De Moivre's theorem and roots of unity in HSC Mathematics Extension 2: powers in polar form, nth roots, the roots of unity and their geometry

What this dot point is asking

NESA wants you to use de Moivre's theorem to raise complex numbers to powers and to find all the $n$th roots of a complex number. You must derive the $n$th roots of unity, plot them as equally spaced points on the unit circle, and use their structure (for example, that they sum to zero). The geometry, $n$ points evenly spread round a circle forming a regular polygon, is the picture that makes every roots question quick, so this page builds it up carefully.

De Moivre's theorem

If $z = r(\cos\theta + i\sin\theta)$ then for every integer $n$,

This follows from the fact that multiplication in polar form multiplies moduli and adds arguments; raising to the $n$th power multiplies the modulus by itself $n$ times and adds the argument $n$ times. The theorem can be proved formally by induction for positive integers, then extended to negative integers via reciprocals.

A typical use is computing high powers. To find $(1 + i)^8$, write $1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$, so

Computing $(1+i)^8$ by repeated Cartesian multiplication is laborious and error-prone; in polar form it is two short lines. That contrast is the whole reason de Moivre's theorem is examined.

Finding nth roots

To solve $z^n = w$ where $w = R(\cos\phi + i\sin\phi)$, write the unknown as $z = r(\cos\theta + i\sin\theta)$. Then $r^n = R$ gives $r = R^{1/n}$ (the positive real root), and $n\theta = \phi + 2\pi k$ for integer $k$. The distinct solutions are

Values of $k$ outside this range repeat earlier roots. Geometrically the $n$ roots lie on a circle of radius $R^{1/n}$, equally spaced by angle $\dfrac{2\pi}{n}$.

Roots of unity

The $n$th roots of unity solve $z^n = 1$. Since $1 = \cos 0 + i\sin 0$, they are

Writing $\omega = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n}$, every root is a power $\omega^k$. They sit at the vertices of a regular $n$-gon inscribed in the unit circle, with one vertex at $1$.

The figure builds the sixth roots of unity ($z^6 = 1$) stage by stage, so you can see the regular polygon emerge.

Stage 1, start on the unit circle. Every root of $z^6 = 1$ has modulus $1$, so all six lie on the unit circle. The simplest root is $k = 0$: $\omega_0 = \cos 0 + i\sin 0 = 1$, sitting on the positive real axis.

Stage 2, step round by $\tfrac{2\pi}{6} = 60^\circ$. Each successive root advances the argument by $\dfrac{2\pi}{n} = 60^\circ$. So $\omega_1 = \operatorname{cis}60^\circ$ and $\omega_2 = \operatorname{cis}120^\circ$. Writing $\omega = \omega_1$, the roots are the powers $1, \omega, \omega^2, \ldots$

Stage 3, all six roots. Continuing around, the six arguments are $0^\circ, 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ$, equally spaced $\dfrac{360^\circ}{6} = 60^\circ$ apart. The root at $180^\circ$ is $-1$, as expected for an even $n$.

Stage 4, join them into a regular polygon. The six roots are the vertices of a regular hexagon inscribed in the unit circle. By symmetry the vertices balance about the origin, which is the geometric reason the roots of unity sum to zero.

Why the roots of unity sum to zero

A key property: the roots of unity sum to zero for $n \ge 2$. There are two ways to see it. Algebraically, $z^n - 1 = (z - 1)(z^{n-1} + \dots + z + 1)$, and the sum of all roots equals the negative of the coefficient of $z^{n-1}$ in $z^n - 1$, which is $0$. Geometrically, the roots are the vertices of a regular polygon centred at the origin, so the position vectors cancel in symmetric pairs (or triples), leaving a resultant of zero, exactly the balance you can see in the hexagon above.

Multiple-angle formulae from de Moivre

One of the most powerful uses of de Moivre's theorem is deriving identities for $\cos n\theta$ and $\sin n\theta$. Expand $(\cos\theta + i\sin\theta)^n$ with the binomial theorem, then equate the real part to $\cos n\theta$ and the imaginary part to $\sin n\theta$. Replacing $\sin^2\theta = 1 - \cos^2\theta$ converts the cosine identity to a polynomial in $\cos\theta$ alone, as in the $\cos 5\theta$ question above. This technique turns a trigonometric identity that would be hard to prove directly into a routine expansion.

Roots of a general complex number

The same method finds the roots of any complex number, not just $1$. To solve $z^n = w$, first write $w$ in modulus-argument form $w = R\operatorname{cis}\phi$. The $n$ roots then sit on a circle of radius $R^{1/n}$, equally spaced by $\frac{2\pi}{n}$, with the first root at argument $\frac{\phi}{n}$. Geometrically the roots of unity are the special case $w = 1$; multiplying every root of unity by one particular root of $z^n = w$ rotates and scales the regular polygon onto the roots of the general equation. This is why drawing the roots of unity well prepares you for every roots question.

Reducing arguments

After raising to a high power the argument often exceeds $2\pi$ or falls below $-2\pi$. Always reduce it into a single revolution by adding or subtracting multiples of $2\pi$ before reading off the Cartesian value, since $\cos$ and $\sin$ are periodic with period $2\pi$. Forgetting this step leaves the answer correct in principle but in an awkward, unmarked form.

How exam questions ask about de Moivre and roots of unity

• "Find $z^n$" or "evaluate $(\,\ldots\,)^n$ in the form $x + iy$": convert to polar form, apply de Moivre, reduce the argument, then convert back.
• "Solve $z^n = w$" or "find the $n$th roots of ...": use the $n$th-roots formula for $k = 0, \dots, n-1$; expect to plot them.
• "Show that the roots ... form a regular polygon" or "sketch on an Argand diagram": state the common modulus $R^{1/n}$ and the equal spacing $\tfrac{2\pi}{n}$.
• "Show that the roots sum to zero" or "$1 + \omega + \omega^2 + \dots = 0$": use the factorisation of $z^n - 1$ or the symmetry of the polygon.
• "Show that $\cos n\theta = \ldots$": expand $(\cos\theta + i\sin\theta)^n$ by the binomial theorem and take the real part.
• "It is known that $u = z + \tfrac{1}{z}$ ...": recognise $z + \tfrac1z = 2\cos\theta$ for a root on the unit circle, then solve the resulting real equation.