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Unformatted text preview: 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 i + b | = | ce i + d | , for all [0 , 2 ) . Squaring both sides and multiplying out, we get | a | 2 + | b | 2 + a be i + abe- i = | c | 2 + | d | 2 + c de i + cde- i . 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 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 , also has | | = 1 , and we have a b c d = 1 0 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 Mobius 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....
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