De Moivre & Roots of Unity

Powers and nth-roots of complex numbers via polar form.

The idea

De Moivre's theorem computes powers of a complex number in polar form, and, read in reverse, produces all of its $n$th roots.

Every nonzero complex number can be written in polar form $z = r(\cos\theta + i\sin\theta)$, where the modulus $r = |z|$ is its distance from the origin and the argument $\theta$ is the angle it makes with the positive real axis. Multiplying two numbers in this form multiplies their moduli and adds their arguments, so a product is a stretch followed by a rotation.

Theorem (De Moivre's theorem).

For every real $\theta$ and every integer $n \ge 1$, $(\cos\theta + i\sin\theta)^{n} = \cos(n\theta) + i\sin(n\theta).$ Consequently, if $z = r(\cos\theta + i\sin\theta)$ then $z^{n} = r^{n}\bigl(\cos(n\theta) + i\sin(n\theta)\bigr)$.

This is the multiplication rule applied $n$ times to one number: raising to a power raises the modulus to that power and multiplies the angle by $n$, with no binomial expansion.

To solve $w^{n} = z$ for a nonzero $z$, reverse the theorem: the modulus $|w| = \sqrt[n]{r}$ is forced, and $\theta/n$ serves as an argument. Adding $360^{\circ}$ to $\theta$ before dividing lands on a different $w$ that still works, so the arguments $\frac{\theta + 360^{\circ}k}{n}$ for $k = 0, 1, \ldots, n-1$ give $n$ distinct $n$th roots, equally spaced around the circle of radius $\sqrt[n]{r}$. Taking $z = 1$ gives the $n$th roots of unity.

Ways to work on it

• Walkthrough. De Moivre's theorem and how many nth-roots a complex number has.
• Proof. Prove De Moivre's theorem by induction using the angle-addition formulas.
• Practice. Take a power of a unit-circle number with De Moivre.
• Hardest. Compute all cube roots of a complex number and identify one.

Not sure where to start? Take the ten-question placement test.