Complex Analysis 11111

7/31/2019 Complex Analysis 11111

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Figure 0.1

COMPLEX MADE SIMPLE

DAVID C. ULLRICH

Solutions provided by Scott Larson.

Contents

Part 1. Complex Made Simple 30. Differentiability and the Cauchy-Riemann Equations 31. Power Series 6

2. Preliminary Results on Holomorphic Functions 83. Elementary Results on Holomorphic Functions 134. Logarithms, Winding Numbers and Cauchys Theorem 195. Counting Zeroes and the Open Mapping Theorem 306. Eulers Formula for sin( z) 336.0. Motivation 336.1. Proof by the Residue Theorem 336.2. Estimating Sums Using Integrals 346.3. Proof Using Liouvilles Theorem 357. Inverses of Holomorphic Maps 368. Conformal Mappings 388.0. Meromorphic Functions and the Riemann Sphere 388.1. Linear-Fractional Transformations, Part I 38

8.2. Linear-Fractional Transformations, Part II 398.3. Linear-Fractional Transformations, Part III 398.4. Linear-Fractional Transformations, Part IV:

The Schwarz Lemma and Automorphisms of the Disk 398.5. More on the Schwarz Lemma 409. Normal Families and the Riemann Mapping Theorem 419.0. Introduction 419.1. Quasi-Metrics 419.2. Convergence and Compactness in C (D ) 42

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Part 1. Complex Made Simple

0. Differentiability and the Cauchy-Riemann Equations

Proposition 0.0. Suppose that V is an open subset of C , that f : V C , and z V . Then f is complex-differentiable at z if and only if it is real-differentiable at z and the real derivative Df (z) is

complex-linear.Proposition 0.1. Suppose that V is an open subset of C , that f : V C , and z V . Then f = u + ivis complex-differentiable at z if and only if it is real-differentiable at z and the real and imaginary parts satisfy the Cauchy-Riemann equations

ux (z) = vy (z), uy (z) = vx (z).

Corollary 0.2. Suppose that V is an open subset of C , that f : V C , and z V ; write f = u + iv.If f is complex-differentiable at z V then u and v satisfy the Cauchy-Riemann equations at z.

Corollary 0.3. Suppose that V is an open subset of C , that f : V C , and z V ; write f = u + iv. If the rst-order partial derivatives of u and v are continuous at z V and satisfy the Cauchy-Riemann equations there then f is complex-differentiable at z.

0.1. Suppose that f is dened in a neighborhood of the complex number z. Show that f is complex-differentiable at z if and only if there exists a complex number a such that f (z + h) = f (z) + ah + o(h)as h 0.

Proof. Suppose that f is complex-differentiable at z so

f (z) = limh 0

f (z + h) f (z)h

.

Now let E (h) be a function such that f (z + h) = f (z) + f (z)h + E (h) for all h in the neighborhoodof z for which f is dened. But E (h) satises E (0) = 0 and lim h 0 |

E (h ) ||h | = 0, so f (z) = a.

Now suppose that there exists a complex number a such that f (z + h) = f (z) + ah + o(h) as h 0.Then

a = limh 0

f (z + h) f (z)

h o(h)

h= f (z).

0.2. Suppose that f is complex-differentiable at z. Show that f is continuous at z.

Proof. Since f is complex-differentiable at z,

f (z) = limh 0

f (z + h) f (z)h

.

Since this limit exists, lim h 0 f (z + h) = f (z) so f is continuous at z.

0.3. Suppose that T : C C is R -linear. Show that T is C -linear if and only if T (iz ) = iT z for allz C .

Proof. Suppose that T is C -linear so that for every c, z C we haveT (cz) = cTz.

Since i, z C we haveT (iz ) = iTz.

Now suppose that T (iz ) = iT z for all z C . Since T is R -linear, for every z, w C and c R wehave

T (z + w) = T z + T w, T (cz) = cTz.


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Proof (iii).

f (z) = limh 0

f (z + h) f (z)h

= limh 0

(z + h)n

zn

h

= limh 0

ni=0

ni z

n i h i zn

h

= limh 0

n

i =1

ni

zn i h i 1

= nz n 1 .

Proof (iv).

f (z) = limh 0

f (z + h) f (z)

h= lim

h 0

(z + h)n zn

h=

0.6. Suppose that f is differentiable at z and that g is differentiable at f (z). Show that the compositiong f is differentiable at z, with

(g f ) (z) = g (f (z))f (z).

Proof.

0.7. Dene f : C C by

f (x + iy) =

0 (x = 0) ,

0 (y = 0) ,1 (otherwise) .

Show that f satises the Cauchy-Riemann equations at the origin although f is not complex differen-tiable at the origin.

Proof. We see that f (C ) R so u((x, y)) = f ((x, y)) and v((x, y)) = 0. Calculating the partialderivatives of u and v at the origin gives us

ux (0) = limh 0

u((h, 0)) u(0)h

= 0 ,

uy (0) = limh 0

u((0, h)) u(0)h

= 0 ,

vx (0) = 0 ,

vy (0) = 0 .Thus the Cauchy-Riemann equations hold trivially. However, f is not continuous at the origin andtherefore must not be complex differentiable at the origin either.

0.8. Dene f : C C by

f (z) = |z |2 sin(1/ |z |) (z = 0) ,

0 (z = 0) .Show that f is complex differentiable at the origin although the partial derivative ux is not continuousat the origin.


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Proof. First we show that if 0 < x < 1 then there exists y [0, 1] such that xy = . Let0 < x < 1. Then we let y = log / log x and notice that the given inequality shows that this iswell dened and y [0, 1]. But

xy = x log / log x

= xlog

x

= .

Next we let a = 0. If we let xn = 1n n > 0 and yn =1n > 0, we observe that lim n xn =

limn yn = 0. But now we see

limn

xynn = limn 1

nn1/n

= limn

1n

= 0 .

Thus this is true for a = 0.Now we let a (0, 1]. So let xn = 1 /n > 0 so that lim n xn = 0. Now let > 0 be given. Observe

that there exists N N such that

0 0 such that lim n yn = 0 with the string of inequalitiessuch that for any n > N ,

|xynn a | =1n

yn

a

= |a 1 a |= 1< .

Therefore for any a [0, 1] there exist sequences xn > 0 and yn > 0 such that lim n xn = lim n =

0 andlim

n xynn = a.

1.2. Suppose the power series n =0 cn zn has radius of convergence R > 0, and dene f : D (0, R) C

by

f (z) =

n =0cn zn .


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8 DAVID C. ULLRICH

We have shown that

f (z) =

n =1ncn zn 1 (z D (0, R));

explain why it would not be quite correct to write

f (z) =

n =0ncn zn 1 (z D (0, R))

instead, although it is perfectly correct to say f = n =0 f n , where f n (z) = cn zn .

Solution. Since 0 D (0, R), this formula is valid for f (0). Expanding the sum we see that this formulawould be dividing by zero.

f (0) = 0 c0 10

+

However,

f 0(0) = limh 0c0h0 0 00

h= 0 .

So f 0 = 0 for all z D(0, R) so that it is perfectly correct to use the formula

f =

n =0f n ,

where f n (z) = cn zn .

2. Preliminary Results on Holomorphic Functions

2.1. Dene f : C C by f (z) = z2 sin(1/z ) for z = 0, f (0) = 0. The last few paragraphs almostseem to give a proof that f is differentiable everywhere but f is not continuous at the origin. This isimpossible where does the proof fail?

Proof. To see where the proof for f C (R ) fails, let us look at the derivative of f at the origin andnotice that h C .

f (0) = limh 0

f (h)

h= lim

h 0

h2 sin(1/h )h

= limh 0

h sin(1/h )

The proof for f C (R ) says that |h sin(1/h )| | h | because |sin(x)| 1 for x R . But this is not truefor |sin(z)| for z C . For example,

|sin( 3i)| =exp(3) exp( 3)

2i

(1 + 3) 1/ (1 + 3)2i

= | (15/ 8)i |= 15 / 8.

Lemma 2.0 (Cauchys Theorem for Derivatives) . Suppose V is an open subset of the plane, f : V Cis continuous, and there exists an F : V C such that f = F in V . Then

f (z) dz = 0


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for any smooth closed curve in V .

Lemma 2.1 (ML Inequality) . Suppose that : [a, b] C is a smooth curve and f is continuous on . If |f | M on and L is the length of then

f (z) dz

ML.

2.2.(i) Show that [ b, a] = [a, b].

(ii) Generalize this: Suppose that 1 is a smooth curve in the plane. How can you dene anothersmooth curve 2 so that 2 = 1?

Proof (i). By denition, [ b, a] is the path 1 : [0, 1] C dened by

1(t) = (1 t)b + ta,

and [a, b] is the path 2 : [0, 1] C dened by

2(t) = (1 t)a + tb.

So calculating with the denition of 1 = 2 , we get

[b,a ] f (z) dz = 1

0f 1(t1) 1(t1) dt1 (Let t2 = 1 t1)

= 0

1f 1(1 t2) (a b) dt2

= 1

0f 2(t2) 2(t) dt2

= [a,b ] f (z) dz.Proof (ii). Let 1 : [a, b] C be a smooth path in the plane. We can generalize (i) by dening asmooth path 2 : [a, b] C such that 2 = 1 to be

2(t) = 1 a + b t , t [a, b].

We check that this is well dened by the following calculation.

1 f (z) dz = b

af ( 1(t1)) 1(t1) dt1 (Let t2 = a + b t1)

= a

bf ( 1(a + b t2)) 1(a + b t2) dt2

=

b

af ( 2(t2)) 2(t2) dt2 ( 2(t) = 1(t))

= 2 f (z) dz2.3. Suppose that p [a, b] and show that [ a, b] = [a, p]+[ p, b].

Proof. Let [a, b] be the path : [0, 1] C given by

(t) = (1 t)a + tb.


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10 DAVID C. ULLRICH

Since p [a, b], there exists t0 [0, 1] such that (t0) = p. Let [a, p] and [ p, b] be the paths 1 , 2 : [0, 1] C given by

1(t) = (1 t)a + tp, 2(t) = (1 t) p + tb.Let t1 = t/t 0 and t2 = ( t t0)/ (1 t0) and notice dt1 = 1 /t 0 dt and dt2 = 1 / (1 t0) dt for the following

calculations.

[a,b ] f (z) dz = 1

0f ( (t)) (t) dt

= t 0

0f ( (t))( b a) dt +

1

t 0f ( (t))( b a) dt

= 1

0f ( (t0 t1)) t0 (b a) dt1 +

1

0f ( (t2(1 t0) + t0))

b a1 t0

dt2

= 1

0f ((1 t1)a + t1 p)( p a) dt1 +

1

0f ((1 t2)a + t2 p)(b p) dt2

=

1

0f ( 1(t1)) 1(t1) dt1 +

1

0f ( 2(t2)) 2(t2) dt2

= [a,p ] f (z) dz + [ p,b ] f (z) dzTherefore [ a, b] = [a, p]+[ p, b].

a

b

c

p

q

r

T 1T

2

T 3

T 4

Figure 2.1. Triangle T = [a,b,c], where T 1 = [a,p,r ], T 2 = [b,q,p], T 3 = [c,r,q ],and T 4 = [ p, q, r].

2.4. Show that T = T 1 + T 2 + T 3 + T 4 for any triangle T .

Proof. Let T be given by gure 2.1 so that T = [a, b]+[ b, c]+[ c, a]. But T 1 = [a, p]+[ p, r ]+[ r, a ],T 2 = [b, q ]+[ q, p]+[ p, b], T 3 = [c, r ]+[ r, q ]+[ q, c], and T 4 = [ p, q ]+[ q, r]+[ r, p ]. Thus calculating

with rules from previous exercises, we getT = [ a, b]+[ b, c]+[ c, a]

= [ a, p]+[ p, b]+[ b, q ]+[ q, c]+[ c, r ]+[ r, a ]

= [ a, p]+[ p, r ] [ p, r ]+[ p, b]+[ b, q ]+[ q, p] [q, p]+[ q, c]+[ c, r ]+[ r, q ] [r, q ]+[ r, a ]= [ a, p]+[ p, r ]+[ r, a ]+[ b, q ]+[ q, p]+[ p, b]+[ c, r ]+[ r, q ]+[ q, c]+[ p, q ]+[ q, r]+[ r, p ]

= T 1 + T 2 + T 3 + T 4 .


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12 DAVID C. ULLRICH

Proof. We dene a function

g(w) =f (w)

w z0and notice that

g(z0 + re it ) = f (z0 + reit

)z0 + re it z0

= f (z0 + reit

)re it

.

By Cauchys integral formula for disks we get the formula

f (z) =1

2i f (w) dww z .Thus we calculate f (z0) as follows:

f (z0) =1

2i g(w) dw=

12i

2

0

g( (t)) (t) dt

=1

2i 2

0g(z0 + re it )ire it dt

=i

2i 2

0f (z0 + re it ) dt

=1

2 2

0f (z0 + re it ) dt.

Theorem 2.6. Suppose V is an open subset of the plane and f : V C is differentiable at every point of V . Suppose that D(z0 , r ) V . Then there exists a sequence of complex numbers (cn ) such that

f (z) =

n =0cn (z z0)n

for every z D (z0 , r ) (and hence the series converges uniformly on every compact subset of D (z0 , r )).Further, the coefficients are given by the formula

cn =1

2i f (w) dw(w z0)n +1 ,where : [0, 2] V is dened by (t) = z0 + re it .

Corollary 2.7 (Cauchys Theorem in a Convex Set) . Suppose that V C is a convex open set and f H (V ). Then

f (z) dz = 0 for any smooth closed curve V .

Corollary 2.8 (Moreras Theorem) . Suppose that V is an open subset of the plane and f : V C is continuous. If T f (z) dz = 0 for every triangle T V then f H (V ).Corollary 2.9. If f H (V ) then f H (V ); hence by induction f is innitely differentiable.


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3. Elementary Results on Holomorphic Functions

Corollary 3.0. Suppose that f n H (V ) for n N and f n f in H (V ). Then f H (V ).

Corollary 3.1 (Cauchy Integral Formula for Derivatives) . Suppose f H (V ) and D (z0 , r ) V .Dene : [0, 2] V by (t) = z0 + re it . Then

f (n ) (z) = n!2i f (w) dw(w z)n +1 (n N )

for all z D(z0 , r ).

Corollary 3.2 (Cauchys Estimates) . Suppose that f is holomorphic in a neighborhood of the closed disk D (z0 , r ) and |f | M in D (z0 , r ). Then

f (n ) (z0) Mn !r n

(n N ).

Lemma 3.3. Suppose that V C is open and K V is compact. Then K is at positive distance from the copmlement of V : There exists > 0 such that

|z w| >

for all z K and w C \ V .Lemma 3.4. Suppose that K C is compact, > 0, and let

K =z K

D (z, ).

Then K is compact.

Proposition 3.5. Suppose that f n H (V ) for n N and f n f in H (V ). Then f n f in H (V ).

3.1. Dene f : R R by f (x) = 1 / (1 + x2). Show that for every x R there exists > 0 such thatf is represented by a power series convergent in ( x , x + ), although the power series at the origindoes not converge on the entire line.

Proof. Let V = C \ { i} and dene F : V C by

F (z) = 11 + z2

.

Since F is differentiable at every point of V , theorem 2.6 states there is a sequence of complex numbers(cn ) such that

F (z) =

n =0cn (z z0)n

for every z D(z0 , |z i| / 2). Since cn = F ( n ) (z0 )

n ! =f ( n ) (z0 )

n ! , the coefficients are real for all z0 R .Now if x0 R is given, we have

f (x) =

n =0cn (x x0)n

for all x (x0 | z i| / 2, x0 + |z i| / 2).Suppose that n =0 cn xn is a power series for f centered at the origin converging to f for all real

numbers, with cn = f ( n ) (0)

n ! . Then ( cn rn ) is bounded for all r (0, ), so R = . But these are the

same coefficients for the power series of f (z) = 11+ z 2 centered at the origin, with R = 1. This is acontradiction, so there is no such power series.

Corollary 3.6 (Liouvilles Theorem) . A bounded entire function must be constant.

Corollary 3.7 (Fundamental Theorem of Algebra) . A complex polynomial of positive degree has a (complex root).


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14 DAVID C. ULLRICH

Corollary 3.8. Suppose that f H (V ) and V is connected. If all the derivatives of f vanish at some point of V then f is constant.

Lemma 3.9. Suppose that f H (V ) and f has a zero of order N at z V . Then there exists g H (V ) with g(z) = 0 such that

f (w) = ( w z)N g(w) for all w V .

3.2. Suppose that A and B are open sets, f H (A), g H (B ), and f = g in A B . Show that thereexists F H (A B ) such that F = f in A and F = g in B .

Proof.

3.3. Give an example of two sets A, B C , a function f analytic on A and a function g analytic onB such that A B = but there does not exist a function F analytic on A B with F = f on A andF = g on B .

Example.

Corollary 3.10. Suppose that f H (V ) and V is connected. If Z (f ) has a limit point in V then f is constant.

Theorem 3.11 (Maximum Modulus Theorem) . Suppose that f H (V ) and V is connected. Then |f | cannot achieve a (local) maximum in V unless f is constant: If f is nonconstant then for every a V and > 0 there exists z V with |f (z)| > |f (a)| and |z a | < .

Proposition 3.12 (Parseval Formula) . Suppose that the power series

f (z) =

n =0cn (z a)n

converges for |z a | < r . Then

12

2

0f (a + eit

2dt =

n =0

|cn |2 2n

for every [0, r ).Lemma 3.13. If a holomorphic function is bounded near an isolated singularity then the singularity is removable.

Proposition 3.14. Suppose that f H (D (z, r )) .(i) f has a pole at z if and only if there exist a positive integer N and complex numbers (cn )n N

such that c N = 0 and

f (w) =

n = N cn (w z)n

for w D (z, r ).(ii) f has an essential singularity at z if and only if f (D (z, )) is dense in C for all (0, r ).

Proposition 3.15. Suppose that f H (D (z0 , r )) has the Laurent series expansion

f (z) =

n = cn (z z0)n .

Then (i) f has a removable singularity at z0 if and only if cn = 0 for all n < 0.

(ii) f has a pole of order N > 0 at z0 if and only if c N = 0 and cn = 0 for all n < N .(iii) f has an essential singularity at z0 if and only if there exist innitely many values of n < 0

such that cn = 0 .


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3.4. Suppose that f H (C ) and |f (z)| eR e z for all z. Show that f (z) = cez for some constant c.

Proof. We begin with the inequality

f (z)ez

=|f (z)|eR e z

1.

This shows f (z)/e z is a bounded entire function so Liouvilles theorem states that f (z)/e z = c forsome constant c. Therefore f (z) = cez .

3.5. Suppose that f H (C ), n is a positive integer and |f (z)| (1 + |z|)n for all z. Show that f is apolynomial.

Proof 1. Since f H (C ) we can write f as

f (z) =

k =0

ck zk ,

for all z C . With some calculations we see that

f (z) =

k =0

ck zk

=n

k =0

ck zk +

k = n +1

ck zk

=n

k =0

ck zk + zn +1

k =0

ck + n +1 zk .

Now let h H (C ) be given by

h(z) =

k=0

ck + n +1 zk .

For clarity note that we now have

(3.1) f (z) =n

k =0

ck zk + zn +1 h(z).

Since h H (C ) there is some a C such that h(0) = a. If z = 0, then we get

h(z) =f (z) nk =0 ck zk

zn +1.

Thus we can calculate for z = 0,

|h(z)| =f (z) nk=0 ck zk

zn +1

1

zn +1|f (z)| +

1zn +1

n

k =0

ck zk

1zn +1

(1 + |z|)n + 1zn +1

n

k =0

|ck | | z|k

=1

|z |1 +

1|z|

n

+n

k=0

|ck | | z |k n 1 .


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16 DAVID C. ULLRICH

This shows lim z h(z) = 0 hence h is bounded and entire. By Liouvilles theorem we can let h(z) = cfor some c C . But notice that lim z h(z) = 0 so c = 0 and thus h(z) = 0 for all z C . Thereforeequation 3.1 shows that

f (z) =n

k =0

ck zk ,

so f is a polynomial.

Proof 2. Since f H (C ) we can write f as

k =0

ck zk ,

for all z C . Corollary 3.2 shows

|ck | (1 + r )n

r k, (r > 0).

If k > n , then(1 + r )m

rk 0, (r ).

So ck = 0 for all k > m .

3.6. Suppose that f, g H (D (z, r )), 1 m n, f has a zero of order n at z and g has a zero of orderm at z. Show that f /g has a removable singularity at z.

Proof. Since f, g H (D (z, r )), there exist F, G H (D (z, r )) such that F (z) = 0 = G(z) and

f (w) = ( w z)n F (w), g(w) = ( w z)m G(w).

Notice that g(z) = 0 implies that f /g has a singularity at z. But for w D (z, r ) we have

f (w)g(w)

=(w z)n F (w)(w z)m G(w)

= ( w

z)n mF (w)

G(w).

Since (w z)n m F (w)/G (w) H (D (z, r )), then f/g has a removable singularity at z by denition.

3.7. Suppose that f H (C ) and f (n) = 0 for all n Z . Show that all the singularities of f (z)/ sin(z )are removable.

Proof. First we show with the help of Appendix 3 that sin( z) = 0 if and only if z Z . So note thatif R e(z) = 0 then eiz = e iz , so that sin( z) = 0. So all the zeros of sin are real. It has been shownin Appendix 3 that there are exactly two zeroes in the interval [0 , 2), that sin(0) = sin( ) = 0, andthat sin has period 2 . Thus sin( z) = 0 if and only if z Z .

Now we see that all singularities of f (z)/ sin(z ) occur when z Z . But notice for n Z , sin (n ) = cos(n ) = . Thus sin( z ) has a zero of order one at every n Z . By exercise 3.6 f (z)/ sin(z ) hasa removable singularity at every n Z .

3.8. Suppose that f H (C ), f (z +1) = f (z) for all z, f (0) = 0, and |f (z)| e | I m z | for all z. Showthat f (z) = c sin(z ) for some constant c.

Proof. Note that f (0) = 0 and suppose f (m) = 0 for 0 m < n . Then f (n) = f ((n 1) + 1) = f (n) = 0. So f (n) = 0 for all n Z + . Similarly suppose f (m) = 0 for n < m 0. Thenf (n) = f ((n + 1) 1) = f (n + 1) = 0 so f (n) = 0 for all n Z . Thus by exercise 3.7 all thesingularities of f (z)/ sin(z ) are removable. Let g be the entire function that agrees with f (z)/ sin(z )


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18 DAVID C. ULLRICH

conclude that I = 0.

Proof. Since f has period 2 we can calculate that

12

2 +

f (t) dt =

12

2

f (t) dt +

2 +

2f (t) dt

= 12

2

f (t) dt +

0f (t) dt

=1

2 2

0f (t) dt

= I.

Since f (s + t) = f (s)f (t) we calculate that

12

2 +

f (t) dt =

12

2

0f (t + ) dt

=1

2 2

0f (t)f () dt

= 12

2

0f (t) dt

= I.

3.18. Suppose that N is a positive integer, < N + 1, and f H (D (z0 , r )) satises

|f (z)| c |z z0 | .

Show that f has either a removable singularity or a pole of order no larger than N at z0 .

Proof 1. Since |f (z)| c |z z0 | , we have

(z z0)N +1 f (z) c |z z0 |N +1 ,

for all z D (z0 , r ). But N +1 > 0 implies c |z z0 |N +1

0 as z z0 . Thus ( z z0)N +1 f (z) 0 as z z0 and hence is bounded near z0 . So there exists ( cn ) such that

(z z0)N +1 f (z) =

n =0cn (z z0)n ,

for all z D (z0 , r ). Since c0 = lim z z 0 (z z0)N +1 f (z) = 0,n =0 cn (z z0)n =

n =1 cn (z z0)n .

Thus for all z D (z0 , r ) we have

f (z) = ( z z0) N 1

n =1cn (z z0)n

=

n =1cn (z z0)n N 1

=

n = N cn + N +1 (z z0)n .

So f has a removable singularity or a pole of order no larger than N at z0 .

Proof 2. Proposition 3.15 states that

f (z) =

n = cn (z z0)n


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19

for all z D (z0 , r ), where

cn =1

2i D (z0 , ) f (z)(z z0)n +1 dzfor any (0, r ). This shows that

|cn | 1

2c

n +1 2 = c n .

Now suppose that n < N is an integer. Then n (N + 1) since n is an integer, so n + ( N + 1) > 0. So we have

|cn | lim0

c n = 0

for all integers n < N , and so

f (z) =

n = N

cn (z z0)n .

3.19 (Not from book) . Suppose that f is entire and |f (z)| | z|3/ 2 for all z. Show that f = 0.

Proof. Since |f (z)| | z |3/ 2 , letting g(z) = f (z)/z 2 for all z C \ { 0} we have

|g(z)| =f (z)z2

| z | 1/ 2 .

Thus |g(z)| 1 for all z C \ D . But|zg(z)| | z |1/ 2 ,

so |z|1/ 2 0 as z 0 shows zg(z) 0 as z 0. Since zg(z) is bounded near 0, we let

zg(z) =

n =0cn zn ,

for all z D \ { 0}. Since 0g(0) = 0, we have n =0 cn zn =n =1 cn z

n . Thus

g(z) = z 1

n =1cn zn =

n =1cn zn 1 =

n =0cn +1 zn ,

for all z D \ { 0}. But now there exists M 0 such that | n =0 cn +1 zn | M for all z D , so| n =0 cn +1 zn | max {1, M } for all z C . Thus corollary 3.6 states there exists c C such that

n =0 cn +1 z

n = c for all z C . So f (z)/z 2 = c for all z C \ { 0} and |f (z)| = cz2 | z|3/ 2 for allz C \ { 0}. So |c| | z | 1/ 2 for all z C \ { 0} shows c = 0. Thus f (z) = 0 for all z C \ { 0} and|f (0) | | 0|3/ 2 = 0 shows f = 0.

4. Logarithms, Winding Numbers and Cauchys Theorem

Theorem 4.0. Suppose that V C is open and f H (V ). We have

f (z) dz = 0 for every smooth closed curve in V if and only if f = F for some F H (V ).

Proposition 4.1. Suppose that V is open and f C (V ) satises ef (z ) = z for all z V . Then f H (V ) (so that f is a branch of the logarithm in V ).

Proposition 4.2. Suppose that V is a connected open subset of the plane, z0 V and L H (V ).The following are equivalent:

(i) L is a branch of the logarithm in V .(ii) eL (z0 ) = z0 and L (z) = 1 /z for all z V .


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20 DAVID C. ULLRICH

Proposition 4.3. Suppose that V is an open subset of the plane. There exists a branch of the logarithm in V if and only if there exists f H (V ) with f (z) = 1 /z for all z V .

Corollary 4.4. If V is a convex open subset of C \ { 0} then there exists a branch of the logarithm in V .

Proposition 4.5. Suppose that V is an open subset of C . There exists a branch of the logarithm in V if and only if

dzz = 0 for all smooth closed curves in V .

Lemma 4.6. Suppose that : [0, 1] C is continuous and 0 C \ .(i) If 0 R and ei 0 = (0) / | (0) | then there exists a unique continuous function : [0, 1] R

such that (0) = 0 and (t) = | (t)| ei ( t )

for all t [0, 1].(ii) For every > 0 there exists > 0 such that if (t) = | (t)| ei ( t ) and also (t) (t) < for

all t [0, 1].(iii) If is a closed curve then ((1) (0)/ 2 is an integer; if is a smooth closed curve then

(1) (0)2

=1

2 dzz .Proposition 4.7. Suppose that is a cycle in C . The function I (a) = Ind( , a) is an integer-valued function on C \ which is constant on the components of C \ and vanishes on the unbounded component.

Proposition 4.5 (Part 2) . Suppose that V is an open subset of C . There exists a branch of the logarithm in V if and only if 0 / V and

Ind( , 0) = 0

for all closed curves in V .

Lemma 4.8. Suppose that V is an open subset of the plane and f H (V ). Dene g : V V C by

g(z, w) =f (z ) f (w)

z w (z = w),f (z) (z = w).

Then g is continuous in V V , and if we dene gw (z) = g(z, w) then gw H (V ) for every w V .

Theorem 4.9 (Cauchys Integral Formula; Homology Version) . Suppose that V is an open subset of the plane and is a cycle in V . If has the property that

Ind( , a) = 0

for all a C \ V then 1

2i f (w) dww z = Ind( , z) f (z) for all f H (V ) and all z V \ .Theorem 4.10 (Cauchys Theorem; Homology Version) . Suppose that V is an open subset of the plane and is a cycle in V . If has the property that

Ind( , a) = 0

for all a C \ V then

f (z) dz = 0 for all f H (V ).


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Theorem 4.10 (Cauchys Theorem, Part 2) . Suppose that V is an open subset of the plane and is a cycle in V . If

f (z) dz = 0 for all f H (V ) of the form f (z) = 1 / (z a) then


4.1. Show directly from the denition that Ind ( D (a, r ), a) = 1 for any a C and r > 0.

Proof. Let D (a, r ) = where : [0, 1] C is dened by

(t) = a + re 2it .

So Ind( D (a, r ), a) = Ind( , a ). But Ind( , a ) is given by

Ind( , a ) =1

2((1) (0))

where : [0, 1] R is a continuous function such that

(t) a = | (t) a | ei ( t ) .

If we dene (t) = 2 t , we note that (t) a = re 2it = | (t) a | ei ( t ) . Calculating we nd that

Ind( , a ) =1

2((1) (0)

=1

2(2)

= 1 .

4.2. Suppose a C and r > 0. Show that

Ind( D (a, r ), z) = 1, |z a | < r,0, |z a | > r.

(What is Ind ( D (a, r ), z) when |z a | = r ?)

Proof. First note that D (a, r ) separates C into two components, D(a, r ) and C \ D (a, r ). Thus propo-sition 4.7 states that there is n Z such that Ind( D (a, r ), z) = n for all z D (a, r ). But exercise 4.1shows Ind ( D (a, r ), a) = 1 so Ind( D (a, r ), z) = 1 for all z D (a, r ). Since C \ D (a, r ) is unbounded,proposition 4.7 also states that Ind ( D (a, r ), z) = 0 for all z C \ D (a, r ).

4.3. Show that theorem 4.9 follows from theorem 4.10, by an argument similar to the proof of theo-rem 2.5 above.

Proof (Theorem 4.9 ). Let f H (V ), V withInd( , a) = 0

for all a C \ V . Since V is open, let D (z, r ) V for z V \ and let : I C be given by (t) = z + re Ind( ,z )2 it . Note that Ind , z = Ind ( , z) Ind( , z ) = 0. Since f (w)w z H (V \ { z}),( ) V \ { z}, and Ind , a = 0 for all z C \ (V \ { z}), theorem 4.10 states

f (w)w z dw f (w)w z dw = f (w)w z dw = 0 .


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22 DAVID C. ULLRICH

But 12 f (w)w z dw = Ind ( , z ) f (z) = Ind( , z) f (z). Therefore

12 f (w)w z dw = Ind( , z) f (z),

for all z V \ .

Theorem 4.11. Suppose that a C and 0 r < R . If f H (A(a,r,R )) then there exists a sequence of complex numbers (cn )n = such that

f (z) =

n = cn (z a)n (z A(a,r,R )) .

The series converges uniformly on compact subsets of A(a,r,R ) and the coefficients are given by

cn =1

2i D (a, ) f (w) dw(w a)n 1 for any (r, R ).

Proposition 4.12. Suppose that 0 and 1 are closed curves in the open set V and 0 V 1 . Then

Ind( 0, a) = Ind(

1, a)

for all a C \ V .

Theorem 4.13 (Cauchys Theorem; Homotopy Version) . Suppose V is an open set in the plane, 1and 2 are smooth closed curves in V , and 1 V 2 . Then

12i 1 f (z) dz = 12i 2 f (z) dz

for all f H (V ).

4.4. Suppose that V is simply connected, is a smooth closed curve in V and a C \ V . Show thatInd( , a ) = 0 .

Proof. If z V , then c(t) = z puts any a C \ V in the unbounded component of . Proposition 4.7

shows that Ind( c , a) = 0 and proposition 4.12 shows that Ind( , a ) = Ind( c , a) = 0.4.5. Show that any convex open set is simply connected.

Proof. Take a loop : [0, 1] V where V is our convex open set and let z V . Since V is convex wecan take the line segment t : [0, 1] V dened by

t (t ) = t (1 t ) + t ,where t = (t). Thus we construct a path homotopy : [0 , 1]2 V given by

(t , t ) = t (t )

from 0 = to 1(t ) = z for all t [0, 1]. Therefore V is simply connected.

Theorem 4.14 (Cauchys Theorem for Simply Connected Sets) . If V is simply connected and is a cycle in V then

f (z) dz = 0 for all f H (V ).Corollary 4.15. Any nonvanishing holomorphic function in a simply connected set has a holomorphic logarithm. That is, if V is simply connected, f H (V ) and f has no zero in V then there exists L H (V ) with

eL = f in V .


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23

4.6. Show that any nonvanishing holomorphic function in a simply connected set has a holomorphicsquare root.

Proof. Let V be a simply connected set and let f H (V ) be our nonvanishing function. Corollary 4.15states that there is L H (V ) with eL = f in V . Dene g = exp( 12 L) so that

g2 = exp( 12

L) exp( 12

L) = exp( L) = f.

4.7.(i) Suppose that f and g are holomorphic near z0 and f has a simple zero at z0 . Find an expression

for the residue of g/f at z0 (and prove that it is correct).(ii) Suppose that f has a simple pole at z0 and g is holomorphic near z0 . Show that

Res( fg ,z 0) = g(z0)Res( f, z 0) .

(iii) Suppose that f is holomorphic in a neighborhood of z0 , set g(z) = f (z)/ (z z0)n , and showthat

Res( g, z0) = f (n 1) (z0)/ (n 1)!.For simplicity we note that if f has a removable singularity at z0 then

Res( f, z 0) = 0 = limz z0

(z z0)f (z).

Thus Res( f, z 0) = lim z z0 (z z0)f (z) if f has a removable singularity or a simple pole.

Proof (i). Since g/f has a removable singularity or simple pole at z0 , we calculate

Res( g/f,z 0) = limz z0(z z0)

g(z)f (z)

= limz z0

(z z0)f (z) f (z0)

g(z)

=g(z0)f (z

0)

.

Proof (ii). Since f has a simple pole at z0 , then fg has either a removable singularity or a simple poleat z0 . Thus we calculate

Res( fg ,z 0) = limz z0(z z0)f (z)g(z)

= Res ( f, z 0) g(z0).

Proof (iii). We have

f (z) =

k =0

ck (z z0)k

near z0 . But then

g(z) =

k=0

ck (z z0)k n

=

k= n

ck+ n (z z0)k .


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24 DAVID C. ULLRICH

So Res(g, z0) = cn 1 by denition of residue. But

cn 1 =f (n 1) (z0)

(n 1)!.

4.8. Suppose that f has an isolated singularity at z0 . Fix r > 0 and n Z , and dene : [0, 2] Cby

(t) = z0 + re int .

Show that Ind( , z0) = n and that

12i f (z) dz = Ind ( , z0)Res( f, z 0)

if r > 0 is small enough (be explicit about how small r must be.)

Proof.

Lemma 4.16. Suppose that V C is open, S V is (relatively) closed in V , and every point of S is isolated (that is, for every z S there exists r > 0 such that S D (z, r ) = {z}). Suppose that is a cycle in V \ S such that Ind( , a) = 0 for all a C \ V . Then

Ind ( , z) = 0

for all but nitely many z S .

Theorem 4.17 (Residue Theorem) . Suppose that V C is open, S V is (relatively) closed in V ,and every point of S is isolated. Suppose that f H (V \ S ). (In other words, f is holomorphic in V except for isolated singularities at the points of S .) If is a cycle in V such that V \ S and Ind ( , a) = 0 for all a C \ V then

12i f (z) dz = z S Ind( , z)Res( f, z ) .

(Note that all but nitely many terms in the sum vanish, by Lemma 4.16. )

4.9. Show that the Cauchy Integral Formula (theorem 4.9) is an immediate consequence of the Residuetheorem.

Proof. Suppose that V is an open subset of the plane and is a cycle in V . Let have the propertythat Ind ( , a) = 0 for all a C \ V . If f H (V ) and z V \ , then let g(w) = f (w)/ (w z) sog H (V \ { z}). Thus the residue theorem states that

1

2i

g(w) dw = Ind ( , z)Res( g, z) = Ind( , z) f (z).

4.10. Show that

dx

1+ x 2 = , using the Residue thoerem ( 4.17).

Proof. Let R > 1, = 1 + 2 where 1 = [ R, R ] and 2 : [0, ] C is given by 2(t) = Re it .Let f (z) = 11+ z 2 and notice that f has isolated singularities at i. Observe theorem 4.17 and letV = {z C | I m(z) > 1/ 2} and notice Ind ( , a) = 0 for all a C \ V . Since V is open and {i} is


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25

relatively closed in V with f |V H (V \ { i}), then

f (z) dz = f |V (z) dz= 2 i

z { i}

Ind( , z)Res( f |V , z)

= 2 i Ind( , z)Res( f |V , i)= 2 i Res( f |V , i) .

Since 1 + z2 = ( z i)(z + i), 1 + z2 has a simple zero at i. Thus Res( f |V , i) = 12i by exercise 4.7 (i).Thus

f (z) dz = 2 i 12i = .But

f (z) dz = 1 f (z) dz + 2 f (z) dz=

[ R,R ]

11 + z2

dz +

0

Rie it

1 + ( Re it )2dt

= R

R

11 + t2

dt + i

0

1e itR + Re it

dt

11 + t2

dt (R ).

4.11. Suppose that P and Q are polynomials with deg( Q) deg(P ) + 2 and let f (z) = P (z)/Q (z).Suppose that Q has no zero on the real axis and show that

f (x) dx is equal to 2i times the sum

of the residues of f in the upper half-plane.

Proof. Let 1 = [ R, R ] and 2 : [0, 1] C be given by

2(t) = Reit

,and let = 1 + 2 . The residue theorem states that if S = {z C | I m(z) > 0, Q(z) = 0 }, then

f (z) dz = 2 i z S Ind( , z)Res( f, z ) = 2 i z S Res( f, z ). But 2 f (z) dz = 2 P (z)Q(z) dz =

1

0

P (Re it )Rie it

Q(Re it )dt,

which goes to 0 as R because deg( Q(z)) deg(zP (z)) + 1. But

f (x) dx = lim

R 1 f (z) dz= lim

R f (z) dz= 2 i

z S

Res( f, z ) .

The next exercise gives a version of Parsevals Formula for Laurent series:

4.12. Suppose that f H (A(z0 , r,R )) has Laurent series f (z) = n = cn (z z0)n . Show that

12

2

0f (z0 + eit )

2dt =

n = 2n |cn |2


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26 DAVID C. ULLRICH

for r < < R .

Proof. Let sN =N n = N cn (z z0)

n , so

1

2 2

0

sN (z0 + eit )2

dt =1

2 2

0

N

n = N

cn n eintN

m = N

cm m eimt dt

=N

m,n = N cn cm m + n

12

2

0ei (n m )t dt

=N

n = N

|cn |2 2n .

Now if < r then sN f uniformly on D (a, ), and hence

12

2

0f (a + eit )

2dt = lim

N

12

2

0sN (a + eit )

2dt

= limN

N

n = N |cn |2 2n

=

n = |cn |2 2n .

4.13. Prove the homology version of the Cauchy Integral Formula for derivatives: Under the samehypotheses as theorem 4.9 we have

n!2i f (w) dw(w z)n +1 = Ind( , z) f (n ) (z)

for n N .

Proof. Let g(w) = f (w)/ (w z)n +1 . Exercise 4.7 shows1

2i g(w) dw = w S Ind( , w)Res( g, w)= Ind ( , z)Res( g, z)

= Ind ( , z)f (n ) (z)

n!.

4.14. Show that exercise 4.4 follows immediately from proposition 4.12.

Proof. Let V be simply connected and a smooth loop in V with a C \ V . Then Ind ( , a ) =Ind( a , a) = 0, where a (t) = a for all t [0, 1].

The next few exercises give an example of using the residue theorem (actually just Cauchys theoremhere) to evaluate an improper integral that is a little more elaborate than exercise 4.10 above.

4.15. Suppose that f has a simple pole at z0 and a < b. For r > 0 dene r : [a, b] C by r (t) = z0 + re it , so that r is an arc on the circle with center z0 and radius r . Show that

limr 0+ r f (z) dz = ( b a)iRes( f, z 0) .

Proof.


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27

4.16. Suppose that f n : [a, b] C is continuous for n N , |f n (t)| M for all n and all t , and supposethat f n 0 uniformly on [a + , b ] for every (0, (b a)/ 2). Show that

ba f n (t) dt 0 as

n .

Proof. Let > 0 be given. Then if < (b a)/ 2, we have

a +

af n (t) dt

a +

a|f n (t)| dt

a +

a=

M . Similarly, b

b f n (t) dt M . Let = min( / 4M, (b a)/ 2). Now by uniform convergence,there is N N such that |f n (t)| < / (2(b a 2 )) for n > N , t [a + , b ]. Therefore if n > N ,

b

af n (t) dt < M + 2(b a 2 )

(b a 2 ) + M = .

4.17. Find

lim0+ ,R

R

+

R

eit

tdt.

Let f (z) = eiz /z . Find f (z) dz and apply the previous two exercises to gure out what happensto two pieces of the integral as 0 and R .Proof. For 0 < < R < let = 1 + 2 3 + 4 , where 1 = [ R, ], 2 : [0, ] C is dened by 2(t) = ei ( t ) , 3 = [, R], and 4 : [0, ] C is dened by (t) = Re it . Exercise 4.7 and exercise 4.15show that

lim0+ 2 e

iz

zdz = i Res e

iz

z, 0

= ie i 0

= i.

limR 4 e

iz

zdz = lim

R

0

eiReit

iRe it

Re itdt

= limR

i

0eiRe

it

dt

=

Therefore the answer is i .

4.18. Consider the integral

limR

R

R

sin( t)t

dt.

(i) Explain why you cannot evaluate this integral using the method we used in the previousexercise.

(ii) Use the result of the previous exercise to evaluate the integral.

(i). The method in the last exercise will not work here because lim R sin(Re it ) is not bounded.


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28 DAVID C. ULLRICH

(ii).

i = R

R

eit

tdt

=

R

R

cos(t) + i sin( t)

tdt

= R

R

cos(t)t

+ i R

R

sin( t)t

dt.

Thus R

Rsin( t )

t dt = .

The next two exercises allow one to give a rigorous proof that various winding numbers are whatthey are without doing any work at all, in most of the cases where the value of the winding number issupposed to be obvious:

4.19. Suppose that 1 : [a, b] C \ { z0} is a smooth closed curve and let 2(t) = 1(t) z0 . Show thatInd( 1 , z0) = Ind( 2 , 0).

Proof.

Ind( 1 , z0) =1

2i 1 dzz z0=

12i

b

a

1(t) 1(t) z0

dt

=1

2i b

a

2(t) 2(t)

dt

=1

2i 2 dzz = Ind ( 2 , 0) .4.20. Suppose that : [a, b] C \ { z0} is a smooth closed curve. Suppose that a < c < b . Showthat if R e( (t)) R e(z0) for all t [a, c], R e( (t)) R e(z0) for all t [c, b], I m( (a)) < I m(z0) andI m( (c)) > I m(z0) then Ind ( , z0) = 1. (The previous exercise shows that you can assume that z0 = 0.Now use the existence of a branch of the logarithm in the right half-plane and a different branch of the logarithm in the left half-plane.)

Proof. Let = 8i =1 i where i : [0, 1] C are given by 1 = |[a,c ], 2(t) = c(1 t) + it , 3(t) =ei (/ 2 t ) , 4(t) = i(1 t)+ (a)t , 5(t) = (a)(1 t) it , 6(t) = ei (3 / 2 t ) , 7(t) = i(1 t)+ (c)t, 8(t) = | [c,b ]. If 1 =

4i =1 i and 2 =

8i =5 i , then 0 = Ind( 1 , 0) + Ind ( 2 , 0) because they

are contained in simply connected domains. But this is Ind( , 0) = Ind( , 0) + Ind 3 + 6 , 0 =Ind( , 0) 1. Thus Ind( , 0) = 1.

4.21. Suppose that P (z) is a polynomial of degree n 2 with n distinct zeroes z1 , . . . , zn . Explainwhy it follows that every zero of P is simple, and show thatn

j =1

1P (zj )

= 0 .

(Exercise 4.7 (i) shows that 1 /P (zj )) is the residue at zj of ...)

Proof. Since P (z) is a polynomial of degree n with n distinct zeroes, the fundamental theorem of algebra states that we can write P (z) = ni =1 (z zi ). This makes it clear that every zero of P


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29

is simple. Thus Res(1 /P (z), zj ) = 1 /P (zj ). Let : [0, 1] C be given by (t) = Re 2it so thatzj D (0, R) for all 1 j n. Then

n

j =1

1P (zj )

=n

j =1

Res1

P (z), zj

= 12i 1P (z) dz

=1

2i 1

0

2iRe 2it

nj =1 (Re 2it zi )dt

1

0

Re 2it

nj =1 (Re 2it zj )dt R

| | 1

0

dtnj =1 |Re 2it zj |

=1

Rn 1 | | 1

0

dtnj =1 1

z jRe 2 it

1

Rn 1 | | 0 (R ).

So nj =1 1/P (zj ) = 0.

In the next exercise we give a sketch of an alternate proof of lemma 4.6 (i). You should study bothproofs and convince yourself that they are really the same proof, or at least based on the same idea,but expressed very differently. The notation in the proof in the exercise is simpler than that in theproof in the text; on the other hand the proof in the exercise is somewhat more abstract, so it is notclear to me which proof you will nd easier to understand. (I suspect this may vary from student tostudent!)

So you should learn both proofs, and then the next time you see an argument that reminds you of the version of the proof you nd harder to understand, you should consider rephrasing it in the styleof the proof you found more transparent:

4.22. Prove part (i) of lemma 4.6 as follows: Let A be the set of all x [0, 1] with the property thatthere exists a continuous function : [0, x] R such that (0) = 0 and

(t) = | (t)| ei ( t ) (t [0, x]).

Show that A is nonempty, closed, and relatively open in [0 , 1], and conclude that A = [0, 1] since [0, 1]is connected.

Proof. Let Arg (0) = 0 so A is nonempty. Now let lim n xn = x for xn a nondecreasing sequencein [0, 1] such that there exists continuous n : [0, 1] R with n (0) = 0 and (t)/ | (t)| = ei n ( t ) fort [0, xn ]. But limn ei n (x n ) = lim n (xn )/ | (xn )| = (x)/ | (x)|, so A is closed.

To show that A is relatively open in [0, 1], let a A and we will show some neighborhood of a iscontained in A. If a = 1 then we are done so suppose a = 1. Since a A there exists a continuousfunction : [0, a] R such that (t) = | (t)| ei ( t ) for t [0, a]. Note since 0 / , there exists > 0such that 0 / D ( (a), ). Let > 0 be a number such that ([a, a + )) D ( (a), ). Thus there existsL H (D ( (a), )) such that eiL (z ) = z, specically eiL ( ( t )) = (t) for all t [a, a + ). So eiL ( (a )) = (a) = | (a)| ei (a ) , thus (a) = R e(L( (a))) + 2 k for some k Z . So dene f : [0, a + ) R by

f (t) = (t), t [0, a],L( (t))) + 2 k, t [a, a + ).

By the pasting lemma, f is continuous. Thus ( a , a + ) [0, 1] A and therefore A is relativelyopen in [0, 1].


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30 DAVID C. ULLRICH

4.23. Derive lemmea 3.13 from theorem 4.9 as follows: Suppose that f H (A(z0 , 0, R)) is bounded.Fix r (0, R), and show that

f (z) =1

2i D (z0 ,r ) f (w) dww z 12i D (z0 , ) f (w) dww zwhenever 0 < < |z z0 | < r . Now x z and consider what happens when tends to 0.Proof. First note if z C \ D(z0 , R) then Ind ( D (z0 , r ), z) = Ind( D (z0 , ), z) = 0 and

Ind D (z0 , r ) D (z0 , ), z0 = 0. So by theorem 4.9 we have

f (z) =1

2i D (z0 ,r ) f (w) dww z 12i D (z0 , ) f (w) dww z .Now let z A(z0 , ,R ) be given so that

lim0+

12i D (z0 , ) f (w) dww z = lim0+ 12i

1

0

f (z0 + e2it ) 2ie 2it

z0 + e2it zdt

1

0

f (z0 + e2it )e2it

(z z0) e2itdt

1

0

f (z0 + e2it )e2it

(z z0) e2itdt

1

0

f (z0 + e2it ) |z z0 |

dt

M/ |z z0 | .

So lim0+ D (z0 , )f (w) dw

w z = 0. Thus f (z) =1

2i D (z0 ,r )f (w) dw

w z for all z A(z0 , 0, R). Deneg : D (z0 , R) C by

g(z) =f (z), z = z0 ,

12i D (z0 ,r )

f (w) dww z0 , z = z0 .

But calculating we nd

limh 0

g(z0 + h) g(z0)

h= lim

h 0

1

h

1

2i D (z0 ,r )

f (w)

w z0 + h f (w)

w z0dw

= limh 0

12i D (z0 ,r ) f (w)(w z0)2 + h(w z0) dw

= 12i D (z0 ,r ) f (w)(w z)2 dw

Thus f has a removable singularity at z0 .

5. Counting Zeroes and the Open Mapping Theorem

Theorem 5.0. Suppose that f H (V ), and f is not constant on any component of V . Suppose that is a smooth curve in V such that Ind( , z ) is either 0 or 1 for all z C \ , and equals 0 for all z C \ V . Suppose that f has no zero on , and let = {z V | Ind( , z ) = 1 }. Then the number of zeroes of f in is given by

12i f (z)f (z) dz.

Theorem 5.1 (The Argument Principle) . Suppose that f , , etc. are as in Theorem 5.0. Dene a curve (t) by (t) = f ( (t)) . Then the number of zeroes of f in is equal to

Ind( , 0) .

Lemma 5.2. If z, w C and |z w| < |z | + |w| then 0 does not lie on the line segment joining z and w.


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31

Proposition 5.3. Suppose that 0 and 1 are two closed curves in C \ { 0} with the same parameter interval. If

| 1(t) 0(t)| < | 1(t)| + | 0(t)| for all t then 0 and 1 are homotopic in C \ { 0}, and in particular we have

Ind( 0 , 0) = Ind ( 1 , 0) .Theorem 5.4 (Rouches Theorem) . Suppose that is a smooth closed curve in the open set V such that Ind( , z ) is either 0 or 1 for all z C \ and equals 0 for all z C \ V , and let ={z V | Ind( , z ) = 1 }. If f, g H (V ) and

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

for all z then f and g have the same number of zeroes in .

5.1. Suppose that n is a positive integer and show that the polynomial zn (z 2) 1 has exactly nroots in the open unit disk D (0, 1).

Proof. Let g(z) = zn (z 2) so that if z D (0, 1)\ { 1}, then

|f (z) g(z)| = 1

< |zn (z 2) 1| + 1< |zn (z 2) 1| + |zn (z 2)|= |f (z)| + |g(z)| .

If z = 1, then |f (1) g(1) | = 1 < 3 = |f (1) | + |g(1) | . So |f (z) g(z)| < |f (z)| + |g(z)| for allz D (0, 1), so Rouches Theorem states that f and g have the same nubmer of zeroes in D (0, 1).Since g clearly has n zeroes in D(0, 1), then the polynomial zn (z 2) 1 also has n roots in D (0, 1).

Theorem 5.5 (The Open Mapping Theorem) . If V is a connected open subset of the plane and f H (V ) is nonconstant then f is an open mapping.

Theorem 5.6 (The Open Mapping Theorem, Explicit Version) . Suppose that f is holomorphic in a neighborhood of D (z0 , r ) and

f (z0 + reit

) f (z0) > 0 for all t R . Then

D (f (z0), ) f (D (z0 , r )) .

Theorem 5.7 (The Open Mapping Theorem, with Bounds) . Suppose that f H (V ) and |f (z)| B for all z V . Suppose that z0 V and f (z0) = 0 . If r > 0 is small enough that D (z0 , r ) V and

also r 0 such that D (f (0), r ) f (D ). But if |u(0) | + |v(0) | = 1, then |u(0) r/ 2| + |v(0) | = 1 which contradicts this belonging to the image of D ,hence f is constant.


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32 DAVID C. ULLRICH

Proof (ii). Since proof to part (i) did not depend on which z D we chose, if there exists z D suchthat |u | + |v| = 1 then f is constant. Thus if f is nonconstant, then |u | + |v| < 1 everywhere.

5.3. Determine the number of zeroes of f (z) = 1 + 6 z3 + 3 z10 + z11 in the annulus A(0, 1, 2).

Proof. First we determine the number of zeroes in D (0, 2). Let g(z) = z10 (z + 3) so that

|f (z) g(z)| = 6z3 + 1= 49

< 1 + 6z3 + 3 z10 + z11 + 2 10

= 1 + 6z3 + 3 z10 + z11 + z10 (z + 3)= |f (z)| + |g(z)| .

So f has ten zeroes in D (0, 2).Now we determine the number of zeroes in D (0, 1). Let g(z) = 6 z3 + 3 z10 + z11 so that

|f (z) g(z)| = 1 < 2 1 + 6z3 + 3 z10 + z11 + 6z3 + 3 z10 + z11

= |f (z)| + |g(z)| ,

where the second inequality comes from the second absolute value combined with the triangle inequal-ity.

The number of zeroes in D (0, 1) is shown to be zero by

|f (z)| = 1 + 6z3 + 3 z10 + z11

6 3 1 1 = 1 .

Therefore the total number of zeroes in A(0, 1, 2) is given by 10 3 = 7.

5.4. Give another proof of Rouches Theorem (Theorem 5.4) as follows: Let W = C \ ( , 0] be theplane with the non-negative real axis removed. Show that there exists a branch of the logarithmL H (W ). (You should write down an explicit formula for L, and also note that the existence of abranch of the logarithm in W follows from one of the general results in Chapter 4.) Now show that thereexists an open set O with O such that f (z)/g (z) W for all z O; hence F = L (f /g ) H (O).Theorem 4.0 shows that F (z) dz = 0; explain how the result follows. Give an analogous proof forProposition 5.3.Proof Theorem 5.4. Suppose that is a smooth loop in W such that Ind ( , z ) is either 0 or 1 for allz C \ and equals 0 for all z C \ W , and let = {z W | Ind( , z ) = 1 }. Let f, g H (W ) and|f (z) g(z)| < |f (z)| + |g(z)| for all z . Since W is open, 0 / W and W is simply-connected,then there exists L H (W ). If z = re it for t ( , ), then we let L(z) = log( r ) + it . Since f, g arenot zero on , each point of has a neighborhood such that f, g are non-zero values. Let O be theunion of all of these neighborhoods, so O with f (z)/g (z) W for all z O. Theorem 5.0 showsthat f /g has no zeroes on . Therefore f and g have the same number of zeroes in .

Proof Theorem 5.3. Suppose that 0 and 1 are two closed curves in C \ { 0} with the same parameterinterval. Let

| 1(t) 0(t)| < | 1(t)| + | 0(t)|for all t. Then 0 / 1 is a closed curve in W , a simply-connected set, so 0 / 1 is homotopic to a point.Multiplying the homotopy by 1 gives rise to a homotopy between 0 and 1 .

5.5. Suppose that f : D D is continuous and f is holomorphic in D . Show that f has a xed pointin D .


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33

Proof. Suppose that f has no xed points in D with |f (z)| = r < 1 so f (z) z = 0 and z = 0 on D . So |f (z) z + z| = |f (z)| < 1 | f (z) z | + | z|. Thus f (z) z has one zero in D . Now letf r (z) = rf (z) so f r has a xed point for every r [0, 1). But then f (zr ) = zr /r for zr the xed pointof f r . So as r 1, z1 D by closure so f (z1) = z1 is a xed point of f in D .

5.6. Prove Hurwitzs Theorem: Suppose that D is an open set, f n H (D ), f n f uniformly oncompact subsets of D , D (z, r ) D , and f has no zero on D (z, r ). Then there exists N such thatf n and f have the same number of zeroes in D(z, r ) for all n > N . In particular, if D is connected,f n H (D ), f n f uniformly on compact sets, and f n has no zero then either f has no zero or f vanishes identically.

Proof.

5.7. Suppose that D is a connected open set, f n H (D ), and f n f uniformly on compact subsets of D . If f is nonconstant and z D then there exists N and a sequence zn z such that f n (zn ) = f (z)for all n > N .

Proof. We can assume that f (z) = 0. Since f is nonconstant then this zero is isolated so takeD (z, ) such that f (w) = 0 for all w D (z, )\ { z}. Let > 0 be given and let D (z, ) D with = min {, , }. By exercise 5.6 there exists N such that both f n and f have one zero in D (z, ) forall n > N . So let zn D (z, ) be the point such that f n (zn ) = 0. But |zn z| < for all n > N and since was arbitrary, zn z as n .

6. Eulers Formula for sin( z)

6.0. Motivation.

6.1. Proof by the Residue Theorem.

Theorem 6.1.0 (Cauchy Integral Formula for Functions with Simple Poles) . Suppose that V C is open and is a cycle in v such that Ind ( , a) = 0 for all z C \ V . Suppose that S is a (relatively)closed subset of V , S = , and every point of S is isolated. Suppose that f H (V \ S ) and f has a simple pole or removable sinularity at every point of S . Then for z V \ (S ) we have

Ind( , z) f (z) =1

2i f (w)w z dw + p S

Ind( , p)Res( f, p )z p .

Lemma 6.1.1. There exists a constant M such that

|cot( z )| M whenever |I m(z)| 1 or R e(z) = n + 1 / 2 ( n Z ).

Theorem 6.1.2. If z C \ Z then

cot( z ) =1z

+

n =1

1z n

+1

z + n.

6.1. Show that the series

n =0

1

z ndiverges for all z C \ Z .

Proof.

6.2. Show that the series

n =1

1z n

+1

z + n

converges absolutely for all z C \ Z .


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34 DAVID C. ULLRICH

Proof.

Lemma 6.1.3. If z1 , . . . , zn C and n 1j =1 |1 zj | < 1/ 2 then

1 n

j =1

zj 2n

j =1

|1 zj | .

Lemma 6.1.4.(i) If z1 , . . . C and j =1 |1 zj | < then

j =1

zj = limn

n

j =1

zj

exists; furthermore, nj =1 zj = 0 unless zj = 0 for some j .(ii) If (f j ) is a sequence of complex-valued funtions on some set S and the sum j =1 |1 f j |

converges uniformly on S then P n =nj =1 f j tends to P =

j =1 f j uniformly on S ; if z S

then P (z) = 0 unless f j (z) = 0 for some j .

Proposition 6.1.5. Suppose that V is a connected open set, f 1 , f 2 , . . . H (V ) and f n f uniformly

on compact subsets of V . Suppose that f is not identically zero. Then L(f n ) L(f )

uniformly on K , if K is any compact subset of V on which f has no zero.

Lemma 6.1.6. Suppose that V is a connected open subset of C , f and g are holomorphic functions in V neither of which vanishes identically, and

L(f ) = L(g)

on the set where neither f nor g vanishes. Then f = cg for some constant c.

6.2. Estimating Sums Using Integrals.

6.3. Suppose that is continuous on [ 1/ 2, 1/ 2].(i) Show that

1/ 2

1/ 2(t) dt (0) = 1

2 1/ 2

1/ 2|t | 1

2

2

(t) dt.

(ii) Deduce that

1/ 2

1/ 2(t) dt (0)

18

1/ 2

1/ 2| (t)| dt.

Proof (i).

12

1/ 2

1/ 2|t |

12

2

(t) dt =12

1/ 2

0t

12

2

( (t) + ( t)) dt

=12

t 12

2

( (t) ( t))1/ 2

0

1/ 2

0( (t) ( t))(2 t 1) dt

= 12

(2t 1)((t) + ( t)) |1/ 20 1/ 2

02((t) + ( t)) dt

= 1/ 2

0((t) + ( t)) dt (0)

= 1/ 2

1/ 2(t) dt (0) .


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35

Proof (ii).

1/ 2

1/ 2

(t) dt (0) =1

2 1/ 2

1/ 2

|t | 1

2

2

(t) dt

12

1/ 2

1/ 2|t | 1

2

2

(t) dt

18

1/ 2

1/ 2| (t)| dt.

Theorem 6.2.0. Suppose that : (0, ) (0, ) is continuous.(i) If is nonincreasing then

1(t) dt

n =1(n)

0(t) dt.

(ii) If is continuous then

1/ 2(t) dt

n =1(n)

18

1/ 2| (t)| dt.

(iii) If is convex then

n =1(n)

1/ 2(t) dt.

6.3. Proof Using Liouvilles Theorem.

6.4. Suppose that P :C

C

is continuous, P (z + 2) = P (z), and |P (z)| e |z |

. Show that thereexists c such that |P (z)| ce | I m( z ) | for all z.

Proof. First notice that |z| | R e(z)| + | I m(z)|, so e |z | e | R e( z ) | e | I m( z ) | . Thus

|P (z)|e | I m( z ) |

e |z |

e | I m( z ) | e

for |R e(z)| 1. But if z C , then z = z0 + 2 k for some |R e(z0)| 1 and k Z . Therefore

|P (z)| = |P (z0)| e e |I m( z ) | .

6.5. Prove Wallis formula:2

=

n =1

4n2

(2n 1)(2n + 1)=

21

23

43

45

.

Proof. Using Eulers formula for sin( z) gives us

1 = sin( / 2) =2

n =11 1

4n2=

2

n =1

(2n 1)(2n + 1)4n2

.


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38 DAVID C. ULLRICH

Note we can use the Cauchy Integral Formula for the last equality because for / ,

Ind( , ) =1

2i 1z dz=

1

2i 2

0

f (a + re it ) ire it

f (a + reit

) dt

=1

2i D (a,r ) f (z)f (z) dt= 0

by Cauchys theorem.

Theorem 7.7. Suppose that f H (V ), z0 V , and f f (z0) has a zero of nite order m at z0 .Then there is an open set with z0 and a one-to-one function g H () such that g(z0) = 0 ,g() = D (0, r ) for some r > 0, and such that

f (z) = f (z0) + ( g(z))m

for all z . (In particular f is exactly m-to-one in \ { z0}.)

Corollary 7.8. Suppose that f H (V ), z0 V , and f f (z0) has a zero of nite order m at z0 .Then there exists an open set with z0 , a number r > 0 and a function H (D(0, r )) such that maps D(0, r ) bijectively onto , and such that

f (z) = f (z0) + zm

for all z D(0, r ).

7.3. Suppose that f H (V ), W is an open set in the plane, and h : W V satisesf (h(z)) = z

for all z W .(i) Give an example showing that it does not follow that h H (W ).

(ii) Suppose in addition that h is continuous and show that h H (W ).

Solution (i).

Proof (ii).

8. Conformal Mappings

8.0. Meromorphic Functions and the Riemann Sphere.

Theorem 8.0.0. If f : C C is holomorphic then there exist polynomials P and Q such that

f (z) =P (z)Q(z)

for all z C .

8.1. Linear-Fractional Transformations, Part I.

Proposition 8.1.0. Suppose that f H (V ), z0 V , and f (z0) = 0 . Suppose that 0 , 1 : [0, 1] V are curves with nonvanishing (right) derivatives at 0, with j (0) = z0 . Let j (t) = f ( j (t)) . Then

arg 1(0) 0(0)

= arg 1(0) 0(0)

(by which we mean that every argument of 1(0) / 0(0) is an argument of 1(0) / 0(0) and conversely).

Theorem 8.1.1. Aut( C ) is the set of all linear-fractional transformations.

Theorem 8.1.2. Aut( C ) is equal to the set of nonconstant affine maps.


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39

8.1. Show directly (without using any results in Chapters 7 or 8) that an entire function with a poleat must be a polynomial.

Proof.

8.2. Give a direct proof of Theorem 8.1.2, without using Theorem 8.1.1.

Proof.

8.3. Show that Theorem 8.1.1 follows from Theorem 8.1.2.

Proof.

8.2. Linear-Fractional Transformations, Part II.

8.4. Give te details of both proofs of the identity A B = AB sketched above: Write out a directproof and also verify carefully that M A = A .

Proof.

Lemma 8.2.0. The group M is generated by the subgroups { | C }, { t | t > 0}, { | R },and { j, I } (here I is the identity: I (z) = z).

Theorem 8.2.1. If C C and M then (C ) C .Theorem 8.2.2. If z1 , z2 , z3 are three distinct elements of C and w1 , w2 , w3 are three distinct ele-ments of C then there xists a unique M with (zj ) = wj ( j { 1, 2, 3}).

8.5. The proof of Theorem 8.2.1 contains a proof that if M xes 0,1, and then is the identity.Show how it follows immediately that if M has three xed points it must be the identity. (Theproof should exploit the fact that M is a group, as in the proof of Theorem 8.2.1, instead of doingcalculations with the coefficients of linear-fractional transformations.)

Proof.

8.6. Suppose that C is a circle. Show that the only subset of C homeomorphic to a circle is C itself.

Proof.

8.3. Linear-Fractional Transformations, Part III.

8.4. Linear-Fractional Transformations, Part IV:The Schwarz Lemma and Automorphisms of the Disk.

Theorem 8.4.1 (Schwarz Lemma) . Suppose that f : D D is holomorphic and f (0) = 0 . Then |f (z)| | z | for all z D and |f (0) | 1. Furthermore, if |f (0) | = 1 or |f (z)| = |z| for some nonzeroz then f is a rotation: f (z) = z for some constant with | | = 1 .

Theorem 8.4.2. Suppose that Aut( D ) and (0) = 0 . Then is a rotation:

(z) = z

for some C with | | = 1 .

Lemma 8.4.3.(i) a (a) = 0 and a (0) = a.

(ii) a is its own inverse. (That is, a is an involution.)(iii) a Aut( D ) for any a D .(iv) For any a, z D we have

1 | a (z)|2 =(1 | a |2)(1 | z |2)

|1 az |2.

(v) |a (0) | = 1 | a |2 and |a (a)| = 1 / (1 | a |

2).


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40 DAVID C. ULLRICH

Theorem 8.4.4. For any Aut( D ) there exists a unique a D and C with | | = 1 such that

(z) = a (z)

for all z D .

8.7. Suppose that Aut( + ). Show that there exist a,b,c,d R , with ad bc = 1, such that

(z) =az + bcz + d

for all z + . Show that a,b,c,d are not unique but are almost unique.

Proof.

8.8. Explain why it is obvious that for any a + there exists Aut( + ) with (a) = i. (Thisfollows from the corresponding fact in the disk. But an example is much more obvious in the upperhalf-plane; pretend you dont know the result for the disk and give the obvious argument for P i + .)

Proof.

8.5. More on the Schwarz Lemma.

Theorem 8.5.0 (Invariant Schwarz Lemma) . Suppose that f : D D is holomorhpic. If f / Aut( D )then

d(f (z), f (w)) < d (z, w) for all z, w D with z = w and

|f (z)|1 | f (z)|2

0 such that

1 | f (w)| c(1 | f (z)|) (w K )

for every holomorphic f : D D .

Proof.


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41

9. Normal Families and the Riemann Mapping Theorem

9.0. Introduction.

9.1. Quasi-Metrics.

Lemma 9.1.0. Suppose that : I R is concave. Suppose that a,b,a , b I , a < b, a < b , a a,and b b. Then

(b ) (a )b a

(b) (a)b a

.

Lemma 9.1.1. Suppose that : [0, ) R is concave and (0) = 0 .

Then (x + y) (x) + (y)

for all x, y 0.

Lemma 9.1.2. Suppose that : [0, ) R is concave and (0) = 0 .

Suppose further that is nondecreasing, (t) > 0 for all t > 0, and is continuous at 0.If d is a quasi-metric on X then d = d is also a quasi-metric on X such that

(i) d(x, y) = 0 if and only if d(x, y) = 0 ,(ii) d(xn , yn ) 0 if and only if d(xn , yn ) 0.

Lemma 9.1.3. Suppose that dj is a quasi-metric on X for j N . Dene

d(x, y) =

j =1

2 jdj (x, y)

1 + dj (x, y)(x, y X ).

Then d is a quasi-metric on X with the property that d(xn , yn ) 0 as n if and only if dj (xn , yn ) 0 for every j .

Furthermore, d is a metric on X if and only if for every x, y X with x = y there exists a positive

integer j such that dj (x, y) > 0 (that is, if and only if the family (dj ) separates points of X ).Lemma 9.1.4. Suppose that cj 0 for j N and

j =1

cj < .

Suppose that (a j , n ) is a double sequence such that

|a j,n | < c j for all j and n, and such that

limn

a j,n = a j for all j . Then

limn

j =1

a j,n =

j =1

a j .

Lemma 9.1.5. Suppose that (dj ) is a sequence of quasi-metrics on X which separates points of X , sothat

d(x, y) =

j =1

2 jdj (x, y)

1 + dj (x, y)(x, y X )

denes a metric on X , as in Lemma 9.1.3 . For x X and r > 0 denes

Bd (x, r ) = {y X | d(x, y) < r } ,


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42 DAVID C. ULLRICH

as usual. For each positive integer N , x X and > 0 dene

BN (x, ) = {y X | dj (x, y) < , j { 1, . . . N }} .(i) Each BN (x, ) is open in the metric space (X, d ).

(ii) For every r > 0 there exists a positive integer N and a number > 0 such that

BN (x, ) Bd (x, r ) for all x X .

(iii) For every > 0 and positive integer N there exists a number r > 0 such that

Bd (x, r ) BN (x, )

for all x X .

Proposition 9.1.6. Suppose that (dj ) is a sequence of quasi-metrics on X which separates points of X (that is, for any x, y X with x = y there exists a j with dj (x, y) > 0). There exists a metric d on X with the following two properties:

(i) For any sequences (xn ) and (yn ) in X we have d(xn , yn ) 0 if and only if dj (xn , yn ) 0 for all j .

(ii) A set S X is totally bounded in the metric space (X, d ) if and only if for every positive

integer N and ever > 0 there xist nitely many points x1 , . . . , x n X such that

S n

k=1

BN (xk , ).

9.1. Prove part (i) of Lemma 9.1.5

Proof.

9.2. Prove Proposition 9.1.6

Proof.

9.2. Convergence and Compactness in C (D ).

Lemma 9.2.0. Suppose that D is an open subset of the plane. There exist compact sets K j ( j N )

such that K j K j +1and

D =

j =1

K j .

Theorem 9.2.1. Suppose that D is an open subset of the plane. Let (K j ) be an exhaustion of D , and dene

d(f, g ) =

j =1

2 jf g K j

1 + f g K j.

Then d is a metric on C (D ), and for a sequence (f n ) C (D ) and a function f C (D ) we have f n f in C (D ) if and only if d(f n , f ) 0.

Lemma 9.2.2. The metric space (C (D ), d) is complete.Lemma 9.2.3. The set S C (D ) is totally bounded if and only if for every compact K D and every > 0 there exist nitely many functions f 1 , . . . f n S such that

S n

k =1

BK (f k , ).

Theorem 9.2.4 (Arzela-Ascoli) . The set S C (D ) is totally bounded if and only if it is bounded and equicontinuous at every point of D .


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43

Theorem 9.2.5 (Arzela-Ascoli) . Suppose that D is an open set in the plane and S C (D ). The following are equivalent:

(i) S is pointwise bounded and pointwise equicontinuous.(ii) Any sequence in S has a subsequence which converges uniformly on compact subsets of D .

9.3. Montels Theorem.Lemma 9.3.0. If D C is open then H (D) is a closed subspace of C (D ).

Theorem 9.3.1 (Montels Theorem) . Suppose D C is open and S H (D ) is closed. Then S is compact if and only if it is uniformly bounded on compact sets (that is, if and only if for every compact K D there exists a number M such that f K M for all f S ).

Theorem 9.3.2 (Montels Theorem) . If F H (D ) is a normal family and (f n ) is a sequence in F then there exists a subsequence f n j such that f n j f H (D ) uniformly on compact subsets of D .

9.3. Suppose that D is a connected open set in the plane. For F H (D ) dene

F = {f | f F} .

(i) Show that if F is a normal family then so is F .

(ii) Give an example showing that the converse of part (i) is false.(iii) Show that the converse of (i) becomes true if you add one more very small hypothesis.

Proof (i). Let K D be compact and let f K M for all f F .

Solution (ii).

Proof (iii).

9.4. Suppose that D is a connected open set, F H (D ) is a normal family, and ( f n ) is a sequence of elements of F . Suppose that S D has a limit point in D and that f H (D ) has the property thatf n (z) f (z) for every z S . Show that f n f in H (D ).

Proof. Let X be a compact metric space, x X , (xn ) X , and xn x. Since xn x, there existsan innite subsequence ( xn j ) such that xn j x > . Since X is compact there exists a subsequenceof xn j that converges to x X . Since xn j x > then x = x .

Now suppose f n f in H (D ). Since F is precompact, there exists a subsequence ( f n j ) H (D )such that f n j g = f in H (D). Now consider the function h(z) = g(z) f (z). Since f n j (z) g(z)and f n j (z) f (z) for all z S , f |S = g|S . Thus h |S = 0. Since Z (h) has a limit point, h vanishes onD . Therefore f = g and hence f n f in H (D ).

9.5. Suppose that D is an open set in the plane, M is a positive number, and let

F = f H (D ) : D |f (z)|2 dxdy M .Show that F is a normal family.

Proof.

9.6. Suppose that D is a connected open set in the plane, ( f n )n =1 H (D ) is a normal family, andeach f n has no zero in D . If there exists z D with f n (z) 0 then f n 0 uniformly on compact

subsets of D .

Proof. Suppose f n 0 in H (D ). The proof of exercise 9.4 shows there exists a subsequence f n j f = 0 in H (D ). Let K be a compact subset of D . Since D is open, let D (z, r ) D with f (w) > for w D (z, r ). Now let J N be a number such that if j > J , then f n j (w) f (w) < for allw D (z, r ). Since

f n j (w) f (w) < < f n j (w) + |f (w)|


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44 DAVID C. ULLRICH

for all w D (z, r ), theorem 5.4 shows that if j > J , then f n j and f have the same number of zeroesin D (z, r ). This is a contradiction so f n 0 in H (D ).

9.7. Show that S C (D) is bounded (in the topological-vetor-space sense as above) if and only if forevery compact K D there exists M such that f K M for all f S .

Proof.9.8. Let L be the space of all absolutely summable real sequences: x L if and only if x = ( xn ) whereeach xn R and n =1 |xn | < . Dene a metric d on L by

d(x, y) =

n =1|xn yn | .

(i) Show that this formula actually denes a complete metric on L.(ii) Show that a subset of L is bounded in the topological-vector-space sense if and only if it is

bounded as a subset of the metric space ( L, d).(iii) Show that the closed ball B (0, 1) is not compact.

9.4. The Riemann Mapping Theorem.

Theorem 9.4.0 (Riemann Mapping Theorem) . Suppose that D C is a nonempty simply connected open set and D = C . Then D is conformally equivalent to the unit disk D .

Lemma 9.4.1. Suppose that f : D D and : D D are holomorphic. If Aut( D ) then H f (z) = H f (z)

for all z D, while if / Aut( D ) then

H f < H f (z)

for all z D with f (z) = 0 . In particular

H f 2 (z) < H f (z)

for all z D with f (z) = 0 .

Theorem 9.4.2 (Riemann Mapping Theorem, restated) . Suppose that D C is open, connected,nonempty, and has the property that any nonvanishing function holomorphic in D has a holomorphic square root. If D = C then D is conformally equivalent to D .

Corollary 9.4.3. Suppose that D is a simply connected open set, D = C , and z0 D . Then there exists a unique conformal equivalence F : D D such that F (z0) = 0 and F (z0) > 0.

9.5. Montels Theorem Again. See theorem 9.3.2.

10. Harmonic Functions

10.0. Introduction.

Lemma 10.0.0. Suppose that D R 2 is open and : D C is twice continuously differentiable in the real sense. Then is harmonic if and only if 2/z z = 0 .

Lemma 10.0.1.(i) A holomorphic function is harmonic and hence the real and imaginary parts of a holomorphic

function are harmonic.(ii) If is harmonic then f = /z is holomorphic.

(iii) If D is simply connected and is a real-valued harmonic function in D then there exists f H (D) with = R e(f ).

Theorem 10.0.2. Suppose that D is a connected open subset of the plane. Then D is simply connected if and only if every real-valued harmonic function in D is the real part of some holomorphic function.


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Theorem 10.0.2 (Part 2) . A connected open subset of the plane is simply connected if and only if every real-valued harmonic function has a harmonic conjugate.

10.1. Poisson Integrals and the Dirichlet Problem.

10.1. Explain why the following argument showing that there is no continuous function in the closed

unit disk which is holomorphic in the interior and equal to e it ate very boundary point eit is wrong: If were such a function then v(z) = (z) 1/z would dene a holomorphic function in D \ { 0} vanishingat every point of the boundary of the disk. But if the zero set of a holomorphic function has a limitpoint the function must vanish identically. Thus (z) = 1 /z for z = 0 and hence is not continuousat the origin.

Solution.

Proposition 10.1.0. If is harmonic in a neighborhood of D (z, r ) then

(z) =1

2 2

0(z + re it ) dt.

Lemma 10.1.1. If 0 r < 1 and t R then

P r (t) =

n = r |n | eint

= R e1 + re it

1 re it

=1 r 2

1 2r cos(t) + r 2

=1 r 2

(1 r cos(t))2 + r 2 sin2(t).

Lemma 10.1.2.(i) P r (t) > 0 for 0 r < 1, t R .

(ii)1

2

2

0 P r (t) dt = 1 for 0 r < 1.(iii) If 0 < then P r (t) 0 as r 1, uniformly for t [ , ] [, ].

Theorem 10.1.3. Suppose that f C ( D ). Dene : D C by

(re i ) = f (eit ), r = 1 ,

P [f ](re i ), 0 r < 1.

Then | D = f , C (D ), and |D is harmonic in D .

10.2. Suppose that : D C is continuous in D and that (re it (eit ) uniformly as r 1. Showthat is continuous in D .

Proof.

10.3. There is actually an application of the inequality

eit eis | t s |

for s, t R hidden in the proof above.(i) Explain where this inequality was used.

(ii) Prove the inequality.

Solution (i).

Proof (ii).


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46 DAVID C. ULLRICH

Lemma 10.1.4. Suppose that D is a bounded open set in the plane, and that C (D ) satises a weak mean value property in D : for every z D there exists = (z) > 0 such that D(z, (z)) Dand

(z) =1

2 2

0(z + re it ) dt

for 0 < r < (z). If vanishes identiacally on D then vanishes identically in D .Lemma 10.1.4 (Part 2) . Suppose that D is a bounded open set in the plane, and that 1 , 2 C (D )satisfy a weak mean value property in D: for every z D there exists = (z) > 0 such that D (z, (z)) D and

j (z) =1

2 2

0j (z + re it ) dt

for j { 1, 2}, 0 < r < (z). If 1 = 2 at every point of D then 1 = 2 everywhere in D .

Theorem 10.1.5. Suppose that C (D ) is harmonic in D . Then

|D = P [| D ].

Theorem 10.1.6. Suppose that C (D ) satises a weak mean value property in D : for every z D

there exists = (z) > 0 such that D (z, (z))D

and (z) =

12

2

0(z + re it ) dt

for 0 < r < (z). Then is harmonic in D .

Proposition 10.1.7. Suppose that D1 and D2 are open subsets of the plane. If f : D1 D2 is holomorphic and : D2 C is harmonic then f is harmonic in D1 .

10.4. Prove proposition 10.1.7.

Proof.

Theorem 10.1.6 (Part 2) . Suppose that C (D (a, R )) satises a weak mean value property in D (z, R): for every z D there exists = (z) > 0 such that D(z, (z)) D (a, R ) and

(z) = 12

2

0(z + re it ) dt

for 0 < r < (z). Then is harmonic in D (a, R ).

Theorem 10.1.8. Suppose that D is an open subset of the plane, C (D ), and satises a weak mean value property in D . Then is harmonic in D .

Theorem 10.1.6 (Part 3) . Suppose that C (D ) satises a very weak version of the mean value property in D : for every z D there exists r (0, 1 | z|) such that

(z) =1

2 2

0(z + re it ) dt.

Then is harmonic in D .

10.5. Suppose that D is an open subset of the plane and : D C .(i) Show that if satises a small-radius mean value property in D it follows that satises a

small-radius mean value property in any open subset of D .(ii) Explain why the same argument does not work for a one-radius mean value property.

Proof (i).

Solution (ii).

10.2. Poisson Integrals and Aut( D ).


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10.3. Poisson Integrals and Cauchy Integrals.

10.4. Series Representations for Harmonic Functions in the Disk.

10.5. Greens Functions and Conformal Mappings.

Lemma 10.5.5. Suppose that D is an open subset of the plane and K is a compact subset of D . There exists a cycle in D such that Ind ( , z) = 0 for all z C \ D , D \ K , Ind( , z) = 0 or 1 for all z D \ , and Ind( , z) = 1 for all z K .

10.6. Intermission: Harmonic Functions and Brownian Motion.

10.7. The Schwarz Reection Principle and Harnacks Theorem.

12. Runges Theorem and the Mittag-Leffler Theorem

12.1. Show that if P is a polynomial then there exists z such that |z | = 1 and P (z) 1z 1.

Proof. Suppose on the other hand that P (z) 1z < 1 for all |z | = 1. Then

D P (z) 1z dz < 2by the ML -inequality. But

D P (z) 1z dz = 2i,which is a contradiction. Therefore there exists z D such that P (z) 1z 1.

Lemma 12.0. Suppose that A C and R is a rational function such that all the nite poles of R are contained in A. Then there exists a polynomial P and complex numbers (ca,j )N j =1 for each a A such that

R(z) = P (z) +a A

Ra (z) = P (z) +a A

N

j =1

ca,j (z a) j .

Note although A may be innite this is really a nite sum; we have ca,j = 0 unless a A actually is a pole of R.

Lemma 12.1. Suppose that K C is compact and D (, r ) C \ K.

Any function f R D (,r ) can be uniformly approximated on K by functions in R { } .

12.2. Suppose that K C is compact. Suppose that ( Rn ) and ( S n ) are sequences of elements of C (K )which converge uniformly on K to f and g, respectively. Show that Rn S n converges to fg uniformlyon K .

Proof. Let > 0 be given. Since f, g C (K ), let |f (K )| M and |g(K )| N . Since Rn , S nconverge uniformly to f, g , then there exists N N such that n > N implies |Rn (K )| M , |Rn f | r }

for some r > 0. Then (i) Any f R A can by uniformly approximated on K by polynomials.

(ii) If | | > 2r then any polynomial can be uniformly approximated on K by functions in R { } .

Lemma 12.3 (Pole-Pushing Lemma) . Suppose that K C is compact and V is a connected open subset of C \ K . If V then any element of R V can by uniformly approximated on K by elements of R { } .

Theorem 12.4. Suppose that K C is compact and , C \ K . Every function in R { } can be uniformly approximated on K by elements of R if and only if and lie in the same component of C \ K .Lemma 12.5. Suppose that D is an open subset of the plane. There exist compact sets K j ( j N )such that K j K j +1 ,

D =

j =1

K j ,

and such that every component of C \ K j contains a component of C \ D .

Lemma 12.6. Suppose that is a smooth curve in the plane and f C ( ). Dene F : C \ Cby

F (z) =

f (w)w z

dw.

Suppose that K C \ is compact and > 0. Then there exist w0 , . . . , wn such that

F (z) n

j =1

f (wj )wj z

(wj wj 1) < (z K ).

Theorem 12.7 (Runges Thoerem for one compact set) . Suppose that K C is compact and AC \ K intersects every component of C \ K . If f is holomorphic in a neighborhood of K and > 0then there exists R R A such that

|f (z) R(z)| < (z K ).

Theorem 12.8 (Runges Theorem) . Suppose that D C is open and A C \ D intersects every component of C \ D. If f H (D ) then there exists a sequence of rational functions (R j ) R A such

that R j f uniformly on compact subsets of D .Corollary 12.9. If D C is an open set such that C \ D is connected and f H (D ) then there exists a sequence of polynomials P j with P j f uniformly on compact subsets of D .

Corollary 12.10. If D C is a simply connected open set and f H (D ) then there exists a sequence of polynomials P j with P j f uniformly on compact subsets of D .

12.4. Show that there exists a sequence of polynomials ( P n ) such that P n (0) = 1 for all n, whileP n (z) 0 as n if z C and z = 0.


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Proof. Let K P n = D(0, n)\ ([( 1/n, 0), (n, 1/n ), (n, 1/n )]) and K n = K P n { 0} [ 1/n,n ]. Now letf n : K n C be given by

f n (z) =0, z = 0 ,1, z K P n [1/n,n ].

Then f n is holomorphic in a neighborhood of K n and C \ K n is connected, so theorem 12.7 showsthere exists a sequence of polynomials ( P n ) such that |P n f n | < 1/n in K n . So |P n | < 1/n onK P n [1/n,n ] and |P n 1| < 1/n on {0}. But if z C \ { 0}, then there exists N N such thatz K P n [1/n,n ] for all n > N . Thus P n (z) 0 for all z C \ { 0} and similarly P n (0) 1.

Since P n (0) 1, then there exists M N such that |P n (0) 1| < 1/ 2 for all n > M . ThusP n (0) = 0 for all n > M , so we can dene P j (z) = P M + j (z)/P M + j (0) for all j N . But now ( P j ) is asequence of polynomials with the desired properties.

12.5. Show that there exists a sequence of polynomials ( P n ) such that P n (0) = 1 for all n, althoughP n (z) 0 as n for all z C \ { 0} and P n (z) 0 for all z C .

Proof. Let K P n be the same as exercise 12.4, K Dn = D (0, d(0, K P n )/ 2), and K n = K P n K Dn [1/n,n ].Now let f n : K n C be given by

f n (z) =z, z K D

n,

0, z K P n [1/n,n ].

Then theorem 12.7 states there exists P n such that |P n f n | < 1/n 2 for all n N . So if z C \ { 0},then there exists N N such that z K P n [1/n,n ] for all n > N . Thus |P n (z)| < 1/n 2 for all n > N .Since z C \ { 0}, there exists r > 0 such that D(z, r ) C \ { 0}. So Cauchys estimates show that|P n (z)| 1n 2 r for all n > N .

Now if z = 0, then |P n (0) f n (0) | = |P n (0) | < 1/n 2 . Also, if z K Dn , then |P n (z) z| < 1/n 2 ,so Cauchys estimates state |P n (0) 1| < n/n 2 = 1 /n . Thus P n (0) = 0 for all n > 2. Now deneP j (z) = P j +1 (z)/P j +1 (0) for all j N . But now ( P j ) is a sequence of polynomials with the desiredproperties.

12.6. Fix a number M < . Show that there does not exist a sequence of polynomials ( P n ) such that

limn P n (z) = 0 (z = 0) ,1 (z = 0) ,

such that|P n (z)| M (|z| 1, n N ).

Proof. Let P = P |D H (D ) for P a polynomial and ( P n ) be a sequence of polynomials with (P n )

M . Since (P n ) M , exercise 9.4 shows that if ( P n ) converges, then the limit function is in H (D ).Therefore there does not exist a sequence of polynomials with the given limit function.

Theorem 12.11 (Mittag-Leffler Theorem) . Suppose that D C is open and E D has no limit point in D . Suppose that for each a E we are given a positive integer N a and a nite sequence of complex numbers (ca,j )N aj =1 . Then there exists a function f meromorphic in D such that f has no poles

except at points of E and such that for each a E , f has a pole at a with principal part N a

j =1

ca,j(z a)j

.

12.7. Assume Runges theorem 12.8 and prove Cauchys theorem 4.10.

Proof. Suppose that V C is open, is a cycle in V , has the property that

Ind( , a) = 0


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50 DAVID C. ULLRICH

for all a C \ V , and f H (V ). Let A intersect every component of C \ V . Suppose that R(z) dz =0 for all rational functions R R A . Then by Runges theorem, we can take a sequence ( R j ) R Asuch that ( R j ) f uniformly on compact subsets of V . Thus

f (z) dz =

limj

R j (z) dz

= limj R j (z) dz

= 0 .

So we can assume that f = R R A . But now as in lemma 12.0,

R(z) dz = P (z) + a AN a

j =1

ca,j (z a) j dz

= P (z) dz + a AN a

j =1 ca,j (z a) j dz=

a A ca, 1(z a)

1dz

with the last equality provided by Cauchys theorem 2.0 for derivatives.But now we have

R(z) dz = a A ca, 1 (z a) 1 dz=

a A

ca, 1 Ind( , a)

= 0

because Ind ( , a) = 0 for all a C \ V . Therefore we have


13. The Weierstrass Factorization Theorem

Lemma 13.0. For z D and n N we have

|1 E n (z)| zn +1 .

Theorem 13.1. Suppose that (zj )j =1 is a sequence of nonzero complex numbers which tend to innity as j ; suppose N is a non-negative integer. If (n j ) is a sequence of non-negative integers such that

j =1

r|zj |

n j +1

<

for all r > 0 then the product

f (z) = zN

j =1E n j

zzj

converges uniformly on compact subsets of the plane to an entire function with a zero of order N at the origin, a zero at each zk and no other zeroes. (The order of the zero of f at zk is equal to the number of times the value zk appears in the sequence (zj ).)

Theorem 13.2 (Weierstrass Factorization Theorem for Entire Functions) . Suppose that f H (C )is nonconstant. Suppose that f has a zero of order N at the origin (0 N < ). Suppose that F has M (0 M ) zeroes away from the origin and suppose that (zj )M j =1 is a sequence containing


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the nonzero zeroes of f , each listed according to its multiplicity. If (n j )M j =1 is a sequence of positive integers such that

M

j =1

r|zj |

n j

<

for all r > 0 then there exists an entire function g such that

f (z) = eg(z ) zN N

j =1

E n jzzj

for all z C ; the product converges uniformly on compact subsets of the plane.

Theorem 13.3 (Weirstrass Factorization Theorem in an Open Set) . Suppose that D C is open,A D has no limit point in D , and m is a positive integer for each A. There exists f H (D )such that f has a zero of order m at each A and no other zeroes.

Theorem 13.3 (Weierstrass Factorization Theorem in an Open Set, Part 2) . Suppose that D C is open, D = C , A D has no limit point in D , and m is a positive integer for each A. There exists f H (D ) such that f has a zero of order m at each A and no other zeroes.

Theorem 13.4. If D is an open subset of the plane and f is meromorphic in D then there exist functions g, h H (D ) such that f =

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