Excercise Sheet Solution 1 - MATH 20142 Complex Analysis...

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MATH 20142 Complex Analysis Solution Sheet 1 1. (a) (3 + 4 i ) 2 = 9 + 24 i 16 = 7 + 24 i ; (b) 2 + 3 i 3 4 i = (2 + 3 i )(3 + 4 i ) 25 = 6 25 + 17 25 i ; (c) 1 5 i 1 + 3 i = 8 5 + 1 5 i ; (d) 1 i 1 + i + 2 i = i + 2 i = 2 2 i. 2. (a) Write z = a + ib ; then a 2 + 2 abi b 2 = 5 + 12 i , whence a 2 b 2 = 5 , 2 ab = 12 . The second equation yields b = 6 /a (with a ̸ = 0), whence a 4 + 5 a 2 36 = 0. Solving this bi-quadratic equation gives a = ± 2, whence b = ± 3. Hence there are two solutions: z = 2+3 i and z = 2 3 i . (b) We have ( z +2) 2 = 8+6 i , whence it is more convenient to write z +2 = a + ib . Similarly, to the previous solution, we have a = ± 1 , b = ± 3, whence the solutions are z = 1 + 3 i and z = 3 3 i . 3. (a) Writing z = x + iy , we obtain Re z = { ( x, y ) : x > 2 } , i.e., a half-plane. (b) Here we have the open strip { ( x, y ) : 1 < y < 2 } . (c) The condition | z | < 3 is equivalent to x 2 + y 2 < 9, hence our set is an open disc of radius 3 centred at the origin. (d) Write z = x + iy ; we have | x 1 + iy | < | x + 1 + iy | , which is equivalent to ( x 1) 2 + y 2 < ( x + 1) 2 + y 2 , i.e., after all cancellations, to x > 0. Hence we have an open half-plane.
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