Roots of complex numbers and roots of unity (WACE Year 12 Mathematics Specialist Unit 3)

What this dot point is asking

SCSA wants you to solve $z^n = w$ for all roots, identify the roots of unity, and plot them as a regular polygon on the Argand plane.

The general root formula

To solve $z^n = w$, write $w = R\,\text{cis}\,\phi$ and let $z = \rho\,\text{cis}\,\alpha$. By de Moivre $z^n = \rho^n\,\text{cis}(n\alpha)$, so matching moduli gives $\rho = R^{1/n}$ and matching arguments gives $n\alpha = \phi + 2\pi k$ for integer $k$. Hence

Taking $k = 0$ to $n-1$ gives $n$ distinct roots; further values of $k$ repeat them. All roots share the modulus $R^{1/n}$, so they lie on a circle of that radius, spaced $\tfrac{2\pi}{n}$ apart. This is why they form a regular $n$-gon.

Why the roots are equally spaced

The equal spacing is a direct consequence of the formula. Consecutive roots differ only in the value of $k$, and increasing $k$ by one adds $\dfrac{2\pi}{n}$ to the argument while leaving the modulus $R^{1/n}$ unchanged. Geometrically, each root is the previous one rotated by $\dfrac{2\pi}{n}$ about the origin. After $n$ such rotations the total turn is $2\pi$, returning to the starting root, which is why exactly $n$ distinct roots arise before they repeat. This is the reason the roots always form a regular $n$-gon inscribed in a circle of radius $R^{1/n}$, a fact that lets you draw all the roots once you have located a single one.

The nth roots of unity

Setting $w = 1 = \text{cis}\,0$ gives the $n$th roots of unity:

They all lie on the unit circle, include $z = 1$, and are equally spaced. Writing $\omega = \text{cis}\tfrac{2\pi}{n}$, every root is a power $\omega^k$.

Roots of unity as a cyclic group

The $n$th roots of unity have a tidy structure: every root is a power of the single primitive root $\omega = \text{cis}\dfrac{2\pi}{n}$, so the full set is $\{1, \omega, \omega^2, \dots, \omega^{n-1}\}$ and multiplying any two of them gives another (with exponents added modulo $n$). This is why the sum of all $n$ roots is zero for $n \ge 2$: they are the roots of $z^n - 1 = 0$, and the coefficient of $z^{n-1}$ in $z^n - 1$ is zero, so by the sum-of-roots relation the total is zero. The factorisation $z^n - 1 = (z - 1)(z^{n-1} + z^{n-2} + \dots + 1)$ shows the same thing: the non-trivial roots satisfy $z^{n-1} + \dots + 1 = 0$.

Roots of any number from the roots of unity

Once you know the $n$th roots of unity, the $n$th roots of any complex number follow by multiplication. If $z_0$ is any one $n$th root of $w$, then every $n$th root of $w$ is $z_0 \omega^k$ for $k = 0, 1, \dots, n-1$, because multiplying by a root of unity rotates without changing the modulus and preserves the property of being an $n$th root. This gives a fast way to write down all the roots once a single one is found.

Plotting strategy

Find one root (usually $k = 0$), then rotate by $\tfrac{2\pi}{n}$ repeatedly to get the rest. You never recompute the modulus.