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P
ROBLEM SET
11
Partial solution
6. Consider the function
h
=
f

g
. Since
f
and
g
are analytic, so is
h
. Since real parts
of
f
and
g
are identical on the boundary
B
, the real part of
h
is equal to 0 on
B
. Let us
denote by
u
the real part, and by
v
the imaginary part of
h
. By the consequence of the
Maximum Modulus Principle, that was given in (4) on page 192 of the textbook, both
the maximum and the minimum on
D
∪
B
of
u
are achieved on
B
. Therefore both the
maximum and the minimum of
u
on
B
∪
D
are equal to
0
. Hence
u
is constant, and
equal to 0 on
D
∪
B
.
By the CauchyRiemann equations
v
y
=
u
x
= 0
and
v
x
=

u
y
= 0
on
D
∪
B
. Hence
v
must be a constant on
D
∪
B
; we denote it by
α
. Therefore
f
=
g
+
h
=
g
+
iα
on
D
∪
B
.
8. Let us consider ﬁrst the case that
f
is equal to zero at some point
w
∈
B
. Then for all
z
∈
D
∪
B
,

f
(
z
)
 ≥ 
f
(
w
)

= 0
and hence

f

achieves its minimum on the boundary
(which is what we wanted to prove).
Now consider the case that
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This document was uploaded on 03/27/2010.
 Spring '09

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