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Unformatted text preview: Roots The nth root of z C , n N , is the solution of the polynomial equation w n = z . We write the numbers w, z C in polar form as z = r (cos + i sin ) , w = R (cos + i sin ) . (498) Then, with De Moivres formula, we obtain w n = R n (cos( n ) + i sin( n )) = r (cos + i sin ) , (499) and thus R n = r, cos( n ) = cos , sin( n ) = sin . (500) There are n solutions, given by R = r 1 /n = n r, k = + 2 k n , k = 0 , . . . , n 1 . (501) Therefore, the nth root of z is nvalued: n z = w k = n radicalbig  z  parenleftbigg cos parenleftbigg arg z + 2 k n parenrightbigg + i sin parenleftbigg arg z + 2 k n parenrightbiggparenrightbigg (502) = n radicalbig  z  parenleftBig cos parenleftBig arg z n parenrightBig + i sin parenleftBig arg z n parenrightBigparenrightBig parenleftbigg cos parenleftbigg 2 k n parenrightbigg + i sin parenleftbigg 2 k n parenrightbiggparenrightbigg , = n radicalbig  z  parenleftBig cos parenleftBig arg z n parenrightBig + i sin parenleftBig arg z n parenrightBigparenrightBig parenleftbigg cos parenleftbigg 2 n parenrightbigg + i sin parenleftbigg 2 n parenrightbiggparenrightbigg k , for k = 0 , . . . , n 1. In particular, for the nth roots of unity, n 1, we obtain, with  1  = 1, arg 1 = 0: n 1 = cos parenleftbigg 2 k n parenrightbigg + i sin parenleftbigg 2 k n parenrightbigg = parenleftbigg cos parenleftbigg 2 n parenrightbigg + i sin parenleftbigg 2 n parenrightbiggparenrightbigg k , (503) for k = 0 , . . . , n 1. These points lie on the unit circle in the complex plane, and they form the vertices of a regular nsided polygon....
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This note was uploaded on 11/17/2011 for the course MATH 529 taught by Professor Staff during the Spring '08 term at UNC.
 Spring '08
 Staff
 Math

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