Chapter 20 - 20 Conformal Mappings and Applications...

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20 Conformal Mappings and Applications Exercises 20.1 1. For w = 1 z , u = x x 2 + y 2 and v = y x 2 + y 2 . If y = x , u = 1 2 1 x , v = 1 2 1 x , and so v = u . The image is the line v = u (with the origin (0 , 0) excluded.) 2. If y = 1, u = x x 2 + 1 and v = 1 x 2 + 1 . It follows that u 2 + v 2 = 1 x 2 + 1 = v and so u 2 +( v + 1 2 ) 2 = ( 1 2 ) 2 . This is a circle with radius r = 1 2 and center at (0 , 1 2 ) = 1 2 i . The circle can also be described by | w + 1 2 i | = 1 2 . 3. For w = z 2 , u = x 2 y 2 and v = 2 xy . If xy = 1, v = 2 and so the hyperbola xy = 1 is mapped onto the line v = 2. 4. If x 2 y 2 = 4, u = 4 and so the hyperbola x 2 y 2 = 4 is mapped onto the vertical line u = 4. 5. For w = Ln z , u = log e | z | and v = Arg z . The semi-circle | z | = 1, y > 0 may also be described by r = 1, 0 < θ < π . Therefore u = 0 and 0 < v < π . The image is therefore the open line segment from z = 0 to z = πi . 6. If θ = π/ 4, then v = θ = π/ 4. In addition u = log e r will vary from −∞ to . The image is therefore the horizontal line v = π/ 4. 7. For w = z 1 / 2 = ( re ) 1 / 2 = r 1 / 2 e iθ/ 2 and θ = θ 0 , w = r e 0 / 2 . Therefore Arg w = θ 0 / 2 and so the image is the ray θ = θ 0 / 2. 8. If r = 2 and 0 θ π 2 , w = 2 e iθ/ 2 . Therefore | w | = 2 and 0 Arg w π/ 4. This image is a circular arc. 9. For w = e z , u = e x cos y and v = e x sin y . Therefore if e x cos y = 1, u = 1. The curve e x cos y = 1 is mapped into the line u = 1. Since v = sin y cos y = tan y , v varies from −∞ to and the image is the line u = 1. 10. If w = z + 1 z and z = e it , w = e it e it = 2cos t . Therefore u = 2cos t and v = 0 and so the image is the closed interval [ 2 , 2] on the u -axis. 11. The first quadrant may be described by r > 0, 0 < θ < π/ 2. If w = 1 /z and z = re , w = 1 r e . Therefore Arg w = θ and so π/ 2 < Arg w < 0. The image is therefore the fourth quadrant. 12. For w = 1 z , u = x x 2 + y 2 and v = y x 2 + y 2 . The line y = 0 is mapped to the line v = 0, and, from Problem 2, the line y = 1 is mapped onto the circle | w + 1 2 i | = 1 2 . Since f ( 1 2 i ) = 2 i , the region 0 y 1 is mapped onto the points in the half-plane v 0 which are on or outside the circle | w + 1 2 i | = 1 2 . (The image does not include the point w = 0.) 13. Since w = e x + iy = e x e iy and π/ 4 y π/ 2, π/ 4 Arg w π/ 2 and | w | = e x . The image is therefore the angular wedge defined by π/ 4 Arg w π/ 2. 14. Since w = e x + iy = e x e iy and 0 x 1, 0 y π , we have | w | = e x and Arg w = y . Therefore 1 ≤ | w | ≤ e and 0 Arg w π . These inequalities define a semi-angular region in the w -plane. 15. The mapping w = z +4 i is a translation which maps the circle | z | = 1 to a circle of radius r = 1 and with center w = 4 i . This circle may be described by | w 4 i | = 1. 821
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Exercises 20.1 16. If w = 2 z 1 and | z | = 1, then, since z = w + 1 2 , w + 1 2 = 1 or | w + 1 | = 2. The image is a circle with center at w = 1 and with radius r = 2. 17. The mapping w = iz is a rotation through 90 since i = e iπ/ 2 . Therefore the strip 0 y 1 is rotated through 90 and so the strip 1 u 0 is the image in the w -plane.
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  • Fall '07
  • Delahanty
  • Boundary value problem, Line segment, arg, Upper half-plane

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