FUNCTIONS OF A COMPLEX VARIABLE in VS .NET

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FUNCTIONS OF A COMPLEX VARIABLE
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By Cauchy s integral formula (Problem 16.15), we have 1 f z dz f a h 2i C z a h By division 1 1 z a h z a 1 h= z a ( ) 1 h h2 hn hn 1 1 z a n z a n z a h z a z a z a 2 Substituting (2) in (1) and using Cauchy s integral formulas, we have 1 f z dz h f z dz hn f z dz Rn f a h 2i C z a 2i C z a 2 2i C z a n 1 f a h f 0 a h2 00 hn f a f n a Rn 2! n! f z dz z a n 1 z a h C
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   f z   Now when z is on C,  z a h @ M and jz aj R, so that by (4), Page 394, we have, since 2R is the length of C jRn j @ jhjn 1 M 2R 2Rn 1
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As n ! 1; jRn j ! 0. Then Rn ! 0 and the required result follows. If f z is analytic in an annular region r1 @ jz aj @ r2 , we can generalize the Taylor series to a Laurent series (see Problem 16.92). In some cases, as shown in Problem 16.22, the Laurent series can be obtained by use of known Taylor series.
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16.22. Find Laurent series about the indicated singularity for each of the following functions. Name the singularity in each case and give the region of convergence of each series.
a ez ; z 1: z 1 2 Let z 1 u: Then z 1 u and
( ) ez e1 u eu e u2 u3 u4 2 e 2 2 1 u 2! 3! 4! u u u z 1 2 e e e e z 1 e z 1 2 2 z 1 2! 3! 4! z 1
z 1 is a pole of order 2, or double pole. The series converges for all values of z 6 1. b z cos 1 ; z 0: z z cos
  1 1 1 1 1 1 1 z z 1 2 4 6 z 2! z 4! z3 6! z5 2! z 4! z 6! z
z 0 is an essential singularity. The series converges for all values of z 6 0.
CHAP. 16]
FUNCTIONS OF A COMPLEX VARIABLE
sin z ; z : z 
Let z  u:
Then z  u and
! sin z sin u  sin u 1 u3 u5 u 3! 5! z  u u u 1 u2 u4 z  2 z  4 1 3! 5! 3! 5!
z  is a removable singularity. The series converges for all values of z. d z ; z 1: z 1 z 2 Let z 1 u. Then
z u 1 u 1 1 u u2 u3 u4 z 1 z 2 u u 1 u 1 2 2u 2u2 2u3 u 1 2 2 z 1 2 z 1 2 z 1 z 1 is a pole of order 1, or simple pole. The series converges for values of z such that 0 < jz 1j < 1. e 1 ; z 0; 2: z z 2 3 Case 1, z 0. Using the binomial theorem, & '  z  3 4  z 2 3 4 5  z 3 1 1 1 1 3 2 2! 2 3! 2 z z 2 3 8z 1 z=2 3 8z 1 3 3 5 2 z z 8z 16 16 32
z 0 is a pole of order 1, or simple pole. The series converges for 0 < jzj < 2. Case 2, z 2. & ' 1 1 1 1 u u2 u3 u4 3 1 2 2 2 2 2u z z 2 3 u 2 u3 2u3 1 u=2 1 1 1 1 1 3 2 u 8u 16 32 2u 4u 1 1 1 1 1 z 2 2 z 2 3 4 z 2 2 8 z 2 16 32 Let z 2 u. Then
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