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Maximum Modulus Principle

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See attached a pdf file.

You can find more Mobius transformation information from John Conway - Functions of One Complex Variable I or Serge Lang - Complex Analysis or other Complex Analysis books.

Problem of Schwarz's Lemma
1. Suppose f: D → ¢ satisfies Re f(z) ≥ 0 for all z in D and suppose that f is analytic and
(a) Show that Re f(z) > 0 for all z in D
(b) By using an appropriate Mobius transformation, apply Schwarz's Lemma to prove
that if f(0) = 1 then |f(z)| ≤  ||
- ||
for |z| < 1. What can be said if f(0) ≠ 1?
(c) Show if f(0) = 1, f also satisfies |f(z)| ≥  ||
 || .
Nots: Re f(z) indicates the residue of f(z)
Theorem, Lemma and Propositon:
Schwarz's Lemma:
Let D = {z: |z| < 1}, suppose f is analytic on D with
(1) |f(z)| ≤ 1 for z in D
(2) f(0) = 0
Then |f'(0)| <= 1 and |f(z)| ≤ |z| for all z in D.
If |f'(0)| = 1 or if |f(z)| = |z| for some z not equal to zero, then there exists a constant c,
|c| = 1, such that f(w) = cw for all w in D.
If |a| < 1 then ϕa is a one-one map of D = {z: |z| < 1} onto itself; the inverse of ϕa is ϕ-a.
Furthermore, ϕa maps ∂D onto ∂D, ϕa(a) = 0, ϕa'(0) = 1 - |a|
, and ϕa'(a) = (1 - |a|
) -1
Let f: D → D be a one-one analytic map of D onto itself and suppose f(z) = 0. Then there
is a complex number c with |c| = 1 such that f = cϕa.
A mapping of the form S(z) = 

is called a linear fractional transformation. If a, b, c,
and d also satisfy ad - bc ≠ 0 then S(z) is called a Mobius transformation.

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Problem #1
(a) We consider the function . Since is analytic and non-constant in , then is also analytic and ...

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