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# ans11 - Math 185 Sample Answers to Problem Set#11 Page 257...

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Math 185. Sample Answers to Problem Set #11 Page 257 1. The function 1 / ( z 2 + 1) is even, is defined everywhere on the real axis, and has only a pole at z = i in the upper half plane. By Theorem 2 on page 243, the residue there is 1 / 2 i . For all z in the semicircle C R (defined on page 253), we have 1 z 2 + 1 1 R 2 - 1 , so Z C R dz z 2 + 1 πR · 1 R 2 - 1 = πR R 2 - 1 for all R > 1 . This expression goes to 0 as R → ∞ , so by the method of Section 71, we have Z 0 dx x 2 + 1 = 1 2 Z -∞ dx x 2 + 1 = 1 2 · 2 πi · 1 2 i = π 2 . 5. We first compute some residues. Let f ( z ) = z 2 ( z 2 + 9)( z 2 + 4) 2 . By Theorem 2 on page 243 (with p ( z ) = z 2 / ( z 2 + 4) 2 ), we have Res z =3 i f ( z ) = (3 i ) 2 2(3 i )((3 i ) 2 + 4) 2 = - 9 6 i ( - 5) 2 = 3 i 50 . By the theorem on page 234, we have Res z =2 i f ( z ) = d dz z 2 ( z 2 + 9)( z + 2 i ) 2 z =2 i = 2 z ( z 2 + 9)( z + 2 i ) 2 - z 2 (2 z ( z + 2 i ) 2 + 2( z 2 + 9)( z + 2 i )) ( z 2 + 9) 2 ( z + 2 i ) 4 z =2 i = 4 i (5)( - 16) - ( - 4)(4 i ( - 16) + 2(5)(4 i )) 5 2 · 16 2 = 4 i ( - 80 - 64 + 40) 80 2 = - 13 i 200 . If R > 3 , then for all z C R (defined on page 253), we have | f ( z ) | ≤ R 2 ( R 2 - 9)( R 2 - 4) 2 , 1

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2 so letting R → ∞ in (3) on page 256, we have lim R
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ans11 - Math 185 Sample Answers to Problem Set#11 Page 257...

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