s8_95-100aSOLUTION

s8_95-100aSOLUTION - ACM95a/100a Last updated December 2,...

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Unformatted text preview: ACM95a/100a Last updated December 2, 2009 Problem Set VIII Solutions Prepared by Faisal Amlani 1. (20 pts.) Find a Mobius transformation mapping the lower half-plane to the disk | w + 1 | < 1. [Hint: Do it in steps.] SOLUTION : The conformal transformation mapping one region to another is never unique. One possible mapping begins with w 1 = f ( z ) = z + i z- i . Checking the image of the boundary of the lower half plane, we see that w 1 (- 1) =- i, w 1 (0) =- 1 , w 1 (1) = i. w 1 maps the line through- 1 , , 1 through a line or circle connecting- i,- 1 , i. Since a line cannot connect these three points, the map can only be a circle = w 1 maps the real axis boundary to the unit circle centered at the origin. Since the function must map the lower half plane either to the interior or exterior of this circle, and since w 1 (- i ) = 0 , we must have that w 1 maps the lower half plane to the interior. Now to re-center at z =- 1 , we merely translate with w = g ( w 1 ) = w 1- 1 . Putting it together gives the map w = g ( w 1 ) = w 1- 1 = f ( z )- 1 = z + i z- i- 1 = 2 i z- i . 2. (20 pts.) Map the region between the two cirles | z | = 2 and | z- 1 | = 1 conformally onto the upper half plane. [Hint: Use a Mobius transformation to map the point 2 to . Show that the image region is a strip . Then use the exponential map, after modifying (translating, rotating) the strip, if necessary.] SOLUTION : Consider a map w 1 = f ( z ) = 1 2- z . This clearly maps the point z = 2 to . To show that the image is a strip, we can check where key points on each circle are mapped: 2 1 / 2 1 + i 1 2 (1 + i ) 1- i 1 2 (1- i )- 2 1 / 4 2 i 1 4 (1 + i )- 2 i 1 4 (1- i ) ....
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This note was uploaded on 01/14/2010 for the course ACM 1 taught by Professor Prof during the Fall '09 term at Caltech.

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s8_95-100aSOLUTION - ACM95a/100a Last updated December 2,...

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