ans7 - Math 185. Sample Answers to Problem Set #7 Page 171...

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Math 185. Sample Answers to Problem Set #7 Page 171 1. Following the suggestion, we note that if | z - z 0 | = R , then | z | ≤ | z 0 | + R , so | f ( z ) | ≤ A ( | z 0 | + R ). By Cauchy’s inequality (page 165) with n = 2, we then have ± ± f 00 ( z 0 ) ± ± 2 A ( | z 0 | + R ) R 2 . Letting R → ∞ for fixed z 0 , we have f 00 ( z 0 ) = 0, so f 0 ( z ) is constant on C by the theorem on page 71. Let a 1 be that constant. Then the function g ( z ) = f ( z ) - a 1 z is entire, and its derivative vanishes everywhere on C , so it too is constant. Say g ( z ) = a 0 . Thus f ( z ) = a 1 z + a 0 . Now applying Cauchy’s inequality with n = 0 at z 0 = 0, we have | a 0 | = | f (0) | ≤ AR . Letting R 0 + gives a 0 = 0, so f ( z ) = a 1 z for all z C . 3. As noted in class, if 1 k n and if | z | ≥ R k q ± ± 2 na n - k a n ± ± , then | z | k 2 n ± ± a n - k a n ± ± , so | a n - k z n - k | ≤ | a n z n | / 2 n . Letting R be the largest of these values, we then have | P ( z ) | = | a n z n + a n - 1 z n - 1 + ··· + a 0 | ≤ | a
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ans7 - Math 185. Sample Answers to Problem Set #7 Page 171...

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