AEM_3e_Chapter_20

# AEM_3e_Chapter_20 - 20 20 Conformal Mappings EXERCISES 20.1...

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20 20 Conformal Mappings EXERCISES 20.1 Complex Functions as Mappings 3. For w = z 2 , u = x 2 y 2 and v =2 xy .I f xy =1 , v = 2 and so the hyperbola xy = 1 is mapped onto the line v =2. 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. 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. 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.) 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 | 18. Since w =(1+ i ) z = 2 e iπ/ 4 z , the mapping is the composite of a rotation through 45 and a magni±cation by α = 2. The image of the ±rst quadrant is therefore the angular wedge 4 Arg w 3 4. 21. We ±rst let z 1 = z i to map the region 1 y 4 to the region 0 y 1 3. We then let w = e iπ/ 2 z 1 to rotate this strip through 90 . Therefore w = i ( z i iz 1 maps 1 y 4 to the strip 0 u 3. 24. The mapping w = iz will rotate the strip 1 x 1 through 90 so that the strip 1 v 1 results. 27. By Example 1, Section 20 . 1, z 1 = e z maps the strip 0 y π onto the upper half-plane y 1 0, or 0 Arg z 1 π . The power function w = z 3 / 2 1 changes the opening of this wedge by a factor of 3 / 2 so the wedge 0 Arg w 3 2 results. The composite of these two mappings is w =( e z ) 3 / 2 = e 3 z/ 2 . 30. The mapping z 1 = ( z πi ) lowers R by π units in the vertical direction and then rotates the resulting region through 180 . The image region R 1 is upper half-plane y 1 0. By Example 1, Section 20 . 1, w =Ln z 1 maps R 1 onto the strip 0 v π . The composite of these two mappings is w = Ln( z ). 301

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EXERCISES 20.2 Conformal Mappings 20.2 Conformal Mappings 3. f 0 ( z )=1+ e z and 1 + e z = 0 for z = ± i ± 2 nπi . Therefore f is conformal except for z = πi ± 2 nπi . 6. The function f ( z )= 1 2 [Ln( z +1)+Ln( z 1)] is analytic except on the branch cut x 1 0or x 1, and f 0 ( z 1 2 µ 1 z +1 + 1 z 1 = z z 2 1 is non-zero for z 6 =0, ± 1. Therefore f is conformal except for z = x , x 1. 9. f ( z ) = (sin z ) 1 / 4 is the composite of z 1 = sin z and w = z 1 / 4 1 . The region π/ 2 x 2, y 0 is mapped to the upper half-plane y 2 0by z 1 = sin z (See Example 2) and the power function w = z 1 / 4 1 maps this upper half-plane to the angular wedge 0 Arg w 4. The real interval [ 2 ,π/ 2] is Frst mapped to [ 1 , 1] and then to the union of the line segments from e i π 4 to 0 and 0 to 1. See the Fgures below.
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## This note was uploaded on 01/03/2011 for the course BIS 511 at Yale.

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AEM_3e_Chapter_20 - 20 20 Conformal Mappings EXERCISES 20.1...

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