# N 0 2 n 1 w n and 1 2 w 1 1 2 n 0 1 n w n therefore f

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Chapter 4 / Exercise 2
Calculus
Stewart
Expert Verified
= n = 0 2 ( n + 1 ) w n and - 1 2 ( w + 1 ) = - 1 2 n = 0 ( - 1 ) n w n . Therefore, f ( z ) = 3 w + n = 0 2 n + 9 2 - ( - 1 ) n 2 w n = 3 z + 1 + n = 0 2 n + 9 2 - ( - 1 ) n 2 ( z + 1 ) n .
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Chapter 4 / Exercise 2
Calculus
Stewart
Expert Verified
82 Chapter 4. Series 9. Write the two Laurent series in powers of z that represent the function f ( z ) = 1 z ( 1 + z 2 ) in certain domains and specify these domains. Solution. Since f ( z ) is analytic at z 6 = 0 , ± i , it is analytic in 0 < | z | < 1 and 1 < | z | < . For 0 < | z | < 1, f ( z ) = 1 z ( 1 + z 2 ) = 1 z 1 1 - ( - z 2 ) = 1 z n = 0 ( - 1 ) n z 2 n = n = 0 ( - 1 ) n z 2 n - 1 and for 1 < | z | < , f ( z ) = 1 z ( 1 + z 2 ) = 1 z 3 1 1 - ( - z - 2 ) = 1 z 3 n = 0 ( - 1 ) n z - 2 n = n = 0 ( - 1 ) n z - 2 n - 3 10. Let f ( z ) = z 2 z 2 - 3 z + 2 Find the Laurent series of f ( z ) in each of the following domains: (a) 1 < | z | < 2 (b) 1 < | z - 3 | < 2 Solution. First, we write f ( z ) as a sum of partial fractions: z 2 z 2 - 3 z + 2 = 1 + 3 z - 2 ( z - 2 )( z - 1 ) = 1 + 4 z - 2 - 1 z - 1 In 1 < | z | < 2, z 2 z 2 - 3 z + 2 = 1 - 2 1 - z / 2 - 1 z 1 1 - 1 / z = 1 - 2 n = 0 2 - n z n - 1 z n = 0 z - n = - 1 - n = 1 2 1 - n z n - n = 1 z - n
4.1 Taylor and Laurent series 83 In 1 < | z - 3 | < 2, z 2 z 2 - 3 z + 2 = 1 + 4 ( z - 3 )+ 1 - 1 2 +( z - 3 ) = 1 + 4 z - 3 1 1 + 1 / ( z - 3 ) - 1 2 1 1 +( z - 3 ) / 2 = 1 + 4 z - 3 n = 0 ( - 1 ) n ( z - 3 ) - n - 1 2 n = 0 ( - 1 ) n 2 - n ( z - 3 ) n = 1 2 + 4 n = 1 ( - 1 ) n + 1 ( z - 3 ) - n - n = 1 ( - 1 ) n 2 - n - 1 ( z - 3 ) n 11. Let f ( z ) = z 2 z 2 - z - 2 Find the Laurent series of f ( z ) in each of the following domains: (a) 1 < | z | < 2 (b) 0 < | z - 2 | < 1 Solution. First, we write f ( z ) as a sum of partial fractions: z 2 z 2 - z - 2 = 1 + z + 2 ( z - 2 )( z + 1 ) = 1 + 4 3 ( z - 2 ) - 1 3 ( z + 1 ) In 1 < | z | < 2, z 2 z 2 - z - 2 = 1 - 2 3 1 1 - z / 2 - 1 3 z 1 1 + 1 / z = 1 - 2 3 n = 0 2 - n z n - 1 3 z n = 0 ( - 1 ) n z - n = 1 3 - 1 3 n = 1 2 1 - n z n + 1 3 n = 1 ( - 1 ) n z - n In 0 < | z - 2 | < 1, z 2 z 2 - z - 2 = 1 + 4 3 ( z - 2 ) - 1 3 1 3 +( z - 2 ) = 1 + 4 3 ( z - 2 ) - 1 9 1 1 +( z - 2 ) / 3 = 1 + 4 3 ( z - 2 ) - 1 9 n = 0 ( - 1 ) n 3 - n ( z - 2 ) n = 8 9 + 4 3 ( z - 2 ) + n = 1 ( - 1 ) n + 1 3 - n - 2 ( z - 2 ) n
84 Chapter 4. Series 12. Find the Laurent series of 1 e z 2 - 1 in z up to z 6 and show the series converges in 0 < | z | < 2 π . Solution. Let f ( z ) = 1 / ( e z - 1 ) . Since e z - 1 = n = 0 z n n ! - 1 = n = 1 z n n ! = z n = 0 z n ( n + 1 ) ! e z - 1 has a zero at 0 of multiplicity one and hence f ( z ) has pole at 0 of order 1 . So the Laurent series of f ( z ) is given by f ( z ) = n = - 1 a n z n = a - 1 z + a 0 + a 1 z + a 2 z 2 + a 3 z 3 + n 4 a n z n in 0 < | z | < r for some r > 0. Since ( e z - 1 ) f ( z ) = 1, we have 1 = a - 1 + a 0 z + a 1 z 2 + a 2 z 3 + a 3 z 4 + n 5 a n - 1 z n ! 1 + z 2 + z 2 6 + z 3 24 + z 4 120 + n 5 z n ( n + 1 ) ! ! . Comparing the coefficients of 1, z , z 2 , z 3 and z 4 on both sides, we obtain a - 1 = 1 a 0 + a - 1 2 = 0 a 1 + a 0 2 + a - 1 6 = 0 a 2 + a 1 2 + a 0 6 + a - 1 24 = 0 a 3 + a 2 2 + a 1 6 + a 0 24 + a - 1 120 = 0 Solving it, we have a - 1 = 1 , a 0 = - 1 / 2 , a 1 = 1 / 12 , a 2 = 0 and a 3 = - 1 / 720 . Hence f ( z ) = 1 z - 1 2 + z 12 - z 3 720 + n 4 a n z n and 1 e z 2 - 1 = f ( z 2 ) = 1 z 2 - 1 2 + z 2 12 - z 6 720 + n 4 a n z 2 n . Note that f ( z ) is analytic in { z : e z - 1 6 = 0 } = { z 6 = 2 n π i } . So it is analytic in 0 < | z | < 2 π . Therefore, f ( z 2 ) is analytic in 0 < | z 2 | < 2 π , i.e., 0 < | z | < 2 π . So the series converges in 0 < | z | < 2 π .
4.2 Classification of singularities 85 4.2 Classification of singularities 1. For each of the following complex functions, do the following: find all its singularities in C ; write the principal part of the function at each singularity;