Page 421 - Schaum's Outline of Theory and Problems of Advanced Calculus

412 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16

(b)Since jzj¼ 10 encloses both poles z ¼ 1 and z ¼ 3
!
e 5e 3 3
the required integral ¼ 2 i ¼ iðe 5e Þ
16 16 8

EVALUATION OF DEFINITE INTEGRALS
M i ð
16.27. If j f ðzÞj @ for z ¼ Re , where k > 1 and M are constants, prove that lim f ðzÞ dz ¼ 0
R k R!1
where is the semicircular arc of radius R shown in Fig. 16-9.
By the result (4), Page 394, we have
ð ð
M M

f ðzÞ dz @ j f ðzÞjjdzj @
R k R þ R k 1

since the length of arc L ¼ R. Then
ð ð

lim f ðzÞ dz ¼ 0 and so lim f ðzÞ dz ¼ 0

R!1 R!1
Fig. 16-9
M
i
16.28. Show that for z ¼ Re , j f ðzÞj @ ; k > 1 if
1 R k
.
f ðzÞ¼ 4
1 þ z

i 1 1 1 2
@ @ if R is large enough (say R > 2, for
If z ¼ Re , j f ðzÞj ¼ 4 4i ¼ 4 4
1 þ R e jR e j 1 R 1 R
4 4i
example) so that M ¼ 2; k ¼ 4.
4 4i
Note that we have made use of the inequality jz 1 þ z 2 j A jz 1 j jz 2 j with z 1 ¼ R e and z 2 ¼ 1.
ð
dx
1
16.29. Evaluate 4 .
0 x þ 1
þ
dz
Consider , where C is the closed contour of Problem 16.27 consisting of the line from R to R
4
C z þ 1
and the semicircle , traversed in the positive (counterclockwise) sense.
4
4
Since z þ 1 ¼ 0 when z ¼ e i=4 ; e 3 i=4 ; e 5 i=4 ; e 7 i=4 ,these are simple poles of 1=ðz þ 1Þ. Only the poles
e i=4 and e 3 i=4 lie within C. Then using L’Hospital’s rule,
1
Residue at e i=4 ¼ lim ðz e i=4 Þ
4
z!e i=4 z þ 1
1 1 3 i=4
¼ lim 3 ¼ e
z!e i=4 4z 4
1
Residue at e 3 i=4 ¼ lim ðz e 3 i=4 Þ
4
z!e 3 i=4 z þ 1
1 1 9 i=4
¼ lim 3 ¼ e
z!e 3 i=4 4z 4
Thus
p ffiffiffi
þ
dz 1 3 i=4 1 9 i=4 2
¼ 2 i 4 e þ e ¼ ð1Þ
4
4
C z þ 1 2