z 2 is a pole of order 3. The series converges for 0 < jz 2j < 2. in .NET framework

Drawing QR Code JIS X 0510 in .NET framework z 2 is a pole of order 3. The series converges for 0 < jz 2j < 2.

z 2 is a pole of order 3. The series converges for 0 < jz 2j < 2.
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RESIDUES AND THE RESIDUE THEOREM 16.23. Suppose f z is analytic everywhere inside and on a simple closed curve C except at z a which is a pole of order n. Then a n a n 1 f z a0 a1 z a a2 z a 2 z a n z a n 1 where a n 6 0. Prove that
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FUNCTIONS OF A COMPLEX VARIABLE
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[CHAP. 16
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f z dz 2ia 1
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b a 1 lim
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1 d n 1 f z a n f z g: n 1 ! dzn 1
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(a) By integration, we have on using Problem 16.13 a n a 1 dz fa0 a1 z a a2 z a 2 g dz f z dz n dz C C z a Cz a C 2ia 1 Since only the term involving a 1 remains, we call a 1 the residue of f z at the pole z a. (b) Multiplication by z a n gives the Taylor series z a n f z a n a n 1 z a a 1 z a n 1 Taking the n 1 st derivative of both sides and letting z ! a, we nd n 1 !a 1 lim from which the required result follows. d n 1 f z a n f z g dzn 1
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16.24. Determine the residues of each function at the indicated poles.
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a z2 ; z 2; i; i: z 2 z2 1 Residue at z 2 is Residue at z i is Residue at z i is These are simple poles. Then: ) z2 4 : z!2 5 z 2 z2 1 ( ) 2 z i2 1 2i lim z i : z!i 10 z 2 z i z i i 2 2i ( ) z2 i2 1 2i : lim z i z! i 10 z 2 z i z i i 2 2i lim z 2 (
1 ; z 0; 2: z z 2 3
z 0 is a simple pole, z 2 is a pole of order 3.
Then:
Residue at z 0 is Residue at z 2 is
1 1 : z z 2 3 8 & ' 1 d2 1 z 2 3 lim z! 2 2! dz2 z z 2 3   2   1 d 1 1 2 1 lim lim : z! 2 2 dz2 z z! 2 2 z3 8
lim z
Note that these residues can also be obtained from the coe cients of 1=z and 1= z 2 in the respective Laurent series [see Problem 16.22(e)]. c zezt ; z 3; a pole of order 2 or double pole. Then: z 3 2 Residue is lim & ' d zezt d z 3 2 lim zezt lim ezt ztezt z!3 dz z!3 dz z 3 2 e3t 3te3t
CHAP. 16]
FUNCTIONS OF A COMPLEX VARIABLE
(d) cot z; z 5, a pole of order 1. Residue is lim z 5
Then: cos z sin z 
z!5
z!5
    z 5 1 lim cos z lim 1 z!5 z!5 cos z sin z
1 1 1 where we have used L Hospital s rule, which can be shown applicable for functions of a complex variable.
16.25. If f z is analytic within and on a simple closed curve C except at a number of poles a; b; c; . . . interior to C, prove that f z dz 2i fsum of residues of f z at poles a; b; c; etc.g
Refer to Fig. 16-8.
By reasoning similar to that of Problem 16.12 (i.e., by constructing cross cuts from C to C1 ; C2 ; C3 ; etc.), we have f z dz f z dz f z dz
C C1 C2
For pole a, a m a 1 a0 a1 z a f z z a m z a hence, as in Problem 16.23,
Fig. 16-8
f z dz 2i a 1 : b n b 1 b0 b1 z b z b n z b f z dz 2i b 1
Similarly for pole b; f z
so that
Continuing in this manner, we see that f z dz 2i a 1 b 1 2i (sum of residues)
16.26. Evaluate
ez dz where C is given by (a) jzj 3=2; 2 C z 1 z 3
' ez e 16 z 1 z 3 2
b jzj 10.
& Residue at simple pole z 1 is lim z 1
Residue at double pole z 3 is & ' d ez z 1 ez ez 5e 3 z 3 2 lim lim 2 z! 3 dz z! 3 16 z 1 z 3 z 1 2 (a) Since jzj 3=2 encloses only the pole z 1, the required integral 2i e 16 ie 8
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