{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture13

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

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}