Mathematics 185 – Intro to Complex Analysis
Fall 2009 – M. Christ
Solutions to Selecta from Problem Set 7
2.5.3
Give an example of a bounded analytic function
f
on a domain
A
, such that
f
is continuous on the closure of
A
, yet sup
z
∈
A

f
(
z
)

>
sup
w
∈
bd
A

f
(
w
)

.
Solution.
Let
A
=
{
z
: Re(
z
)
>
0
}
and consider
f
(
z
) =
e
z
.
Then bd(
A
) is the imaginary axis, and

f
(
w
)

= 1 for all
w
∈
bd(
A
), yet
f
is not even bounded on
A
.
2.5.4(a)
Let
R >
0 and suppose that

z
0

< R
. Let
D
=
D
(0
, R
) and
D
0
=
D
(0
,
1). Define
T
(
z
) =
R
(
z

z
0
)
R
2

¯
z
0
z
. Show that
T
maps
D
bijectively to
D
0
, and that
T
(
z
0
) = 0.
Solution.
Certainly
T
(
z
0
) = 0. First suppose that
R
= 1, to simplify the formulas; see
below for the general case. Note that the denominator does not vanish for
z
in question:

1

¯
z
0
z
 ≥
1
 
z
0
 · 
z
 ≥
1
 
z
0

>
0
so long as

z
 ≤
1.
Now if

z

= 1 then since
z

1
= ¯
z
,

z

z
0


1

¯
z
0
z

=

z

z
0


z
 · 
¯
z

¯
z
0
z

=

z


1

z

z
0


¯
z

¯
z
0
z

= 1
since

z

z
0

=

¯
z

¯
z
0

.
In other words,

T
(
z
)
 ≤
1 for all
z
in the closure of
D
.
By the maximum principle,

T
(
z
)
 ≤
1 whenever

z
 ≤
1.
The quickest way to show that
T
is a bijection is to find its inverse algebraically: Solving
the equation
T
(
z
) =
w
gives
z
=
˜
T
(
w
) =
w
+
z
0
1+
w
¯
z
0
.
˜
T
is of the same form as
T
, but with
z
0
replaced by

z
0
; so what we have proved above shows that
˜
(
D
0
)
⊂
D
0
. Thus
T
:
D
0
→
D
0
is a bijection, with inverse
˜
T
.
Now for the general case, just write
T
(
z
) =
(
z/R
)

(
z
0
/R
)
1

(¯
z/R
)(
z/R
)
to reduce to the case
R
= 1.
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 Fall '08
 OGUS
 Math, Continuous function, Inverse function, Logarithm, Compact space, mean value property

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