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Question related to conformal mappings

  1. Jun 26, 2011 #1
    1. P(z) = a_0 + a_1*z + a_2*z^2 + ... + a_n*z^n is 1-1 on the open unit disk D = {z: |z|<1}. Prove that |a_n| < |a_1|\n.


    3. P(z) is analytic and 1-1, and therefore conformal, so |P'(z)|>0 for every z in D.
    So 0 < |P'(z)|=|a_1 + 2a_2*z +...+ na_n*z^n-1| < |a_1| + ...+|na_n|.

    Also the inverse function P^-1(z) is conformal in P(D) and |P^-1(z)|<1.
    We also know that the derivative d\dz P^-1(z) = 1\P'(P^-1(z)) and that |d\dz P^-1(z)|>0.

    Additionally, we know that a_k = (P^(k) (0)\ k!) .

    The only other things I thought of that could be useful are the Cauchy inequality |a_k| < M(r)\ r^k, where M(r) = max{f(z): |z| = r }, and maybe Shwarz's Lemma, though since P(z) doesn't map 0 to 0 and doesn't map the unit disc to itself, I don't see how it would be relevant.

    Would appreciate any help very much ... I don't really know where to go from here.
  2. jcsd
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