Page 884 - Advanced_Engineering_Mathematics o'neil

864 Answers to Selected Problems

2
2
2
2
9. w = u + iv = 2(x − y + 4xyi,so u = 2(x − y and v = 4xy. The vertical line x = 0 maps to the negative u-axis.
2
2
2
Other vertical lines x = x 0
= 0 map onto parabolas u = 2x − v /8x . The horizontal line y = 0 maps onto the positive
0 0
2
2
2
u-axis. Other horizontal lines y = y 0
= 0 map onto parabolas u = (v /8y ) − 2y .
0 0
11. The circle of radius 1/2 centered at (0,−1/4)
13. The circle
2 2

1 7 1
u + + v + =
8 8 32
15. the circle
2

19 377
2
(u − 1) + v + =
4 16
(1 + 4i)z − (3 + 8i)
17. w =
(2 + 3i)z − (4 + 7i)
(33 + i)z − (48 + 16i)
19. w =
5(z − 4)
(3 + 22i)z + (4 − 75i)
21. w =
(2 + 3i)z − (21 − 4i)
23. The mapping does not preserve angles. As an example, look at the images of the unit circle and the real axis, and their
point of intersection at z = 1.
25. Suppose T (z 0 ) = (az 0 + b)(cz 0 + d) = z 0 . This yields a quadratic equation for z 0 , having one or two solutions (fixed
points of T ). This argument fails if T (z) = z + a + bi, a translation.
27. Use Theorem 23.4 to explicitly write the bilinear transformation mapping z 2 → 1, z 3 → 0, z 4 →∞.
Section 23.2 Construction of Conformal Mappings
1. w = 2z + 1 − i 3. w = (3z + 2 + 6i)/(z + 2i)
5. w = 4(1 + z)/(1 − z) 7. w = (1/3)z −2/3
9. f (0) = 0 is immediate. Next
1
2
f (1) = 2i (ξ − 1) −1/2 −1/2 dξ
ξ
0
2 −1/2
1 (1 − ξ ) 1
2 −1/2 −1/2
= 2i ξ −1/2 dξ = 2 (1 − ξ ) ξ dξ.
0 i 0
Let ξ = u 1/2 to obtain f (1) = 0 1 (1 − u) −1/2 −3/4 du, and this is B(1/4,1/2),where B is the beta function (see Problem
u
40, Section 15.3). This in turn is (1/4) (1/2)/ (3/4). Call this number c. Similarly evaluate f (−1) = ic and
f (∞) = (1 + i)c.
Section 23.3 Conformal Mappings and Solutions of the Dirichlet Problem
y ∞ g(t)
1. u(x, y) = dt for y > 0
2
π −∞ (x − t) + y 2
3.
1 2π
u(x, y) = g(x 0 + R cos(t), y 0 + R sin(t))K(x, y,t)dt
2π 0
where
2
2
R − (x − x 0 ) − (y − y 0 ) 2
K(x, y,t) =
2
2
2
R + (x − x 0 ) + (y − y 0 ) − 2R(x − x 0 )cos(t) − 2R(y − y 0 )sin(t)
1 2π r(cos(ξ) − sin(ξ))(1 −r )
2
5. u(r cos(θ),r sin(θ)) = dξ
2
2π 0 1 +r − 2r cos(ξ − θ)
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October 14, 2010 17:50 THM/NEIL Page-864 27410_25_Ans_p801-866