Review Exam II
Complex Analysis
Underlined Definitions: May be asked for on exam
Underlined Propositions or Theorems: Proofs may be asked for on exam
Chapter 5.3
Definition. Let G be a region. A function f is meromorphic on G if . . .
Argument Principle Let G be a region and let f be meromorphic on G with poles { p1 , p2 , , pm } and zeros
{z1 , z2 , , zn } , counted according to multiplicity. Let
through any of the points p j and

be a closed rectifiable curve in G which does not pass
zk and let   0 in G. Then,
n
m
1
f ( z )
dz   n( ; zk )  n( ; p j ) .
2 i  f ( z )
k 1
j 1
Corollary Let G be a region and let f be meromorphic on G with poles { p1 , p2 , , pm } and zeros
{z1 , z2 , , zn } , counted according to multiplicity. Let

be a simple closed rectifiable curve in G which is
positively oriented which does not pass through any of the points p j and
zk and let   0 in G. Let
int( ) denote the bounded component of  \{ } . Let Pf , and Z f , denote the cardinality of
{ p1 , p2 , , pm }  int( ) and {z1 , z2 , , zn }  int( ) , respectively . Then,
1
f ( z )
dz  Z f ,  Pf , .
2 i  f ( z )
Corollary Let G be a region and let f be analytic on G with zeros {z1 , z2 , , zn } , counted according to
multiplicity. Let  be a simple closed rectifiable curve in G which is positively oriented which does not pass
through any of the points
zk and let   0 in G. Let int( ) denote the bounded component of  \{ } . Let
Z f , denote the cardinality of {z1 , z2 , , zn }  int( ) . Then, 1  f ( z) dz  Z f , .
2 i  f ( z)
Problems about counting the number of zeros of a given function in a given region using Argument Principle
Rouche’s Theorem (Ver #1) Let G be a region and let f and g be mereomorphic on G. Let
B (a, r )  G such that f and g have no zeros or poles on C (a, r ) . Let Z f , Z g , Pf , Pg denote the cardinality
of the zeros of f and g and the cardinality of the poles of f and g inside C ( a, r ) , respectively. If
| f ( z )  g ( z ) || f ( z ) |  | g ( z ) | on C (a, r ) , then Z f  Pf  Z g  Pg .
Corollary Let G be a region and let f and g be analytic on G. Let B ( a, r )  G such that f and g have no
zeros on C ( a, r ) . Let Z f , Z g denote the cardinality of the zeros of f and g inside C ( a, r ) , respectively. If
| f ( z )  g ( z ) || f ( z ) |  | g ( z ) | on C (a, r ) , then Z f  Z g .
Rouche’s Theorem (Ver #2) Let G be a region and let f and g be mereomorphic on G. Let
B (a, r )  G such that f has no zeros or poles on C (a, r ) . Let Z f , Z f  g , Pf , Pf  g denote the cardinality of
the zeros of f and g and the cardinality of the poles of f and f+g inside C ( a, r ) , respectively. If
| g ( z)|  | f ( z)|   on C ( a, r ) , then Z f  Pf  Z f  g  Pf  g .
Corollary Let G be a region and let f and g be analytic on G. Let B ( a, r )  G such that f has no zeros on
C (a, r ) . Let Z f , Z g denote the cardinality of the zeros of f and g inside C (a, r ) , respectively. If
| g ( z)|  | f ( z)| on C (a, r ) , then Z f  Z f  g .
Problems about counting the number of zeros of a given function in a given region using Rouche’s Theorem
Chapter 6.1
Maximum Modulus Theorem (Ver #1) Let G be a region and let f A(G) . If there exists a  G such that
| f (a) || f ( z ) | for all z  G , then f is constant.
Maximum Modulus Theorem (Ver #2) Let G be a bounded region and let f  A(G) C (G) . Then,
max | f ( z ) | max | f ( z ) | .
z G
z  G
Definition. Let G be a region and let f : G   . Let a  G or a   . Then, lim sup f ( z )  . . .
za
Definition. Let G be a region. Then,   G  . . .
Maximum Modulus Theorem (Ver #3) Let G be a region and let f  A(G ) . Suppose there exists a constant
M such that lim sup | f ( z ) |  M for all a    G . Then, | f ( z ) |  M for all z  G .
za
Supplement Maximum Modulus
Definition (for Supplement). Let D  {z :| z |  1} and for 0  r  1 let Dr  {z :| z |  r}
Theorem For f  A( D) let
a)
M (r , f )  max | f ( z ) |
A(r , f )  max Re f ( z )
b)
| z|  r
c)
1
I p (r , f ) 
2
| z|  r

 | f (re
i
) | p d

Then,
(i) M ( r , f ) is a strictly increasing function of r on (0,1) unless f is constant.
(i) A(r, f ) is a strictly increasing function of r on (0,1) unless f is constant.
(i) I p (r, f ) is a strictly increasing function of r on (0,1) unless f is constant, for p  

Parseval’s Identity Let f A( D) , f ( z)   an z n . For 0  r  1 ,
n0


I 2 (r, f )  1  | f (rei )|2 d   | an |2 r 2n
2 
n 0
Chapter 6.2
Let D denote the open unit disk (centered at 0) and let Dr denote open disk of radius r (centered at 0).
Schwarz’s Lemma (Ver #1) Let f  A( D) such that (a) | f ( z ) |  1 for all z  D and (b) f (0)  0 .
| f ( z ) | | z | for all z  D and (ii) | f (0) |  1 . Moreover, equality occurs in (i) or (ii) if and only
if f ( z )   z for some  , |  | 1 .
Then, (i)
Schwarz’s Lemma (Ver #2) Let f  A( D) such that (a)
Then, (i) f ( Dr )  Dr and (ii)
f ( D)  D and (b) f (0)  0 .
| f (0) |  1 . Moreover, equality occurs in (i) or (ii) if and only if
f ( z )   z for some   D .
za
. Then, a  A( D), a : D  D and,  a is one-to-one and onto.
1  az
2
2 1
Further, for | z | 1 , |  a ( z ) | 1 and  a (0)  1 | a | ,  a ( a )  (1 | a | ) .
Proposition. For a  D let
a ( z ) 
Proposition Let f  A( D), f : D  D . For a  D let f ( a )   . Then, | f ( a ) | 
occurs if and only if f ( z)   (a ( z)) for some  D .
1 |  |2
. Equality
1 | a |2
Theorem Let f  A( D ), f : D  D such that f is one-to-one and onto. Then, there exists
a  D and   D such that f   a .
Supplement Subordination
Definition (for Supplement). Let D  {z :| z |  1} and for 0  r  1 let Dr  {z :| z |  r}
Definition Let f , g  A( D) . Then, f is subordinate to g ( f  g ) if . . .
Theorem. Let f  g , f ( z ) 

a
n0
i)
ii)
iii)
iv)
v)
| a
k 0
vii)
n
z , g ( z )   bn z n . Then,
n0
f (0)  g (0), | f (0) || g (0) |
f ( D)  g ( D)
f ( Dr )  g ( Dr )
M (r , f )  M (r , g )
I p (r , f )  I p (r , g )
m
vi)

n
k
m
|   | bk |2 , m  0, 1, 2, 3, 
2
k 0
max (1 | z |2 ) | f ( z ) | max (1 | z |2 ) | g ( z ) |
z  Dr
z  Dr
Proposition. Let f , g  A( D) . Suppose (a) f (0)  g (0) , (b)
f ( D)  g ( D) , (c) g is one-to-one. Then,
f  g.
Example. Let P  { f  A( D):Re f ( z)  0 for all z  D, f (0)  1}. Let p ( z ) 
f P implies f  p .
1 z
. Then,
1 z
Chapter 10.1
Definition. Let G be a region and let f : G   . f is harmonic on G if . . .
Definition. Let G be a region and let f : G   . f satisfies the Mean Value Property (MVP) on G if . . .
Proposition. Let G be a region. Let f be harmonic on G. Then, f satisfies the MVP on G.
Maximum Principle (Ver. #1) Let G be a region and let u : G  R satisfy the MVP on G. If there exists
a  G such that u (a )  u ( z ) for all z  G , then u is constant.
u, v : G   be bounded functions satisfying the
MVP on G. Suppose for each a    G that lim sup u ( z )  lim sup v( z ) , then either
Maximum Principle (Ver. #2) Let G be a region and let
za
za
u ( z )  v( z ) for all z  G or else u  v .
Minimum Principle (Ver. #1) Let G be a region and let u : G  R satisfy the MVP on G. If there exists
a  G such that u (a )  u ( z ) for all z  G , then u is constant.
Chapter 10.2
Definition. The Poisson kernel Pr ( )  . . .
Proposition Pr ( )  Re
1  rei
1 r2
ei  r


Re
1  rei 1  2r cos( )  r 2
ei  r
Proposition. The Poisson kernel satisfies
(i)
1
2

 P ( ) d  1 ,
r
0  r 1

(ii)
Pr ( )  0 , Pr ( )  Pr ( )
(iii)
Pr ( )  Pr ( ) for 0   |  |  , i.e., Pr ( ) is strictly decreasing on (0,  )
(iv)
lim Pr ( )  0 for each  , 0    
r  1
Theorem Let f  C (D) , f : D   . Then, there exists u  C ( D)  Har ( D) such that
(i)
(ii)
u (ei )  f (ei )
u is unique
1
Corollary Let u  C ( D)  Har ( D) . Then, u ( re ) 
2
i

 P (  t )u(e
r
it
) dt , 0  r  1

Corollary Let h C (C (a, R)) , h : C (a, R)   . Then, there exists w  C ( B (a, R))  Har ( B (a, R)) such that
w( z )  h( z ) on C (a, R ) .
Corollary Let u  C ( B (a, R))  Har ( B (a, R)) . Then,
u(a  rei ) 

1 P (  t )u(a  reit ) dt , 0  r  R
2  r / R
Harnack’s Inequality Let u  C ( B (a, R))  Har ( B (a, R)) , u ( z )  0 . Then,
R  r u(a)  u(a  rei )  R  r u(a) , 0  r  R
Rr
Rr
Chapter 7.1
Definition. Let G be a region and let (, d ) be a complete metric space. Then, C (G, ) = . . .
Proposition. Let G be a region. Then there exists a sequence of subsets {K n } of G such that
(i)
K n  G
(ii)
K n  int( K n 1 )

(iii)
K
n
G
n 1
(iv)
K  G implies K  K n for some n  
Lemma If ( S , d ) is a metric space, then ( S ,  ) is a metric space, where  ( s, t ) 
d ( s, t )
. A set O is open
1  d ( s, t )
in ( S , d ) if and only if O is open in ( S ,  ) .
Definition. For K  G and f , g C (G , ) , let  K ( f , g )  ,  K ( f , g )  , BK ( f , )  .
Definition. For {K n } a compact exhaustion of a region G and for f , g C (G, ) let
( f , g) 
...
Proposition. (C (G, ),  ) is a metric space.
Lemma 1.7
Lemma 1.10
  0 there exists   0 and K  G such that for f , g  C (G, )
K ( f , g )     ( f , g )  
(ii) Given   0 and K  G there exists   0 such that
 ( f , g )    K ( f , g )  
(i) A set O  C (G , ) is open if and only if for each f  O there exists   0 and
K  G such that O  B ( f ,  )
(i) Given
K
(ii) A sequence { f n }  C (G , ) converges to f (in the  metric) if and only if for each
K  G { f n } converges to f in the  K metric.
Proposition (C (G , ),  ) is a complete metric space.
Definition. A set F  C (G , ) is normal . . .
Proposition. A set F  C (G , ) is normal if and only if F is compact.
Proposition. A set F  C (G , ) is normal if and only if for each
  0 and K  G there exist functions
n
f1 , f 2 ,  , f n  F such that F   B K ( f k ,  ) .
k 1
Definition. A set F  C (G , ) is equicontinuous at a point
z0  G if . . .
Definition. A set F  C (G , ) is equicontinuous on a set
E  G if . . .
Proposition. Suppose a set F  C (G , ) is equicontinuous at each point of G. Then, F is equicontinuous
on each K  G .
Arzela-Ascoli Theorem A set F  C (G , ) is normal if and only if
(a) for each z  G the orbit of z under F, i.e.,
{ f ( z ) : f  F } , has compact closure and
(b) F  C (G , ) is equicontinuous at each point of G.