MATH 316 Lecture : Homework solution 2

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Problem 1 (cid:19)7 (cid:18) 4 + 4i (cid:19)7 (cid:18)4 4i (cid:19)7. The second term is the conjugate of the rst thus it is equal to 2 re(cid:0) 4 4i (cid:1)7 = 2 re(128i) = 0. Therefore we have the equation equal to 2 re(cid:0) 4 4i (cid:18)4 4i (cid:18)1 i. This is clear since z zz = z z z = zz z. We have z1 = + i , z1 = i , so |z1 + z2| = 2| | = a. 2 + 2 = (cid:112) (cid:113) 1 a2 a2 which gives us (cid:114) 1 a2 a2 and z2 = a. 2 (cid:114) 1 a2 a2 z1 = a. 2 substituting back we get = 1. 6 so we have ( (cid:18)5 + 12k (cid:18) (cid:19) 5 ) for k = 0, 1, 2, 3, 4 equivalently f or k = 0, 1, 2, 3, 4.

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