Excercise Sheet Solution 7

# Excercise Sheet Solution 7 - MATH 20142 Complex Analysis...

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MATH 20142 Complex Analysis Solution Sheet 7 1. Let f ( z ) = + m = -∞ c m z m denote the Laurent series of f at z = 0. (a) This function has no singularity at 0, whence its Laurent series is none other than its Taylor series, i.e., 1 z 3 = 1 3( z/ 3 1) = 1 3 · ( 1 + z 3 + z 2 9 + z 3 27 + · · · ) The annulus here is in fact the disc { z : | z | < 3 } . (b) Here 1 1 z = 1 + z + z 2 + · · · , whence 1 z (1 z ) = 1 z + 1 + z + z 2 + z 3 + · · · The annulus here is the punctured { z : 0 < | z | < 1 } . (c) We have for z ̸ = 0, e 1 /z = 1 + 1 z + 1 2 z 2 + 1 6 z 3 + 1 24 z 4 + · · · , whence z 3 e 1 /z = z 3 + z 2 + 1 2 z + 1 6 + 1 24 z + · · · Our annulus is the punctured complex plane { z C : 0 < | z | < ∞} . (d) We have for z ̸ = 0, cos(1 /z ) = 1 1 2 z 2 + 1 4! z 4 1 6! z 6 + · · · Here the annulus is { z C : 0 < | z | < ∞} as well. 2. (a) z 2 + 1 = ( z i )( z + i ), whence the poles are z = ± i , both simple. (b) The poles are the roots of the polynomial z 4 + 16. To compute them, write 16 = 16 e , whence our roots are as follows: x n = 2 e 4 + inπ 2 , n = 0 , 1 , 2 , 3 .

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