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# 814hw3sol - Math 814 HW 3 p 54 9 14 18 24 25 26 p.54...

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Math 814 HW 3 October 16, 2007 p. 54: 9, 14, 18, 24, 25, 26 p.54, Exercise 9. If Tz = az + b cz + d , find necessary and sufficient conditions for T to preserve the unit circle. T preserves the unit circle iff | ae + b | = | ce + d | , for all θ [0 , 2 π ) . Squaring both sides and multiplying out, we get | a | 2 + | b | 2 + a ¯ be + ¯ abe - = | c | 2 + | d | 2 + c ¯ de + ¯ cde - . Comparing coefficients, we get two equations: | a | 2 + | b | 2 = | c | 2 + | d | 2 , a ¯ b = c ¯ d. Dividing the first by | d | 2 and using the second, we get | c | 2 | b | 2 + | b | 2 | d | 2 = | c | 2 | d | 2 + 1 , which can be written | c | 2 - | b | 2 | b | 2 = | c | 2 - | b | 2 | d | 2 . If the numerators are zero, this means | c | = | b | , hence | a | = | d | . If the numerators are nonzero, we have | b | = | d | , hence | a | = | c | . 1

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In the first case, there are u, v with | u | = | v | = 1 such that c = ub, d = va. Then we have a ¯ b = c ¯ d = ub ¯ v ¯ a, so ub ¯ b = va ¯ a . This last number, call it λ , also has | λ | = 1 , and we have a b c d = 1 0 0 λ a b ¯ b ¯ a . (1) In the second case, there are u, v with | u | = | v | = 1 such that d = ub, c = va. This time, we have a ¯ b = c ¯ d = va ¯ u ¯ b, so v ¯ u = 1 . But ¯ u = u - 1 , so v = u , and we have ad - bc = aub - bua = 0 , which is illegal for a M¨obius transformation. So the second case does not occur, and all such transformations T are of the form (1). p.54, Exercise 14. Let G be the region between two circles inside one another, tangent at the point a . Map G conformally to the open unit disk. The map Tz = ( z - a ) - 1 sends a to , hence the circles go to lines , . If and are not parallel, they must meet at some point z C . Then T - 1 z would be a point on both circles other than a . Since there is no such point, the lines , are parallel. The original region G
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