Mathematics 185 – Intro to Complex Analysis
Fall 2009 – M. Christ
Solutions to Selecta from Problem Set 8
3.1.2(a)
Let
c
∈
C
and
z
1
=
c
.
Define
z
n
+1
=
z
2
n
+
c
.
Show that if

c

>
2 then
lim
n
→∞

z
n

=
∞
.
Comment.
Our text says “lim
n
→∞
z
n
=
∞
”. The most reasonable interpretation is the
one I have given, replacing
z
n
by

z
n

, so that tending to infinity becomes welldefined.
Solution.

z
n
+1

=

z
2
n
+
c
 ≥ 
z
n

2
 
c

. Define
r
=

c
 
1. We will show by induction on
n
that

z
n
 ≥ 
c

r
n

1
for all
n
≥
1. This holds for
n
= 1 since
z
1
=
c
.
If

z
n
 ≥ 
c

r
n

1
then

z
n
+1
 ≥ 
c

2
(
r
n

1
)
2
 
c

=

c

(

c

r
2
n

2

1)
=

c

((
r
+ 1)
r
2
n

2

1)
=

c

(
r
2
n

1
+
r
2
n

2

1)
≥ 
c

r
2
n

1
≥ 
c

r
n
since
r >
1 and 2
n

1
≥
n
. This is the induction step.
3.1.4
Show that the series
∑
∞
n
=0
(
n
2
+
z
2
)

1
converges at every point of
C
\
i
Z
, and that
the convergence is uniform on every closed disk in this domain.
Solution.
The solution is motivated by the observation that (
n
2
+
z
2
)

1
behaves like
n

2
for large
n
, so long as
z
is held constant. Note that (
n
2
+
z
2
)

1
is defined if and only if
n
2
+
z
2
6
= 0. Given
z
, the
n
th term of the series is defined for every
n
≥
0 if and only if
z /
∈
i
Z
.

n
2
+
z
2
 ≥
n
2
 
z

2
by the triangle inequality. We cannot say that

(
n
2
+
z
2
)


1
≤
n

2
;
this is false if
z
is imaginary (unless

z

is sufficiently large). But we can hope to say that
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 Fall '08
 OGUS
 Math, Power Series, Taylor Series, Mathematical Series, Analytic function

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