prob01_sol

# prob01_sol - Solution Problem Set 1 1 Evaluate(8 2i(1 i(2...

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Solution: Problem Set 1 1. Evaluate (8 + 2 i ) - (1 - i ) (2 + i ) 2 2 - 3 i 1 + 2 i - 8 + i 6 - i 2. Let z be a complex number such that Im z > 0, show that Im(1 /z ) < 0. Proof. Let z = x + yi with both x and y being real numbers, and y > 0. Then 1 z = x | z | 2 - y | z | 2 , so Im 1 z = - y | z | 2 < 0 . 3. Show that if | z | = 1( z 6 = 1), then Re 1 / (1 - z ) = 1 / 2. Proof. Let z = x + yi with x, y R and x 2 + y 2 = 1. Then 1 1 - z = 1 - x - yi (1 - x ) 2 + y 2 = 1 - x - yi x 2 + y 2 - 2 x = 1 - x - yi 2 - 2 x . So Re 1 1 - z = 1 - x 2 - 2 x = 1 2 . 4. Find the principal value of the argument for following complex numbers: - 1 2 , - 3 + 3 i, ( 3 - i ) 4 , - 1 + 3 i 2 + 2 i , (1 - i )( - 3 + i ) Answer. π, 3 π 4 , - 2 π 3 , 7 π 12 . 5. Given two points P = (1 , 2) and Q = (2 , 0), find the equation for the line trough P and Q in terms of z and ¯ z . Answer. (1 + 2 i ) z - (1 - 2 i z = 8 i . 6. Solve the equation ( z + 1) 5 = z 5 . Solution. It is easy to see that z = 0 is not a solution, so we can divide both side of the equation by z 5 to get (1 + z ) 5 z 5 = (1 + 1 z ) 5 = 1 . Hence 1 + 1 z = e 2 5 i , k = 0 , 1 , 2 , 3 , 4 . (1)

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so z = 1 e 2 5 i - 1 , k = 1 , 2 , 3 , 4 .
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