Solution 5 if z1 0 and z2 0 then thus arg z1

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Unformatted text preview: y when z and w are antiparallel (i.e., |Arg(z ) − Arg(w)| = π ) and |w| ≤ |z |. Solution 4 Recall the definition arcsin(z ) = −i log iz + (1 − z 2 )1/2 . z = arcsin(w), w = −i 1 1 = log iw + 1 − w2 2 , w2 = −1 i √ 1 1 1 = log 1 + 2 2 , 2 2 = ± 2 i √ √ √ √ 1 1 = Log 1 + 2 + iArg 1 + 2 + i2πk , Log 1 − 2 + iArg 1 − 2 + i2πk i i √ √ 1 1 = Log 1 + 2 + i0 + i2πk , Log 2 − 1 + iπ + i2πk i i √ √ = 2πk − iLog 1 + 2 , π + 2πk − iLog 2−1 , k ∈Z , k∈Z (37) Problem 5 (10 points) Show that if (z1 ) > 0 and (z2 ) > 0 then Log(z1 z2 ) = Log(z1 ) + Log(z2 ) using the convention −π < Argz ≤ π . Solution 5 If (z1 ) > 0 and (z2 ) > 0, then − Thus, −π < Arg (z1 ) + Arg (z2 ) < π (39) π π < Arg (zj ) < , 2 2 j = 1, 2 (38) 6 −→ Arg (z1 ) + Arg (z2 ) = Arg (z1 z2 ) Therefore, Log (z1 ) + Log (z2 ) = [Log (|z1 |) + iArg (z1 )] + [Log (|z2 |) + iArg (z2 )] = Log (|z1 z2 |) + iArg (z1 z2 ) = Log (z1 z2 ) , and the statement is proved. (40) (41) Problem 6 (10 points) 6. (10 pts) Find a mapping w = f (z ) that maps the exterior of the circle |z − i| = 1 exactly twice to the interior of the circle |z − 1| = 1. Solution 6 we can construct the mapping from the following procedure, translation u = z − i 1 map exterior of the circle |u| = 1 to its interior v = u map interior of the circle |v | = 1 exactly twice to itself g = v 2 translation ω = g + 1 i.e. 1 1 ω = g + 1 = v2 + 1 = 2 + 1 = +1 u (z − i)2 Therefore, f (z ) = 1 (z −i)2 + 1 is one possible mapping. Problem 7 (4×5 points) Consider the mapping w = f (z ) = z 1/3 and the inverse mapping z = g (w) = w3 . a) Describe the multiple-valuedness of f (z ). b) Describe a region of the w...
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This note was uploaded on 02/22/2010 for the course ACM 95A taught by Professor Nilesa.pierce during the Fall '06 term at Caltech.

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