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    Maximum Modulus Principle

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    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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    Solution Preview

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

    Solution Summary

    The maximum modulus principle is integrated in this solution.