Chapter 20

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 .I f 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 . It follows that u 2 + v 2 = 1 x 2 = 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. w = z 2 , u = x 2 y 2 and v =2 xy f 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. w =Ln z , u = log e | z | and v = Arg z . The semi-circle | z | , y> 0 may also be described by r , 0 <θ<π . Therefore u = 0 and 0 <v<π . The image is therefore the open line segment from z =0to z = πi . 6. If θ = π/ 4, then v = θ = 4. In addition u = log e r will vary from −∞ to . The image is therefore the horizontal line v = 4. 7. w = z 1 / 2 re ) 1 / 2 = r 1 / 2 e iθ/ 2 and θ = θ 0 , w = 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 4. This image is a circular arc. 9. w = e z , u = e x cos y and v = e x sin y . Therefore if e x cos y , 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 t and v = 0 and so the image is the closed interval [ 2 , 2] on the u -axis. 11. The ±rst quadrant may be described by r> 0, 0 <θ<π/ 2. If w /z and z = , w = 1 r e . Therefore Arg w = θ and so 2 < Arg w< 0. The image is therefore the fourth quadrant. 12. 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 1is 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 4 y 2, 4 Arg w 2 and | w | = e x . The image is therefore the angular wedge de±ned by 4 Arg w 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 de±ne 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 | 821
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Exercises 20.1 16. If w =2 z 1 and | z | = 1, then, since z = w +1 2 , ¯ ¯ ¯ ¯ w 2 ¯ ¯ ¯ ¯ =1or | w | = 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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This note was uploaded on 02/13/2008 for the course ENG 342 taught by Professor Delahanty during the Fall '07 term at TCNJ.

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Chapter 20 - 20 Conformal Mappings and Applications...

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