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Material Type: Assignment; Class: COMPLEX ANALYSIS; Subject: Mathematics; University: University of Washington - Seattle; Term: Autumn 2008;
Typology: Assignments
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Math 534 Homework # Autumn 2008
Let D = {z : |z| < 1 }.
anzn^ =
n=−∞
bnzn
for r < |z| < R then an = bn for all n. Convergence of the series on the region is part of the assumption.
z (z^2 + 4)(z − 3)^2 (z − 4)
and state where each expansion converges.
∂D
ζ ζ − z dζ for |z| 6 = 1. In the above integral, ∂D is oriented in the usual “positive” direction. It is much easier to do this problem without explicitly parameterizing the curve.
(Challenge problem) Suppose f is a function defined on D with the property that given any three points a, b, c ∈ D, there is an analytic function g (possibly depending on a, b, c) so that |g| ≤ 1 on D and g(a) = f (a), g(b) = f (b) and g(c) = f (c). Prove that f has a complex derivative at each point of D and |f | ≤ 1. Hint: Use the two point version to prove f is continuous first.