This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 185. Sample Answers to Problem Set #15 Page 316 4. Recall that (6) on page 314 says w = e i z z z z . The condition that 0 is to be mapped to 1 is equivalent to z z = e i , which is equivalent to z 2 =  z  2 e i , hence to z = z  e i/ 2 (the other root  z  e i/ 2 has argument / 2 ( , 0), which means it is in the lower half plane). Let r =  z  > 0; then z = re i/ 2 . Now assume that 0 is mapped to 1, and consider the condition that 1 is mapped to e i/ 2 . This is equivalent to e i 1 + re i/ 2 1 + re i/ 2 = e i/ 2 ; 1 + re i/ 2 = e i/ 2 (1 + re i/ 2 ) ; 1 + re i/ 2 = e i/ 2 + r . Taking imaginary parts gives r sin / 2 = sin / 2, so r = 1 since sin / 2 6 = 0. Conversely, if r = 1 then the above condition holds. Thus 0 7 1 and 1 7 e i/ 2 if and only if the transformation can be written w = e i z + e i/ 2 z + e i/ 2 . Page 350 1. When z = 2 + i , we have f ( z ) = 2 z = 4 + 2 i , which has argument 30 = / 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 / 3. Comparing that with z ( t ) = i , so z (1) = i has argument 90 = / 2. So = + (in the notation of Section 94) because = 2 / 3, = / 2, and = / 6. To check the scale factors, we note that  z (1)  = 1 and  w (1) = 20 = 2 5....
View
Full
Document
 Fall '07
 Lim
 Math

Click to edit the document details