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Math 185. Sample Answers to Problem Set #7
Page 171
1. Following the suggestion, we note that if

z

z
0

=
R
, then

z
 ≤ 
z
0

+
R
, so

f
(
z
)
 ≤
A
(

z
0

+
R
). By Cauchy’s inequality (page 165) with
n
= 2, we then have
±
±
f
00
(
z
0
)
±
±
≤
2
A
(

z
0

+
R
)
R
2
.
Letting
R
→ ∞
for ﬁxed
z
0
, we have
f
00
(
z
0
) = 0, so
f
0
(
z
) is constant on
C
by the
theorem on page 71. Let
a
1
be that constant. Then the function
g
(
z
) =
f
(
z
)

a
1
z
is entire, and its derivative vanishes everywhere on
C
, so it too is constant. Say
g
(
z
) =
a
0
. Thus
f
(
z
) =
a
1
z
+
a
0
.
Now applying Cauchy’s inequality with
n
= 0 at
z
0
= 0, we have

a
0

=

f
(0)
 ≤
AR .
Letting
R
→
0
+
gives
a
0
= 0, so
f
(
z
) =
a
1
z
for all
z
∈
C
.
3. As noted in class, if 1
≤
k
≤
n
and if

z
 ≥
R
≥
k
q
±
±
2
na
n

k
a
n
±
±
, then

z

k
≥
2
n
±
±
a
n

k
a
n
±
±
, so

a
n

k
z
n

k
 ≤ 
a
n
z
n

/
2
n
. Letting
R
be the largest of these values, we then have

P
(
z
)

=

a
n
z
n
+
a
n

1
z
n

1
+
···
+
a
0

≤ 
a
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 Fall '07
 Lim
 Math

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