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where C is the contour of Problem 16.31. The integrand has a pole of order 3 at z 0 and a simple pole z 1 within C. 2 ( ) 2 6 1 d z 1 21 Residue at z 0 is lim z3 3 : z!0 2! dz2 8 z 2z 1 z 2 ( Residue at z
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& ' z6 1 1 21 65  as required. dz 2i 2i 8 24 12 z3 2z 1 z 2 C
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16.33. If j f z j @
M for z Rei , where k > 0 and M are constants, prove that Rk lim eimz f z dz 0
R!1
where is the semicircular arc of the contour in Problem 16.27 and m is a positive constant.
If z Rei ; eimz f z dz
eimRe f Rei iRei d:
Then
       i i  eimRe f Rei iRei d @  eimRe f Rei iRei  d     0 0      eimR cos  mR sin  f Rei iRei  d 0  e mR sin  j f Rei j R d
M Rk 1
e mR sin  d
2M Rk 1
=2
e mr sin  d
Now sin  A 2= for 0 @  @ =2 (see Problem 4.73, 4). Then the last integral is less than or equal to 2M =2 2mR= M e d 1 e mR k 1 R mRk 0 As R ! 1 this approaches zero, since m and k are positive, and the required result is proved.
CHAP. 16]
FUNCTIONS OF A COMPLEX VARIABLE
1 16.34. Show that
eimz dz where C is the contour of Problem 16.27. z2 1 C The integrand has simple poles at z i, but only z i lies within C. & ' eimz e m : Residue at z i is lim z i z!i z i z i 2i Consider Then R or R i.e., and so R 2
cos mx  dx e m ; m > 0. 2 x2 1
 m  eimz e dz 2i e m 2 2i Cz 1
eimx dx 2 R x 1 R
eimz dz e m 1 eimz dz e m z2 1
cos mx dx i 2 R x 1
sin mx dx 2 R x 1
cos mx dx x2 1
eimz dz e m z2 1
Taking the limit as R ! 1 and using Problem 16.33 to show that the integral around approaches zero, we obtain the required result.
1 16.35. Show that
sin x  dx . x 2
The method of Problem 16.34 leads us to consider the integral of eiz =z around the contour of Problem 16.27. However, since z 0 lies on this path of integration and since we cannot integrate through a singularity, we modify that contour by indenting the path at z 0, as shown in Fig. 16-11, which we call contour C 0 or ABDEFGHJA.
Fig. 16-11 Since z 0 is outside C 0 , we have eiz dz 0 C0 z R
or r eix dx R x
eiz dz z
eix dx x
BDEFG
eiz dz 0 z
FUNCTIONS OF A COMPLEX VARIABLE
[CHAP. 16
Replacing x by x in the rst integral and combining with the third integral, we nd, iz R ix e e ix e eiz dx dz dz 0 x z z r
HJA BDEFG
or R 2i
sin x dx x
eiz dz z
BDEFG
eix dz z
Let r ! 0 and R ! 1. By Problem 16.33, the second integral on the right approaches zero. The rst integral on the right approaches lim 0 eire irei d lim i r!0  re
ieire d i
since the limit can be taken under the integral sign. Then we have R sin x dx i lim 2i R!1 x r r!0
1 or
sin x  dx x 2
MISCELLANEOUS PROBLEMS 16.36. Let w z2 de ne a transformation from the z plane (xy plane) to the w plane uv plane). Consider a triangle in the z plane with vertices at A 2; 1 ; B 4; 1 ; C 4; 3 . (a) Show that the image or mapping of this triangle is a curvilinear triangle in the uv plane. (b) Find the angles of this curvilinear triangle and compare with those of the original triangle.
(a) Since w z2 , we have u x2 y2 ; v 2xy as the transformation equations. Then point A 2; 1 in the xy plane maps into point A 0 3; 4 of the uv plane (see gures below). Similarly, points B and C map into points B 0 and C 0 respectively. The line segments AC; BC; AB of triangle ABC map respectively into parabolic segments A 0 C 0 ; B 0 C 0 ; A 0 B 0 of curvilinear triangle A 0 B 0 C 0 with equations as shown in Figures 16-12(a) and (b).
Fig. 16-12
CHAP. 16]
FUNCTIONS OF A COMPLEX VARIABLE   dv  2 1   . du  3;4 v  3;4 2  dv  u 3. The slope of the tangent to the curve u2 2v 1 at 3; 4 is m2  du  3;4 0 Then the angle  between the two curves at A is given by tan  3 1 m2 m1 2 1; and  =4 1 m1 m2 1 3 1 2
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