Problem 2
Solution 2
Problem 4
Solution 4
Define new function h(z)
Define
.
h is continuous on the closure of D
Since
on
, then by the Maximum Modulus Principle,
is not zero in
.
Hence, since
and
are analytic on
and
on
, then
is analytic on
which implies
is continuous on
h is analytic on D
This follows from above
Case 1: h(z) non-constant on D
If
is not constant on
, then by Maximum Modulus Principle,
achieves its maximum value on the boundary of
.
But since
on
(by the hypothesis), then
on
.
In particular
, or equivalently
Case 2: h(z) constant on D
Suppose that
is constant. Then
where
Then from hypothesis we have for all
,
which implies
Hence, by maximum modulus principle, for all
i.e.
Since
, we also have
Problem 6
Solution 6