# ans13 - Math 185 Sample Answers to Problem Set#13 Page 285...

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Math 185. Sample Answers to Problem Set #13 Page 285 1. a . The function z 2 has a double zero at z = 0 and no poles, so Z = 2 and P = 0, and therefore Δ C arg f ( z ) = 2 π (2 - 0) = 4 π . b . The function has a simple pole at z = 0, but no zeroes inside the unit circle (they all have absolute value 3 2), so Z = 0, P = 1, and therefore Δ C arg f ( z ) = 2 π (0 - 1) = - 2 π . c . The function has a triple pole at z = 0 and seven distinct simple zeroes with absolute values 1 / 7 2, so Z = 7, P = 3, and therefore Δ C arg f ( z ) = 2 π (7 - 3) = 8 π . 3. Let θ be such that arg w = θ describes a ray emanating from w = 0 which does not intersect Γ. Then there is a branch of the function arg w on the complement of that ray, which is continuous and satisﬁes θ < arg w < θ + 2 π for all w not in the ray. Using that branch of arg w in the derivation of Δ C arg f ( z ), we have θ < φ 0 < θ +2 π and θ < arg w < θ + 2 π for all w Γ. Therefore θ < φ 1 < θ + 2 π , and hence | Δ

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ans13 - Math 185 Sample Answers to Problem Set#13 Page 285...

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