For
"
u
we have that
"
b
! "
u
! "
a
on
@D
. Assume that
"
u
(
z
0
) =
c
0
>
"
a
for seme
z
0
2
D
. Then by the maximum principle
"
u
/
c
0
on
D
, but this
contradicts that
"
u
is continuously extended to
@D
with
"
u
! "
a
on
@D
.
So
"
u
! "
a
on
D
, equivalently
a
!
u
!
b
on
D
.
This show that
a
!
u
!
b
on
D
.
29

III.5.2
Fix
n
4
1
,
r >
0
and
W
=
Ve
i'
.
What is the maximum modulus of
z
n
+
W
over the disk
fj
z
j !
r
g
? Where does
z
n
+
W
attain its maximum
modulus over the disk?
Solution (A. Kumjian)
We claim that the maximum modulus of
z
n
+
W
over the disk
f
z
2
C
:
j
z
j !
r
g
is
r
n
+
V
and this is attained at
X
=
re
i8
where
/
= (2
k,
+
'
)
=n
for
k
=
0
;
1
; : : : ; n
"
1
. By the maximum modulus principle the maximum modulus
occurs on the boundary
f
z
2
C
:
j
z
j
=
r
g
. For
z
2
C
with
j
z
j
=
r
we have
by the triangle inequality
j
z
n
+
W
j ! j
z
n
j
+
j
W
j
=
r
n
+
V
=
5
5
(
r
n
+
V
)
e
i'
5
5
=
j
X
n
+
W
j
:
so the claim is proven.
30

III.5.3
Use the maximum principle to prove that the fundamental theorem
of algebra, that any polynomial
p
(
z
)
of degree
n
4
1
has a zero, by
applying the maximum principle to
1
=p
(
z
)
on a disk of large radius.
Solution
It is su¢cient to show that any
p
(
z
)
has one root, for by division we can
then write
p
(
z
) = (
z
"
z
0
)
g
(
z
)
, with
g
(
z
)
of lower degree.
Note that if
p
(
z
) =
a
n
z
n
+
a
n
!
1
z
n
!
1
+
# # #
+
a
0
;
then as
j
z
j ! 1
,
j
p
(
z
)
j ! 1
. This follows as
p
(
z
) =
z
n
5
5
5
a
n
+
a
n
!
1
z
+
# # #
+
a
0
z
n
5
5
5
:
Assume
p
(
z
)
is non-zero everywhere. Then
1
p
(
z
)
is bounded when
j
z
j 4
R
.
Also,
p
(
z
)
6
= 0
, so
1
p
(
z
)
is bounded for
j
z
j !
R
by continuity. Thus,
1
p
(
z
)
is a
bounded entire function, which must be constant. Thus,
p
(
z
)
is constant, a
contradiction which implies
p
(
z
)
must have a zero.
Solution (K. Seip)
If
p
(
z
)
has no zeros, then
1
=p
(
z
)
is an entire function. Also, if we denote by
m
(
R
)
the maximum of
1
=p
(
z
)
on the circle
j
z
j
=
R
, then
m
(
R
)
!
0
when
R
! 1
, unless
p
(
z
)
is a constant. By the maximum principle
j
1
=p
(
z
)
j !
m
(
R
)
when
j
z
j !
R
, which means
1
=p
(
z
) = 0
. This is impossible, and so
1
=p
(
z
)
is not an entire function.
31

III.5.4
Let
f
(
z
)
be an analytic function on a domain
D
that has no zeros
on
D
. (a) Show that if
j
f
(
z
)
j
attains its minimum on
D
, then
f
(
z
)
is
constant. (b) Show that if
D
is bounded, and if
f
(
z
)
extends con-
tinuously to the boundary
@D
of
D
, then
f
(
z
)
attains its minimum
on
@D
.
Solution
(a)
If
j
f
(
z
)
j
attains its minimum on
D
, then
j
1
=f
(
z
)
j
attains its maximum on
D
, which can only happens if
f
(
z
)
is a constant, by the maximum principle.
(b)
Since
D
[
@D
is compact and
f
(
z
)
is continuous it follows that
j
f
(
z
)
j
attains
both maximum and minimum value on
D
[
@D
. Assume that
j
f
(
z
)
j
does
not attain its minimum on
@D
. Then
j
f
(
z
)
j
attains its minimum on
D
and
it follows by (a) that
f
is constant so
j
f
(
z
)
j
attains its minimum on
@D
.
32

III.5.5
Let
f
(
z
)
be a bounded analytic function on the right half-plane.
Suppose that
f
(
z
)
extends continuously to the imaginary axis and
satisÖes
j
f
(
iy
)
j !
M
for all points
iy
on the imaginary axis. Show
that
j
f
(
z
)
j !
M
for all
z
in the right half-plane.

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