) =  −℘(z) +  12 ℘(z℘) −(z℘)(u) −  (℘ (℘(z ) − ℘(u))2 We seek an expression for  ℘  (z) in terms of  ℘ of  ℘((z ). For z For  z   = 2 , 2  , + + M , 2  g 2 ℘ (z )2 = 4℘(z )3 − g2 ℘(z ) − g2 ⇒ 2℘  2 ℘ (z )℘ (z ) = 12℘ 12℘(z )2 ℘ (z ) − g2 ℘ (z ) ⇒ ℘ (z ) = 6℘(z )2 − 2  1 ζ (z + u) =  ζ (  ζ (z ) + ζ (u) + 2

℘ (z ) ℘(z )





ω1  ω 2  ω 1

We conclude from continuity that  ℘  (z ) = 6℘(z )2  1 ℘(z ) + 2

−℘(z + u) =  −



g2

2  .

−

ω2

Substituting this identity in,

6℘(z )2 g22 ℘(z ) ℘(u)

−  −  (℘  ( ℘ (z ) − ℘ (u))℘ ))℘ (z ) − (℘(z ) − ℘(u))2



Applying the same arguments as above except taking  u  to be variable, we obtain that  1 ℘(u) + 2

−℘(z + u) =  −

−

6 p(  p(u)2 g22  (℘  ( ℘ (z ) ℘ (u))℘ ))℘ (u) +  p(  p(z )  p(  p(u) (℘(z ) ℘(u))2

−

−

−

−



Hence, 2 − ℘(u)2) −  (℘  ( ℘ (z ) − ℘ (u))2 1 = 2℘ 2 ℘(z )+2℘ )+2℘(u)− 2 − ℘(u) (℘(z ) − ℘(u)) 2 2  1 ℘ (z ) − ℘ (u) ⇒ ℘(  ℘ (z  + u) =  −℘(z ) − ℘(u) + 4 ℘(z ) − ℘(u)

1 ℘(z )+ ℘(u)+ 2

−2℘(z+u) =  −

−





6(℘ 6(℘(z )2 ℘(z )





7.3.3 Exercise 5 Using the identity obtained in the previous exercise, we have by the continuity of  ℘ of  ℘  that ℘(2z (2z ) = lim ℘(z  + u) = lim u

= lim u

→z

 −

→z

u

℘(u) −

where we use the continuity of  w  w

 1 ℘(z ) + 4

→z





−℘(z) −

℘ (z ) z ℘(z ) z

−℘ (u) −u −℘(u) −u

 1 ℘(u) + 4

 



  

− ℘(u) − ℘(u)

2

 1  = −2℘(z ) + 4

 → w 2 to obtain the last expression.

31

℘ (z ) ℘(z )

2

℘ (z ) ℘ (z )

2



℘ (z ) ℘(z )

− ℘(u) − ℘(u)



2

7.3.3 Exercise 7 Fix u Fix  u,v ,v /  M  such  M  such that u = v , and define a function  f  : C M 

∈

f ( f (z ) = det

 

\  →  → C by

 | |   | |

   − − −        −    −

℘(z ) ℘(u) ℘(v )

℘ (z ) 1 ℘ (u) 1  = ℘ (v) 1

℘ (z )(℘ )(℘(u)

= (℘ ( ℘ (u) + ℘ (v )) ℘(z ) + (℘ ( ℘(v ) A

℘(v)) + ℘ (u)(℘ )(℘(z )

℘(u)) ℘ (z ) +

− ℘(v)) + ℘ (v)(℘ )(℘(z ) − ℘(u))

(℘ (u)℘(v) + ℘ (v )℘(u))



B

 

C 

where we use Laplace expansion expansion for determina determinants nts.. By our choice choice of  u, v  and the fact that the Weierstrass function is elliptic of order 2, B = 0. Hence, Hence, f   f ((z ) is an elliptic function of order 3 with poles at the lattice points of  M .  M . Since the determinant of any matrix with linearly dependent rows is zero,  f  has  f  has zeroes at u, at  u, v . Since f  Since  f  has  has order 3, it has a third zero  z,  z , and by Abel’s Theorem (Ahlfors p. 271 Theorem 6),

 

−

u

− v + z ≡ 0

We conclude that det

mod  M 

 ⇒  z  = v  ⇒  =  v − u

 

 

℘ (z ) 1  ℘ (u) 1  = 0 ℘ (u + z ) 1

℘(z ) ℘(u) ℘(u + z )

−

7.3.5 Exercise 1 Since λ Since  λ  is invariant under Γ(2) and Γ Γ(2) is generated by the linear fractional transformations  τ   τ  +  τ  + 1 1 1 − − and τ  τ  , it suffices to show that J (τ  +   + 1) = J (τ ) τ ) and J ( τ  ) = J (τ ). τ ). Recall Recall that that λ  satisfies the functional equations λ(τ ) τ ) 1 λ(τ  +  + 1) = and λ and  λ  = 1 λ(τ ) τ ) λ(τ ) τ ) 1 τ 

\

  → −

−

− 

−

So, J (τ  +  + 1) =

  →

−

4 (1 λ(τ  +  + 1) + λ(τ  +  + 1)2 )3 4 (1 λ(τ )( τ )(λ λ(τ ) τ ) 1)−1 + λ(τ ) τ )2 (λ(τ ) τ ) 1)−2 )3  (λ  ( λ(τ ) τ ) = 2 2 2 27 λ(τ  +  + 1) (1 λ(τ  +  + 1)) 27 λ(τ ) (λ(τ ) τ ) τ )2 (λ(τ ) τ ) 1)−2 (1 λ(τ )( τ )(λ λ(τ ) τ ) 1)−1 )

−

−



τ ) 4 (λ(τ ) = 27 and J 

−

− −

−

  −

·

− 1)6 − 1)6

3 − 1)2 − λ(τ )( τ )(λ λ(τ ) τ ) − 1) + λ(τ ) τ )2 4 (1 − λ(τ ) τ ) + λ(τ ) τ )2 )3 = =  J (  J (τ ) τ ) λ(τ ) τ )2 (λ(τ ) τ ) − 1)2 27 λ(τ ) τ )2 (λ(τ ) τ ) − 1)2

−   −

Observe that

−

−

3 − λ(τ )) τ ))2 4 = 2 − − λ(τ ))) τ ))) 27

τ )) + (1 1 4 1 (1 λ(τ ))  = 2 τ  27 (1 λ(τ )) τ )) (1 (1

−

−

−



τ ) + λ(τ ) τ )2 4 1 λ(τ ) J (τ ) τ ) = 27 λ(τ ) τ )2 (1 λ(τ )) τ ))2

 −  π

So, J (τ ) τ ) assumes the value 0 on  λ −1 e±i 3 of order 3. We proved in Problem Set 8 that g2  =

3



1 λ(τ ) τ ) + λ(τ ) τ )2 λ(τ ) τ )2 (1 λ(τ )) τ ))2

−

π

3

=  J (  J (τ ) τ )

π

4 (λ(τ ) τ ) ei 3 )3 (λ(τ ) τ ) e−i 3 )3 = 27 λ(τ ) τ )2 (1 λ(τ )) τ ))2

−

−

. Since λ  is a bijection on Ω

−

∪ Ω, J (τ ) τ ) has two zeroes, each

 −4(e 4(e1 e2  + e1 e3  + e2 e3 ) = 0

2π

2π

for τ  for  τ  =  e i 3  . So using the identit identity y for J (τ ) τ ) proved below, J (ei 3  ) = 0. Using the invarian invariance ce of  J (  J (τ ) τ ) under π i3 Γ, we see that J  that  J ((e  ) = 0. the  λ i  are the roots of degree 6 the polynomial J (τ ) τ ) assumes the value 1 on  λ −1 ( λ1 ,  , λ6 ), where the λ

{ ···

}

 p(  p(z ) = 4(1 It is easy to check that e3  = ℘  =  ℘

− z + z2)3 − 27 27zz 2 (1 − z )2

   −   1+i ; i  = 2

℘

i+1 ; i  = 2

32

 −e3 ⇒ e3 = 0

Since e Since  e 1  + e2  + e3  = 0 (see below for argument), we have  e 1  = λ(i) =

e3 e1

 −e2  and therefore

 − e2 = 1  − e2 2

Since each point in H+ is congruent modulo 2 to a point in Ω Ω , λ   maps this fundamental conformally onto C 0, 1 , and J  and  J ((τ ) τ ) is invariant under Γ, we conclude  J (  J (τ ) τ ) assumes the value 1 at  τ  =  i, 1 + i, +  i, i+1 2 . I claim that these these are these are the only possible points up to modulo 2 congruenc congruence. e. Suppose J (τ ) τ ) = 1 for  i +1 τ / i, 1 + i, 2 . If we let S  let  S 1 ,  , S 6  denote the complete set of mutually incongruent transformations, then

∪

\{ }

∈

   ∈ iπ 3

···

i 23π

since τ since  τ / e , e  (otherwise J   (otherwise  J ((τ ) τ ) = 0), S  0),  S 1 τ,  , S 6 τ   Ω Ω  are distinct, hence the  λ(  λ (S i τ ) τ ) are distinct roots of  p(  p(z), and we obtain that p that  p((z ) has more than 6 roots, a contradiction. Moreover, this argument shows that the polynomial  p(  p (z ) has three roots, which by inspection, we see are given by 1, 21 , 2 . I claim that  J (  J (τ ) τ ) assumes the value 1 with order 2 at  τ  =  i, 1 + i, +  i,  i +1 2  . We need to show that the zeroes of   p(  p(z ) are each of order 2. Indeed, one can verify that

 ∈  ∈ ∪

− 

 p(  p(z ) = 4(1 Substituting λ Substituting  λ =  =

···

e3 e1

− z + z2)3 − 27 27zz 2 (1 − z )2 = (z ( z − 2)2 (2z (2z − 1)2 (z  + 1) 2

−e2 , we have −e2

 −  −  −   − −

4 = 27 Since 4z 3





3  − e2)−1 + (e (e3 − e2 )2 (e1 − e2 )−2  − e2)−2(e1 − e3)2(e1 − e2)−2 3 )(e1 − e2 ) + (e ( e3 − e2 )2 4 (e1 e2 )2 (e3 − e2 )(e = 27 (e3 − e2 )2 (e1 − e3 )2 (e1 − e2 )2 3 e21 − 2e1 e2  + e22 − e3 e1  + e3 e2  + e2 e1 − e22  + e23 − 2e3 e2  + e22 (e3 − e2 )2 (e1 − e3 )2 (e1 − e2 )2

4 1 (e3 J (τ ) τ ) = 27 (e3

e2 )(e )(e1 e2 )2 (e1





− g2z − g3  = 4(z 4(z − e1 )(z )(z − e2 )(z )(z − e3 ) = 4(z 4(z 2 − (e1  + e2 )z  + e1 e2 )(z )(z − e3 ) = 4(e 4(e1  + e2  + e3 )z 2 + · · ·

we have that e that  e 1  + e2  + e3  = 0 and so, 0 = (e ( e1  + e2  + e3 )2 =  e 21  + e22  + e23  + 2e 2 e1 e2  + 2e 2 e1 e3  + 2e 2 e2 e3

 ⇒ e21 + e22 + e33  =  −2 (e1e2 + e1e3 + e2e3)

Substituting this identity in, 3

4 ( 2(e 2(e1 e2  + e2 e3  + e1 e3 ) (e1 e2  + e2 e3  + e1 e3 )) J (τ ) τ ) = = 27 (e3 e2 )2 (e1 e3 )2 (e1 e2 )2

−

 −

−  −

 −

 −

33

3

(e1 e2  + e2 e3  + e1 e3 ) 4 (e3 e2 )2 (e1 e3 )2 (e1 e2 )2

 −

 −

 −