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# solut4 - Solutions to homework 4 II.27(a Res f(i = lim z...

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Unformatted text preview: Solutions to homework # 4 II.27. (a) Res f (i) = lim z → i ( z- i) f ( z ) = lim z → i ( z- i) z- 2 ( z- i)( z + i) = lim z → i z- 2 z + i = i- 2 2i . (b) Since e 1 /z- 1 = e- 1 e 1 /z , we get e 1 /z- 1 = e- 1 (1 + 1 1! z + 1 2! z 2 + ··· ) . The coefficient c- 1 of 1 /z in this Laurent series, i.e., Res f (0), is equal to e- 1 . (c) Using the Taylor expansion of sin z at the origin, we get: sin z z 2 = 1 z 2 z- z 3 3! + z 5 5!- ··· ! = 1 z- z 3! + z 3 5!- ··· The coefficient c- 1 of 1 /z in this Laurent series is equal to 1, and this is nothing but Res f (0). (d) Res f π 6 = lim z → π 6 cos z d dz (1 / 2- sin z ) = lim z → π 6 cos z (- cos z ) =- 1 . II.28. (a) The circle | z | = 2 encloses the only singularity of the function f ( z ) := cos z z , namely, its pole of order 1 at the origin. Since Res f (0) = lim z → zf ( z ) = lim z → cos z = 1 , we conclude I C cos z z d z = 2 π i Res f (0) = 2 π i ....
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solut4 - Solutions to homework 4 II.27(a Res f(i = lim z...

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