Hence f z 1 z 1 2 z 12 z 3 720 n 4 a n z n and 1 e z

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Calculus
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Chapter 15 / Exercise 23
Calculus
Stewart
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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 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; for each singularity, determine whether it is a pole, a removable singu- larity, or an essential singularity; compute the residue of the function at each singularity. (a) f ( z ) = 1 ( cos z ) 2 Solution. f ( z ) is singular at cos z = 0, i.e., z = n π + π /2. Let w = z - n π - π /2. Then 1 ( cos z ) 2 = 1 ( cos ( w + n π + π /2 )) 2 = 1 ( sin w ) 2 .
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Calculus
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Chapter 15 / Exercise 23
Calculus
Stewart
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68 4. Series Since sin w has a zero of multiplicity one at w = 0, f ( z ) has a pole of order 2 at z = n π + π /2. So 1 ( sin w ) 2 = a - 2 w 2 + a - 1 w + n 0 a n w n . Since ( sin w ) 2 = n = 0 ( - 1 ) n w 2 n + 1 ( 2 n + 1 ) ! ! 2 = w 2 + n = 4 b n w n we have 1 = a - 2 w 2 + a - 1 w + n 0 a n w n ! w 2 + n = 4 b n w n ! . Comparing the coefficients of 1 and w on both sides, we obtain a - 2 = 1 and a - 1 = 0. So the principal part of f ( z ) at z = n π + π /2 is 1 ( z - n π - π /2 ) 2 with residue 0. (b) f ( z ) = ( 1 - z 3 ) exp 1 z Solution. Since e z = n = 0 z n / n !, ( 1 - z 3 ) exp 1 z = ( 1 - z 3 ) n = 0 1 n ! z n = n = 0 1 n ! z n - n = 0 1 n ! z n - 3 = n = 0 1 n ! z n - n = 4 1 n ! z n - 3 - 3 n = 0 z 3 - n n ! = 1 + n = 1 1 n ! z n - n = 1 1 ( n + 3 ) ! z n - 3 n = 0 z 3 - n n ! = n = 1 1 n ! - 1 ( n + 3 ) ! 1 z n + 1 - 3 n = 0 z 3 - n n ! Therefore, f ( z ) has an essential singularity at z = 0 with principal part n = 1 1 n ! - 1 ( n + 3 ) ! 1 z n
4.2. Classification of singularities 69 and residue Res z = 0 f ( z ) = 1 1! - 1 4! = 23 24 . (c) f ( z ) = sin z z 2010 Solution. Since sin z z 2010 = 1 z 2010 n = 0 ( - 1 ) n z 2 n + 1 ( 2 n + 1 ) ! = n = 0 ( - 1 ) n z 2 n - 2009 ( 2 n + 1 ) ! = 1004 n = 0 ( - 1 ) n z 2 n - 2009 ( 2 n + 1 ) ! + n = 1005 ( - 1 ) n z 2 n - 2009 ( 2 n + 1 ) ! f ( z ) has a pole of order 2009 at z = 0 with principal part 1004 n = 0 ( - 1 ) n z 2 n - 2009 ( 2 n + 1 ) ! and with residue Res z = 0 sin z z 2010 = ( - 1 ) 1004 ( 2 · 1004 + 1 ) ! = 1 2009! . (d) f ( z ) = e z 1 - z 2 Solution. Since 1 - z 2 = ( 1 - z )( 1 + z ) , f ( z ) has poles of order 1 at 1 and - 1. Therefore, Res z = 1 e z 1 - z 2 = e z ( 1 - z 2 ) 0 z = 1 = - e 2 and Res z = - 1 e z 1 - z 2 = e z ( 1 - z 2 ) 0 z = - 1 = 1 2 e . And the principal parts of f ( z ) at z = 1 and z = - 1 are - e 2 ( z - 1 ) and 1 2 e ( z + 1 ) respectively.
70 4. Series (e) f ( z ) = ( 1 - z 2 ) exp 1 z Solution. The function has a singularity at 0 where ( 1 - z 2 ) exp 1 z = ( 1 - z 2 ) n = 0 1 ( n ! ) z n = n = 0 1 ( n ! ) z n - n = 0 1 ( n ! ) z n - 2 = 1 + n = 1 1 ( n ! ) z n - n = 3 1 ( n ! ) z n - 2 - ( z 2 + z + 1 2 ) = - z 2 - z + 1 2 + n = 1 1 ( n ! ) z n - n = 1 1 ( n + 2 ) ! z n = - z 2 - z + 1 2 + n = 1 1 n ! - 1 ( n + 2 ) ! z - n So the principal part is n = 1 1 n ! - 1 ( n + 2 ) ! z - n the function has an essential singularity at 0 and Res z = 0 f ( z ) = 1 1! - 1 3! = 5 6
4.2. Classification of singularities 71 (f) f ( z ) = 1 ( sin z ) 2 Solution. The function has singularities at k π for k Z . At z = k π , we let w = z - k π and then 1 ( sin z ) 2 = 1 ( sin w ) 2 = n = 0 ( - 1 ) n w 2 n + 1 ( 2 n + 1 ) !

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