Roots of complex numbers and roots of unityAQA A-Level Further Maths: Revision notes

The equation $z^n=r\mathrm{e}^{\mathrm{i}\theta}$ has exactly $n$ distinct solutions. Since $r\mathrm{e}^{\mathrm{i}\theta}=r\mathrm{e}^{\mathrm{i}(\theta+2k\pi)}$ for any integer $k$, de Moivre's theorem gives $z_k=r^{1/n}\mathrm{e}^{\mathrm{i}(\theta+2k\pi)/n},\quad k=0,1,\dots,n-1.$ Method: write the number in exponential form, include $2k\pi$ in the argument, take the $n$th root of the modulus, divide the argument by $n$, then list $n$ values of $k$ (values of $k$ beyond $n-1$ repeat the roots). Example: $z^4=-8+8\sqrt3\,\mathrm{i}=16\mathrm{e}^{\mathrm{i}(2\pi/3+2k\pi)}$ gives $z=2\mathrm{e}^{\mathrm{i}(\pi/6+k\pi/2)}$, with arguments $\frac{\pi}{6},\frac{2\pi}{3},\frac{7\pi}{6},\frac{5\pi}{3}$ (or $-\frac{5\pi}{6}$ and $-\frac{\pi}{3}$ in $(-\pi,\pi]$).

All $n$ roots have the same modulus $r^{1/n}$, so they lie on a circle centred at the origin. Their arguments differ by $\frac{2\pi}{n}$, so they are equally spaced: they form the vertices of a regular $n$-gon. Multiplying a root by $\mathrm{e}^{2\pi\mathrm{i}/n}$ rotates it to the next root, so once you have one root you can find the rest. For $n\ge2$ the roots sum to $0$, because the polygon is centred at the origin (and the sum of the roots of $z^n-c=0$ is minus the coefficient of $z^{n-1}$, which is $0$).

The solutions of $z^n=1$ are the $n$th roots of unity: $\omega^k,\quad k=0,1,\dots,n-1,\quad \omega=\mathrm{e}^{2\pi\mathrm{i}/n}.$ They lie on the unit circle, and $1$ is always one of them. Key properties: $\omega^n=1$; $\omega^{n-k}=\omega^{-k}$ is the conjugate of $\omega^k$; and $1+\omega+\omega^2+\dots+\omega^{n-1}=0\quad(n\ge2),$ because it is a geometric series equal to $\frac{\omega^n-1}{\omega-1}=0$ with $\omega\ne1$. Also $z^n-1=(z-1)(z-\omega)\cdots(z-\omega^{n-1})$, so $z^{n-1}+\dots+z+1=\prod_{k=1}^{n-1}(z-\omega^k)$.

Roots of unity turn geometry about regular polygons into algebra.

• Side length: the distance between adjacent vertices $1$ and $\omega$ is $|\omega-1|=2\sin\frac{\pi}{n}$, and for circumradius $R$ it is $2R\sin\frac{\pi}{n}$.
• Area: $n$ triangles with two sides $R$ and angle $\frac{2\pi}{n}$ give area $\frac12nR^2\sin\frac{2\pi}{n}$.
• Distances from one vertex: from $z^n-1=(z-1)\prod(z-\omega^k)$, putting $z=1$ in $z^{n-1}+\dots+1$ gives $\prod_{k=1}^{n-1}(1-\omega^k)=n$. So the product of the distances from one vertex of a regular $n$-gon of circumradius $R$ to the others is $nR^{n-1}$.
• Trigonometric sums: take real parts of $1+\omega+\dots+\omega^{n-1}=0$.

The fifth roots of unity are $1,\omega,\omega^2,\omega^3,\omega^4$ with $\omega=\mathrm{e}^{2\pi\mathrm{i}/5}$. Taking real parts of $1+\omega+\omega^2+\omega^3+\omega^4=0$: $1+\cos72^\circ+\cos144^\circ+\cos216^\circ+\cos288^\circ=0$. Since $\cos216^\circ=\cos144^\circ$ and $\cos288^\circ=\cos72^\circ$, $1+2(\cos72^\circ+\cos144^\circ)=0$, so $\cos72^\circ+\cos144^\circ=-\frac12$.