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# ans10 - Math 185 Sample Answers to Problem Set#10 Page 233...

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Math 185. Sample Answers to Problem Set #10 Page 233 1. a . From the example on page 233, we have z exp 1 z = z + 1 + 1 (2!) z + 1 (3!) z 2 + . . . with principal part 1 (2!) z + 1 (3!) z 2 + . . . . This is an essential singularity. b . The isolated singular point is z = - 1 . At that point the Laurent series is z 2 1 + z = ((1 + z ) - 1) 2 1 + z = (1 + z ) 2 - 2(1 + z ) + 1 1 + z with principal part 1 / 1 + z . This is a (simple) pole. c . We have sin z z = X n =0 ( - 1) n z 2 n (2 n + 1)! , with principal part 0 . This is a removable singularity. d . We have cos z z = X n =0 ( - 1) n z 2 n - 1 (2 n )! , with principal part 1 /z . This is a pole. e . The function 1 / (2 - z ) 3 has principal part - 1 / ( z - 2) 3 , which means that it is a pole. 4. Let a be a positive real number, and (as the problem asks) write f ( z ) = 8 a 3 z 2 ( z 2 + a 2 ) 3 = φ ( z ) ( z - ai ) 3 , where φ ( z ) = 8 a 3 z 2 ( z + ai ) 3 . 1

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2 This is so because z 2 + a 2 = ( z - ai )( z + ai ) . The function φ ( z ) is analytic at z = ai , so it has a Taylor series there: φ ( z ) = X n =0 φ ( n ) ( ai ) n ! ( z - ai ) n , so the Laurent series of f ( z ) at z = ai is f ( z ) = X n =0 φ ( n ) ( ai ) n !
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