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Unformatted text preview: Math 185. Sample Answers to Problem Set #15 Page 316 4. Recall that (6) on page 314 says w = e iα z z z z . The condition that 0 is to be mapped to 1 is equivalent to z z = e iα , which is equivalent to z 2 =  z  2 e iα , hence to z = z  e iα/ 2 (the other root  z  e iα/ 2 has argument α/ 2 ∈ ( π, 0), which means it is in the lower half plane). Let r =  z  > 0; then z = re iα/ 2 . Now assume that 0 is mapped to 1, and consider the condition that 1 is mapped to e iα/ 2 . This is equivalent to e iα 1 + re iα/ 2 1 + re iα/ 2 = e iα/ 2 ; ⇐⇒ 1 + re iα/ 2 = e iα/ 2 (1 + re iα/ 2 ) ; ⇐⇒ 1 + re iα/ 2 = e iα/ 2 + r . Taking imaginary parts gives r sin α/ 2 = sin α/ 2, so r = 1 since sin α/ 2 6 = 0. Conversely, if r = 1 then the above condition holds. Thus 0 7→ 1 and 1 7→ e iα/ 2 if and only if the transformation can be written w = e iα z + e iα/ 2 z + e iα/ 2 . Page 350 1. When z = 2 + i , we have f ( z ) = 2 z = 4 + 2 i , which has argument 30 ◦ = π/ 6. To check this, let z ( t ) = 2 + it , which passes through 2 + i when t = 1. We have w ( t ) = z ( t ) 2 = (4 t 2 ) + 4 it , so w ( t ) = 2 t + 4 i and w (1) = 4 i 2. This has argument 120 ◦ = 2 π/ 3. Comparing that with z ( t ) = i , so z (1) = i has argument 90 ◦ = π/ 2. So φ = ψ + θ (in the notation of Section 94) because φ = 2 π/ 3, ψ = π/ 2, and θ = π/ 6. To check the scale factors, we note that  z (1)  = 1 and  w (1) = √ 20 = 2 √ 5....
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 Fall '07
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 Math, arg, cauchyriemann equations, z0 deﬁned

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