C OMPLEX A NALYSIS M ATH 220B
Midterm Sample Exam
Problem 1.
Let L C be the line L = cfw_z = x + iy | x = y . Assume that f : C C is an entire function
such that for any z L we have f (z ) L. Assume that f (1) = 0. Prove that f (i) = 0.
Problem 2.
Let u b
C OMPLEX A NALYSIS M ATH 220B
Final Sample Exam
Problem 1.
Prove that the product
entire function.
k=1
zn
n!
+ exp ( 2zn ) converges uniformly on compact sets to an
Problem 2.
Let f : C C be an entire function such that
f (z + 1) = f (z ), and |f (z )| e|
Solution of Homework 7
Problem (8.3):
z
Solution: Suppose w = R , then |w| < 1. If we want to prove
n
n
R2 + z 2
R
(
)=
n
R2
Rz
n=0
Just need to prove that
n
(1 + w2 ) =
n=0
.
1
1w
Since
ln(1 w) +
= ln(
n
ln(1 + w2 )
n=0
n
(1 + w2 )(1 w)
n=0
N
= ln( lim (
Solution of Homework 2
Problem (7.1):
Solution:
Since u is harmonic in C, there is a harmonic conjugate v such that u + iv is
entire. Dene h(z ) = eu(z )+iv(z ) , then h is also entire. And |h(z )| = eu(z ) .
Since u is bounded, so is h. By Liouvilles the
Solution of Homework 5
Problem (7.40):
Solution:
Yes. If u1 u2 is a decreasing sequence of harmonic functions on
a connected open set U C, then either uj uniformly on compact
sets or there is a harmonic function u on U such that uj u uniformly on
compact
Solution of Homework 4
Problem (7.24):
Solution: Observe that if u is harmonic function, then
2u
2u
2u
2u
2u
= (a b) 2 + c
L(u) = a 2 + b 2 + c
x
y
xy
x
xy
In particular, since is a conformal mapping, u is a harmonic function
and
2 (u )
2 (u )
L(u
Solution of Homework 3
Problem (7.14):
Solution: Since u is nonvanishing, so
1
1 u
( )= 2
x u
u x
2 1
1 2u
2 u
( ) = 2 2 + 3 ( )2
x2 u
u x
u x
1
1 u
( )= 2
y u
u y
2
2
1
1u
2 u
( ) = 2 2 + 3 ( )2
y 2 u
u y
u y
Since
1
u
is harmonic,
0
1
1 2u 2u
2 u
u
( )
Solution of Homework 6
Problem (7.45):
Solution:
Let P U . By example 7.7.12, it is sufcient to prove that locally each
boundary point has a barrier. Since the domain lies only on one side of
the curve at P , there is a line segment emanating from the bou