380_hw_5_solns_08march[1]

380_hw_5_solns_08march[1] - Math 380 Solutions to hw 5...

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Math 380 Solutions to hw 5 3.2.20 Let R = f x + iy : 1 x 1 and 0 y g and A = w : e 1 w e 1 and 0 Arg ( w ) ± : There are three things to show: (1) z 2 R ) e z 2 A , (2) every point in A is the image of some point in R , and (3) e z is 1-1 on R . (1) Let z = x + iy belong to R . Then ± 1 x 1 and 0 y Now e z = e x e iy . Thus e 1 e x = j e z e 1 and 0 y = Arg ( e z ) This means that e z lies in A: (2) Suppose w lies in A . Then w = re with e 1 r e and 0 ± Set z = ln r + i±: Then z belongs to R and w = e z : (3) Since R is contained in a horizontal strip of width less than 2 , the exponential map is 1-1 on R: 3.2.23 This problem is not deep, but I think it is an amusing alternative proof of the Pythagorean identity (8) from the last assignment. Let f ( z ) = sin 2 z + cos 2 z: ( a ) f is di/erentiable on C plane because it is formed from sums, products, and constant mulitples of e iz and e iz ; both of which are di/erentiable on
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380_hw_5_solns_08march[1] - Math 380 Solutions to hw 5...

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