Page 423 - Schaum's Outline of Theory and Problems of Advanced Calculus

414 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16

i 2 2 1
Residue at i=3 ¼ lim z þ ¼ lim ¼ by L’Hospital’s rule.
2
z! i=2 3 3z þ 10iz 3 z! i=2 6z þ 10i 4i

2 dz 1
þ
Then 2 ¼ 2 i ¼ ,the required value.
C 3z þ 10iz 3 4i 2
ð 2 cos 3
16.32. Show that d ¼ .
0 5 4 cos 12
3
i z þ z 1 e 3i þ e 3i z þ z 3
2 2 2
If z ¼ e ,cos ¼ ; cos 3 ¼ ¼ ; dz ¼ iz d .
3
3
ð 2 cos 3 þ ðz þ z Þ=2 dz
Then d ¼ !
0 5 4cos C z þ z 1 iz
5 4
2
6
1 þ z þ 1
dz
3
¼
2i C z ð2z 1Þðz 2Þ
where C is the contour of Problem 16.31.
1
The integrand has a pole of order 3 at z ¼ 0 and a simple pole z ¼ within C.
2
( )
6
1 d 2 3 z þ 1 21
Residue at z ¼ 0is lim z ¼ :
3
z!0 2! dz 2 z ð2z 1Þðz 2Þ 8
( 6 )
z þ 1 65
1 is lim 1 :
2 2 3
Residue at z ¼ ðz Þ ¼
z!1=2 z ð2z 1Þðz 2Þ 24
6
1 þ z þ 1 1 21 65
as required.
Then 3 dz ¼ ð2 iÞ ¼
2i 2i 8 24 12
C z ð2z 1Þðz 2Þ
M i
16.33. If j f ðzÞj @ for z ¼ Re , where k > 0 and M are constants, prove that
R k
ð
lim e imz f ðzÞ dz ¼ 0
R!1
where is the semicircular arc of the contour in Problem 16.27 and m is a positive constant.
ð ð i
i
i
i
If z ¼ Re ; e imz f ðzÞ dz ¼ e imRe f ðRe Þ iRe d :
0
ð ð
imRe i i i imRe i i
Then e f ðRe Þ iRe d @ f ðRe Þ iRe d
i

e
0 0
ð

imR cos mR sin i
i
f ðRe Þ iRe d
¼ e
0
ð
mR sin i
e j f ðRe Þj Rd
¼
0
M ð mR sin 2M ð =2 mr sin
@ e d ¼ e d
R k 1 0 R k 1 0
Now sin A 2 = for 0 @ @ =2 (see Problem 4.73, Chapter 4). Then the last integral is less than or
equal to
2M ð =2 2mR = M
e d ¼ ð1 e mR Þ
R k 1 0 mR k
As R !1 this approaches zero, since m and k are positive, and the required result is proved.