Page 431 - Schaum's Outline of Theory and Problems of Advanced Calculus

422 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16

2
x sin x e sin x
ð ð 2
1 1
16.82. 16.84.
2 2 dx ¼ 4 x 2 dx ¼ 2
0 ðx þ 1Þ 0
ð ð 3
1 sin x 1 sin x 3
16.83. ð2e 3Þ 16.85.
2 2 dx ¼ 4e x 3 dx ¼ 8
0 xðx þ 1Þ 0
ð þ iz
cos x e
1
16.86. dx ¼ . Hint: Consider dz, where C is a rectangle with vertices at ð R; 0Þ,
0 cosh x 2coshð =2Þ C cosh z

ðR; 0Þ; ðR; Þ ð R; Þ. Then let R !1:
r
MISCELLANEOUS PROBLEMS
i
16.87. If z ¼ e and f ðzÞ¼ uð ; Þþ ivð ; Þ, where and are polar coordinates, show that the Cauchy-Rie-
mann equations are
@u 1 @v @v 1 @u
;
@ ¼ @ @ ¼ @
16.88. If w ¼ f ðzÞ, where f ðzÞ is analytic, defines a transformation from the z plane to the w plane where z ¼ x þ iy
and w ¼ u þ iv,prove that the Jacobian of the transformation is given by
@ðu; vÞ 2
0
¼j f ðzÞj
@ðx; yÞ
2
2
@ F @ F
16.89. Let Fðx; yÞ be transformed to Gðu; vÞ by the transformation w ¼ f ðzÞ. Show that if þ ¼ 0, then at
@x 2 @y 2
2
2
@ G @ G
all points where f ðzÞ 6¼ 0, þ ¼ 0.
0
@u 2 @v 2
az þ b
, where ad bc 6¼ 0, circles in the z plane are trans-
16.90. Show that by the bilinear transformation w ¼ cz þ d
formed into circles of the w plane.
16.91. If f ðzÞ is analytic inside and on the circle jz aj¼ R,prove Cauchy’s inequality, namely
n!M
j f ðnÞ ðaÞj @
R n
where j f ðzÞj @ M on the circle. [Hint: Use Cauchy’s integral formulas.]
16.92. Let C 1 and C 2 be concentric circles having center a and radii r 1 and r 2 , respectively, where r 1 < r 2 .If a þ h is
any point in the annular region bounded by C 1 and C 2 , and f ðzÞ is analytic in this region, prove Laurent’s
theorem that
1
X n
a n h
f ða þ hÞ¼
1
1 þ f ðzÞ dz
where
nþ1
a n ¼
2 i C ðz aÞ
C being any closed curve in the angular region surrounding C 1 .
þ þ
1 f ðzÞ dz 1 f ðzÞ dz 1
Hint: Write f ða þ hÞ¼ and expand in two different
z a h
2 i C 2 z ða þ hÞ 2 i C 1 z ða þ hÞ

ways.
z
which converges for 1 < jzj < 2 and
16.93. Find a Laurent series expansion for the function f ðzÞ¼
diverges elsewhere. ðz þ 1Þðz þ 2Þ

z 1 2 1 1
Hint: Write ¼ þ ¼ þ :
z þ 1 z þ 2 1 þ z=2
ðz þ 1Þðz þ 2Þ zð1 þ 1=zÞ