De Moivre’s Theorem and Roots of Unity

Definition Let $z = r\operatorname{cis}\theta$ with $r > 0$ and $\theta$ real. For every integer $n$,

For $n \ge 1$ this follows from the rule for products in polar form, since each new factor multiplies the modulus by $r$ and adds $\theta$ to the argument. For $n = 0$ both sides equal $1$. For a negative integer $n = -m$ the power means $z^{-m} = 1/z^m$, and the quotient rule gives

The theorem is about integer exponents. A fractional exponent such as $z^{1/3}$ has several candidate values, and those are the roots in the sections below.

Definition Let $n$ be a positive integer and $w$ a complex number. An $n$th root of $w$ is a complex number $z$ with $z^n = w$. If $w = R\operatorname{cis}\varphi$ with $R > 0$, the $n$th roots of $w$ are exactly the numbers

where $R^{1/n}$ is the positive real $n$th root of $R$. Here is why. Write a candidate as $z = r\operatorname{cis}\theta$ with $r > 0$. By De Moivre’s theorem $z^n = r^n\operatorname{cis}(n\theta)$, and this equals $R\operatorname{cis}\varphi$ exactly when $r^n = R$ and $n\theta = \varphi + 2k\pi$ for some integer $k$. The first condition forces $r = R^{1/n}$ and the second gives $\theta = (\varphi + 2k\pi)/n$. Integers $k$ outside $0, 1, \ldots, n - 1$ give no new numbers, as the next section shows. The only $n$th root of $0$ is $0$ itself.

Definition Let $w \ne 0$ and let $n$ be a positive integer. Then $w$ has exactly $n$ distinct $n$th roots, the numbers $z_0, z_1, \ldots, z_{n-1}$ above. They all have the modulus $R^{1/n}$, and their arguments

differ from each other by more than $0$ and less than $2\pi$, so no two of them are coterminal. Any other integer $k$ repeats one of these, because replacing $k$ by $k + n$ adds exactly $2\pi$ to the argument. For $n \ge 2$ a nonreal $w$ has $n$ equally good roots, so the symbol $\sqrt[n]{w}$ is ambiguous for it unless a particular root is singled out.

Definition For a positive integer $n$, the $n$th roots of unity are the solutions of $z^n = 1$. Since $1 = \operatorname{cis}0$, they are

With $\omega = \operatorname{cis}\frac{2\pi}{n}$, De Moivre’s theorem gives $\omega_k = \omega^k$, so every $n$th root of unity is a power of $\omega$. All of them have modulus $1$ and lie on the unit circle. For $n \ge 2$ their sum is $0$, because

and $\omega \ne 1$.

Definition Let $w \ne 0$ and let $n$ be a positive integer. Consecutive $n$th roots $z_k$ and $z_{k+1}$ of $w$ have the same modulus and arguments that differ by exactly $\frac{2\pi}{n}$, so $z_{k+1} = \omega z_k$ with $\omega = \operatorname{cis}\frac{2\pi}{n}$. Consequently, if $u$ is any one $n$th root of $w$, the full list of roots is

Each of these is a root, because $(u\omega^k)^n = u^n(\omega^n)^k = w$. They are $n$ different numbers, because $u \ne 0$ and the powers $\omega^0, \ldots, \omega^{n-1}$ are different, so no root is missing.

Definition Let $w \ne 0$ and let $n \ge 3$ be an integer. The $n$th roots of $w$ are the vertices of a regular $n$-gon centered at $0$ and inscribed in the circle of radius $\rho = \lvert w \rvert^{1/n}$. All vertices are at distance $\rho$ from the center and neighboring vertices are $\frac{2\pi}{n}$ apart, so every side is the base of an isosceles triangle with legs $\rho$ and apex angle $\frac{2\pi}{n}$, and its length is

For $n = 2$ the two square roots are the endpoints of a diameter of that circle, and each is the negative of the other.