nth Roots of Unity

This lesson covers finding the nth roots of complex numbers, with special attention to the roots of unity — the solutions of $z^n = 1$zn=1. This is one of the most elegant applications of De Moivre's Theorem.

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The nth Roots of a Complex Number

To solve $z^n = w$zn=w where $w = R(\cos\phi + i\sin\phi)$w=R(cosϕ+isinϕ):

Write $z = r(\cos\theta + i\sin\theta)$z=r(cosθ+isinθ). By De Moivre's Theorem:

$z^n = r^n(\cos(n\theta) + i\sin(n\theta)) = R(\cos\phi + i\sin\phi)$zn=rn(cos(nθ)+isin(nθ))=R(cosϕ+isinϕ)

Comparing moduli: $r^n = R$rn=R, so $r = R^{1/n}$r=R1/n (the positive real nth root).

Comparing arguments: $n\theta = \phi + 2k\pi$nθ=ϕ+2kπ for integer $k$k, so:

$\theta = \frac{\phi + 2k\pi}{n}$θ=nϕ+2kπ​

Taking $k = 0, 1, 2, \ldots, n - 1$k=0,1,2,…,n−1 gives exactly n distinct roots.

$z_k = R^{1/n}\left(\cos\frac{\phi + 2k\pi}{n} + i\sin\frac{\phi + 2k\pi}{n}\right), \quad k = 0, 1, \ldots, n - 1$zk​=R1/n(cosnϕ+2kπ​+isinnϕ+2kπ​),k=0,1,…,n−1

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The nth Roots of Unity

The nth roots of unity are the solutions of $z^n = 1$zn=1.

Since $1 = \cos 0 + i\sin 0$1=cos0+isin0 (so $R = 1$R=1, $\phi = 0$ϕ=0):

$z_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}, \quad k = 0, 1, \ldots, n - 1$zk​=cosn2kπ​+isinn2kπ​,k=0,1,…,n−1