De Moivre's theorem and roots of complex numbers: VCE Specialist Mathematics Unit 3

What this dot point is asking

VCAA wants you to apply De Moivre's theorem to raise complex numbers to integer powers, to find all $n$ distinct $n$th roots of a complex number and plot them on the Argand plane, and to use complex roots when factorising polynomials. The key idea is that polar form turns powers into multiplication of angles, and roots into evenly spaced points on a circle.

De Moivre's theorem

For a complex number in polar form $z = r\,\mathrm{cis}\,\theta$ and any integer $n$,

This follows directly from the multiplication rule for polar form (multiply moduli, add arguments) applied $n$ times: the modulus is raised to the power $n$ and the argument is multiplied by $n$. It holds for negative integers too, so $z^{-1} = r^{-1}\,\mathrm{cis}(-\theta)$.

De Moivre's theorem also generates trigonometric identities. Expanding $(\cos\theta + i\sin\theta)^n$ by the binomial theorem and equating real and imaginary parts with $\cos(n\theta) + i\sin(n\theta)$ yields multiple-angle formulas, for example $\cos(2\theta) = \cos^2\theta - \sin^2\theta$ from $n = 2$.

Finding the $n$th roots

To solve $z^n = w$ where $w = r\,\mathrm{cis}\,\theta$, write $w$ in polar form but allow the argument to differ by full turns: $w = r\,\mathrm{cis}(\theta + 2k\pi)$. Then

These $n$ values are distinct, and $k = n$ repeats the $k = 0$ root, so exactly $n$ roots exist. Geometrically:

• all roots share the modulus $r^{1/n}$, so they lie on a circle of that radius;
• consecutive arguments differ by $\frac{2\pi}{n}$, so the roots are evenly spaced;
• together they form the vertices of a regular $n$-sided polygon centred at the origin.

The $n$th roots of unity (solutions of $z^n = 1$) are the special case $r = 1$, $\theta = 0$, giving $z = \mathrm{cis}\!\left(\frac{2k\pi}{n}\right)$.

Factorising polynomials over the complex numbers

The fundamental theorem of algebra guarantees that a degree-$n$ polynomial has exactly $n$ complex roots, counted with multiplicity, so it factorises completely into $n$ linear factors over $\mathbb{C}$. For a polynomial with real coefficients, the conjugate root theorem says that if $a + bi$ is a root then so is $a - bi$. Multiplying a conjugate pair gives a real quadratic factor:

This is how a real cubic with one real root and a complex conjugate pair factorises into a real linear factor times a real irreducible quadratic.

Examples in context

Example 1. A fourth power. $(\sqrt{2}\,\mathrm{cis}\,\frac{\pi}{8})^4 = (\sqrt{2})^4\,\mathrm{cis}\!\left(4 \cdot \frac{\pi}{8}\right) = 4\,\mathrm{cis}\,\frac{\pi}{2} = 4i$.

Example 2. Conjugate factor. If $2 + 3i$ is a root of a real polynomial, then $2 - 3i$ is also a root, and the corresponding real quadratic factor is $z^2 - 4z + 13$ since $a^2 + b^2 = 4 + 9 = 13$.

Try this

Q1. Use De Moivre's theorem to evaluate $(1 + i)^6$. [3 marks]

• Cue. $1 + i = \sqrt{2}\,\mathrm{cis}\,\frac{\pi}{4}$, so $(\sqrt{2})^6\,\mathrm{cis}\,\frac{6\pi}{4} = 8\,\mathrm{cis}\,\frac{3\pi}{2} = -8i$.

Q2. Find all solutions of $z^4 = 16$. [3 marks]

• Cue. Modulus $2$, arguments $\frac{2k\pi}{4}$: $z = 2, 2i, -2, -2i$.

Q3. A real cubic has roots $-1$ and $1 + 2i$. State its third root and a real quadratic factor. [2 marks]

• Cue. Third root $1 - 2i$; quadratic $z^2 - 2z + 5$.