# For each singularity determine whether it is a pole a

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for each singularity, determine whether it is a pole, a removable singularity, 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 . 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.
4.2 Classification of singularities 71 (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 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! .
72 Chapter 4. Series (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. (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 73 (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 ) ! ! - 1 = 1 w 2 n = 0 ( - 1 ) n w 2 n ( 2 n + 1 ) ! ! - 1 = 1 w 2 1 - n = 1 ( - 1 ) n + 1 w 2 n ( 2 n + 1 ) ! ! - 1 = 1 w 2 m = 0 n = 1 ( - 1 ) n + 1 w 2 n ( 2 n + 1 ) ! ! m = 1 w 2 1 + n = 2 a n w n ! So the principal part at k π is 1 ( z - k π ) 2 the function has a pole of order 2 at k π and Res z = k π f ( z ) = 0 (g) f ( z ) = 1 - cos z z 2 Solution. The function has a singularity at 0 where 1 - cos z z 2 = 1 z 2 1 - n = 0 ( - 1 ) n z 2 n ( 2 n ) ! ! = 1 z 2 n = 1 ( - 1 ) n + 1 z 2 n ( 2 n ) ! = n = 1 ( - 1 ) n + 1 z 2 n - 2 ( 2 n ) ! So the principal part is 0, the function has a removable singularity at 0 and Res z = 0 f ( z ) = 0
74 Chapter 4. Series (h) f ( z ) = e z z ( z - 1 ) 2 Solution. The function has two singularities at 0 and 1. At z = 0, e z z ( z - 1 ) 2 = 1 z n = 0 z n n !
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