Page 422 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 413

i.e.,
R dx dz 2
ð ð p ffiffiffi
4
4
R x þ 1 þ z þ 1 ¼ 2 ð2Þ
Taking the limit of both sides of (2)as R !1 and using the results of Problem 16.28, we have
R dx 1 dx 2
ð ð p ffiffiffi
lim ¼ ¼
4
4
R x þ 1 1 x þ 1 2
R!1
p ffiffiffi
ð ð
1 dx 1 dx 2
Since ¼ 2 ; the required integral has the value :
4
4
1 x þ 1 0 x þ 1 4
ð 2
x dx 7
1
16.30. Show that :
2 2 2 ¼ 50
1 ðx þ 1Þ ðx þ 2x þ 2Þ
z 2
The poles of enclosed by the contour C of Problem 16.27 are z ¼ i of order 2 and
2 2 2
ðz þ 1Þ ðz þ 2z þ 2Þ
z ¼ 1 þ i of order 1.
( )
d z 2 9i 12
Residue at z ¼ i is lim 2 :
z!i dz ðz iÞ 2 2 2 ¼ 100
ðz þ iÞ ðz iÞ ðz þ 2z þ 2Þ
z 2 3 4i
Residue at z ¼ 1 þ i is lim ðz þ 1 iÞ ¼
z! 1þi 2 2 25
ðz þ 1Þ ðz þ 1 iÞðz þ 1 þ iÞ
z dz 9i 12 3 4i 7
þ 2
Then ¼ 2 i
2 2 2 100 þ 25 ¼ 50
C ðz þ 1Þ ðz þ 2z þ 2Þ
2
2
ð R x dx ð z dz 7
or
2 2 2 þ 2 2 2 ¼ 50
R ðx þ 1Þ ðx þ 2x þ 2Þ ðz þ 1Þ ðz þ 2z þ 2Þ
Taking the limit as R !1 and noting that the second integral approaches zero by Problem 16.27, we
obtain the required result.
ð 2 d
16.31. Evaluate .
0 5 þ 3 sin
i i 1
e e z z
i
i
Let z ¼ e . Then sin ¼ ¼ , dz ¼ ie d ¼ iz d so that
2i 2i
ð 2 d þ dz=iz þ 2 dz
2
0 5 þ 3 sin ¼ C z z 1 ! ¼ C 3z þ 10iz 3
5 þ 3
2i
where C is the circle of unit radius with center at the origin, as shown in Fig. 16-10 below.
2
The poles of are the simple poles
2
3z þ 10iz 3
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
100 þ 36
10i
6
z ¼
10i 8i
6
¼
¼ 3i; i=3:
Only i=3 lies inside C. Fig. 16-10