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dy v x: dx vy
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When evaluated at x0 ; y0 , these represent respectively the slopes of the two curves at this point of intersection. By the Cauchy Riemann equations, ux vy ; uy vx , we have the product of the slopes at the point x0 ; y0 equal to    u v x x 1 uy vy so that any two members of the respective families are orthogonal, and thus the two families are orthogonal. (b) If f z z2 , then u x2 y2 ; v 2xy. The graphs of several members of x2 y2 C1 , 2xy C2 are shown in Fig. 16-4. Fig. 16-4
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16.9. In aerodynamics and uid mechanics, the functions  and in f z  i , where f z is analytic, are called the velocity potential and stream function, respectively. If  x2 4x y2 2y, (a) nd and (b) nd f z .
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(a) By the Cauchy-Riemann equations, 1 Method 1. Integrating (1), Integrating (2), @ @ @ @ ; . @x @y @x @y 2 Then
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@ 2x 4 @y
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@ 2y 2 @x
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2xy 4y F x . 2xy 2x G y . Thus,
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These are identical if F x 2x c; G y 4y c, where c is a real constant. 2xy 4y 2x c.
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Method 2. Integrating (1), 2xy 4y F x . Then substituting in (2), 2y F 0 x 2y 2 or F 0 x 2 and F x 2x c. Hence, 2xy 4y 2x c. a From a ; f z  i x2 4x y2 2y i 2xy 4y 2x c x2 y2 2ixy 4 x iy 2i x iy ic z2 4z 2iz c1 where c1 is a pure imaginary constant.
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CHAP. 16]
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FUNCTIONS OF A COMPLEX VARIABLE
" " z z z z ,y . 2 2i
" This can also be accomplished by nothing that z x iy; z x iy so that x " The result is then obtained by substitution; the terms involving z drop out.
INTEGRALS, CAUCHY S THEOREM, CAUCHY S INTEGRAL FORMULAS 2 4i 16.10. Evaluate
z2 dz
(a) along the parabola x t; y t2 where 1 @ t @ 2, (b) along the straight line joining 1 i and 2 4i, (c) along straight lines from 1 i to 2 i and then to 2 4i.
We have 2 4i
2;4 z2 dz
1;1 2;4
2;4 x iy 2 dx i dy x2 y2 dx 2xy dy i x2 y2 2ixy dx i dy
1;1 2;4
1;1
2xy dx x2 y2 dy
1;1
Method 1. (a) The points 1; 1 and 2; 4 correspond to t 1 and t 2, respectively. Then the above line integrals become 2 2 86 f t2 t4 dt 2 t t2 2t dtg i f2 t t2 dt t2 t4 2t dtg 6i 3 t 1 t 1 (b) The line joining 1; 1 and 2; 4 has the equation y 1 2
4 1 x 1 or y 3x 2. 2 1
Then we nd
2 x 3x 2 2 dx 2x 3x 2 3 dx 2 86 2x 3x 2 dx x2 3x 2 2 3 dx 6i i 3 x 1
From 1 i to 2 i [or 1; 1 to 2; 1 ], y 1; dy 0 and we have 2 2 4 x2 1 dx i 2x dx 3i 3 x 1 x 1 From 2 i to 2 4i [or 2; 1 to 2; 4 ], x 2; dx 0 and we have 4 4 4y dy i 4 y2 dy 30 9i
y 1 y 1
4 3
3i 30 91
86 3
6i.
Method 2. By the methods of 10 it is seen that the line integrals are independent of the path, thus accounting for the same values obtained in (a), (b), and (c) above. In such case the integral can be evaluated directly, as for real variables, as follows: 2 4i
z2 dz
 z3 2 4i 2 4i 3 1 i 3 86  6i 3 3 1 i 3 3
16.11. (a) Prove Cauchy s theorem: f z dz 0.
If f z is analytic inside and on a simple closed curve C, then
FUNCTIONS OF A COMPLEX VARIABLE
[CHAP. 16
P2 (b) Under these conditions prove that
a
f z dz is independent of the path joining P1 and P2 .
v dx u dy
f z dz
u iv dx i dy
u dx v dy i
By Green s theorem ( 10),   @v @u u dx v dy dx dy; @x @y C
v dx u dy

 @u @v dx dy @x @y
where r is the region (simply-connected) bounded by C. @u @v @v @u ; (Problem 16.7), and so the above integrals are zero. Since f z is analytic, @x @y @x @y Then