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Lecture13 - Roots The n-th root of z C n N is the solution...

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Roots The n -th 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 ϕ + i sin ϕ ) , w = R (cos ϑ + i sin ϑ ) . (498) Then, with De Moivre’s formula, we obtain w n = R n (cos( ) + i sin( )) = r (cos ϕ + i sin ϕ ) , (499) and thus R n = r, cos( ) = cos ϕ, sin( ) = sin ϕ. (500) There are n solutions, given by R = r 1 /n = n r, ϑ k = ϕ + 2 n , k = 0 , . . . , n 1 . (501) Therefore, the n -th root of z is n -valued: n z = w k = n radicalbig | z | parenleftbigg cos parenleftbigg arg z + 2 n parenrightbigg + i sin parenleftbigg arg z + 2 n parenrightbiggparenrightbigg (502) = 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 , = n radicalbig | z | parenleftBig cos parenleftBig arg z n parenrightBig + i sin parenleftBig arg z n parenrightBigparenrightBig parenleftbigg cos parenleftbigg 2 π n parenrightbigg + i sin parenleftbigg 2 π n parenrightbiggparenrightbigg k , for k = 0 , . . . , n 1. In particular, for the n -th roots of unity, n 1, we obtain, with | 1 | = 1, arg 1 = 0: n 1 = cos parenleftbigg 2 n parenrightbigg + i sin parenleftbigg 2 n parenrightbigg = parenleftbigg cos parenleftbigg 2 π n parenrightbigg + i sin parenleftbigg 2 π 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 n -sided polygon.
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