Page 758 - Advanced_Engineering_Mathematics o'neil

738 CHAPTER 22 Singularities and the Residue Theorem

Case 5 If γ encloses 0 and i but not −i,
1

f (z)dz = 2πi −1 + 1 + cos(i) = πi cos(i).
2
γ
Case 6 If γ encloses 0 and −i but not i,

1 1
f (z)dz = 2πi −1 − 1 + cos(i) = 2πi −2 + cos(i) .
γ 2 2
Case 7 If γ encloses i and −i but not 0,
1 1

f (z)dz = 2πi 1 + cos(i) − 1 + cos(i) = 2πi cos(i).
2 2
γ
Case 8 If γ encloses all three singularities,
1 1

f (z)dz = 2πi −1 + 1 + cos(i) − 1 + cos(i) = 2πi(−1 + cos(i)).
2 2
γ
EXAMPLE 22.15
Let
sin(z)
f (z) = .
2
z (z + 4)
2
f has a simple poles at 0 and ±2i. Suppose γ is a closed path enclosing 0 and 2i but not −2i.
Compute the residues of f at 0 and 2i. In doing this, the corollary does not apply for the
residue of f at 0 because sin(0) = 0. We can, if we wish, use the corollary for the residue at 2i.
Compute
sin(z) 1 1
Res( f,0) = lim zf (z) = lim =
2
z→0 z→0 z z + 4 4
and
sin(z)
Res( f,2i) = lim
z→2i z (z − 2i)(z + 2i)
2
sin(2i) i
= = sin(2i).
(−4)(4i) 16
Then

1 i

f (z)dz = 2πi + sin(2i) .
4 16
γ
EXAMPLE 22.16
1/z
We will evaluate e dz where γ is a closed path enclosing the origin.
γ
The Laurent expansion of e 1/z about 0 is
∞
1 1
1/z
e = ,
n! z n
n=0
so 0 is an essential singularity. There is no simple formula for residues at essential singularities.
Because we have the Laurent expansion of e 1/z about 0, we can read that the coefficient of 1/z is 1,
1/z
so Res( f,0) = 1 and e dz = 2πi.
γ

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

October 14, 2010 15:37 THM/NEIL Page-738 27410_22_ch22_p729-750