#4 - Solution of Homework 1 Problem(6.6 Solution The domain...

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Solution of Homework 1 Problem (6.6): Solution: The domain Ω is conformally equivalent to D (0 , 1) \{ 0 } . f ( z ) = 1 /z is a conformal map from D (0 , 1) \{ 0 } to Ω . Suppose g is a conformal mapping from Ω to Ω . Then f - 1 g f is a biholomorphc-self map of D (0 , 1) \{ 0 } . And Aut ( D (0 , 1) \{ 0 } ) = { e z | ψ R } So we know that g = e - z . Above all: Aut (Ω) = { e z | ϕ R } ¥ Problem (6.11): Solution: Observe that φ 1 φ - 1 2 : D D is a conformal map. According to Aut ( D (0 , 1)) = { e a - z 1 - ¯ az | a D (0 , 1) , θ [0 , 2 π ) } So there exist a θ [0 , 2 π ) , ϕ Aut ( D (0 , 1)) such that φ 1 = ϕ φ 2 . Then we know φ 1 ( z ) = e a - φ 2 ( z ) 1 - ¯ 2 ( z ) ¥ Problem (6.17) Solution: Since φ ( P ) = P , φ ( P ) = 1 , from the hint in the book, we know that φ ( z ) = P + ( z - P ) + h ( z ) , where h ( z ) contains the higher order terms. 1
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Suppose φ ( z ) 6 = z . then h ( z ) has anonzero term, say a n ( z - P ) n . Since φ maps Ω to itself, so | φ ( z ) | ≤ M , where M = sup {| z | , z Ω } . By cauchy estimate, | a n | ≤ M r n . And by induction φ j ( z ) = φ φ φ ···◦ φ = P +( z - P )+ ja n ( z - P ) n + ···· But φ j maps Ω to itself. So the same cauchy estimate gives
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  • Spring '09
  • Compact space, Möbius transformation, Conformal geometry, Conformal map, linear fractional transformation, cauchy estimate

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