Find the nth roots of a complex number using polar form and de Moivre's theorem, and represent them geometrically on the Argand plane.

What this dot point is asking

You need to find the $n$th roots of a complex number using de Moivre's theorem, solve equations of the form $z^{n} = w$, and represent the roots geometrically on the Argand plane.

The formula for nth roots

To solve $z^{n} = w$, write $w$ in polar form $w = r(\cos\theta + i\sin\theta)$. Because adding any whole number of full turns leaves $w$ unchanged, write the argument as $\theta + 2\pi k$. Taking the $n$th root with de Moivre's theorem gives

Only $k = 0$ to $n-1$ give distinct roots; $k = n$ repeats $k = 0$ (the angle increases by a full $2\pi$).

Geometry on the Argand plane

Because all $n$ roots share the modulus $r^{1/n}$, they lie on a circle of radius $r^{1/n}$ centred at the origin. The equal angular spacing of $\dfrac{2\pi}{n}$ means they form the vertices of a regular $n$-sided polygon. Sketching this circle and marking one root lets you read off the rest by rotation.

Roots of unity

The special case $z^{n} = 1$ gives the $n$th roots of unity:

They lie on the unit circle, one of them is always $z = 1$, and their sum is $0$ for $n \ge 2$ (the vertices of a regular polygon centred at the origin balance out). If $\omega$ denotes the root with smallest positive argument, the full set is $1, \omega, \omega^2, \ldots, \omega^{n-1}$.

Solving polynomial equations with roots

Finding $n$th roots is the engine for solving binomial polynomial equations such as $z^4 + 16 = 0$. Rearrange to $z^4 = -16$, write $-16$ in polar form as $16\operatorname{cis}\pi$, and take fourth roots: the modulus of each root is $16^{1/4} = 2$ and the arguments are $\dfrac{\pi + 2\pi k}{4}$ for $k = 0, 1, 2, 3$. This yields four complex roots that pair into conjugates, consistent with the fact that a real polynomial has complex roots in conjugate pairs. Whenever a polynomial equation reduces to $z^n = w$, the $n$th-roots formula gives every solution directly.

A worked tip in practice

Because the roots are equally spaced, the fastest method is to compute the first root carefully and then add $\dfrac{2\pi}{n}$ to the argument repeatedly, keeping the modulus fixed. For sketching, draw the circle of radius $r^{1/n}$, mark the first root at its argument, then step around by $\dfrac{2\pi}{n}$ to place the rest at the vertices of a regular polygon. This both saves time and guards against the common error of recomputing each root from scratch and slipping on the arithmetic.