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Unformatted text preview: ASSIGNMENT 1 SOLUTIONS MAT 572 A FALL 2007 Problem 1 ( cf. [1, Exercise I.4.5]) . Let z = cis(2 /n ) for an integer n 2. Show that 1 + z + z 2 + + z n 1 = 0 . Proof. Let s denote the value of the sum. Since z n = 1, we have zs = z + z 2 + + z n 1 + z n = z + z 2 + + z n 1 + 1 = s. Thus (1 z ) s = s zs = 0. Since n 2, we have z 6 = 1, and it follows that s = 0. Incidentally, for a geometric interpretation of this trick, note that the set { 1 ,z,z 2 ,...,z n 1 } of n points evenly distributed around the unit circle is invariant under the rigid motion of rotation by 2 /n radians; that is, under the transformation w 7 zw . Therefore the sum over all the elements in the set must also be invariant. But the only complex number invariant under a nontrivial rotation is 0. Problem 2 ( cf. [1, Exercise II.1.11]) . (a) Show that the set S = { cis k  k N } is dense in T = { z C   z  = 1 } ....
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This document was uploaded on 09/08/2011.
 Spring '09

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