prob02_sol - Solution: Problem Set 2 1. Consider following...

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Solution: Problem Set 2 1. Consider following sets: a) A = { z : | Arg z | < π/ 4 } b) B = { z : - 1 < Im z < 1 } c) C = { z : | z | ≥ 1 } d) D = { z : (Re z ) 2 1 } Which of these sets are open? Which are closed? Which are bounded? Which are domains? (No need to prove your statements.) Answer. a and b are open, c and d are closed. None of them are bounded. a and b are domains. 2. Suppose u ( x,y ) is a real-valued function defined in a domain D . If ∂u ∂x = y and ∂u ∂y = x at all points of D , show that u ( x,y ) = xy + c for some constant c . Proof. Let v ( x,y ) = u ( x,y ) - xy , then ∂v ∂x = ∂u ∂x - y = y - y = 0 , and ∂v ∂y = ∂u ∂y - x = x - x = 0 . So v ( x,y ) must be a constant, hence u ( x,y ) = xy + v ( x,y ) = xy + c for some constant c . 3. Prove that if | z 0 | < 1, then z n 0 0 as n → ∞ . Proof. For any ε > 0, let N = log | z 0 | ε, then for any n > N , because | z 0 | < 1, so the function | z 0 | x is monotonic decreasing, so | z x 0 - 0 | = | z x 0 | = | z 0 | x ≤ | z 0 | N = ε. Hence lim n →∞ z n 0 = 0 .
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4. Find each of the following limits. (a) lim z 2+3 i ( z - 5 i ) 2 (b) lim z 3 i z 2 + 9 z - 3 i (c) lim z 1+2 i | z 2 - 1 | (d) lim z i z 2 + i z 4 - 1 Answer. - 8 i, 6 i, 4 2 and ‘not exists’. 5. Show that the function Arg z is discontinuous at each point on the nonpositive real
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prob02_sol - Solution: Problem Set 2 1. Consider following...

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