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

To solve $z^n=w$, write $w=r\mathrm{e}^{\mathrm{i}\theta}$ and then allow for the repeating argument: $w=r\mathrm{e}^{\mathrm{i}(\theta+2k\pi)}$ for any integer $k$. Taking the $n$th root of both sides (de Moivre) gives $z=r^{1/n}\,\mathrm{e}^{\mathrm{i}\left(\frac{\theta+2k\pi}{n}\right)},\qquad k=0,1,\dots,n-1.$ There are exactly $n$ distinct roots: the values $k=n,n+1,\dots$ only repeat them. Every root has the same modulus $r^{1/n}$, and the arguments increase in equal steps of $\frac{2\pi}{n}$. Example: $z^3=8\mathrm{i}=8\mathrm{e}^{\mathrm{i}\pi/2}$ gives $z=2\mathrm{e}^{\mathrm{i}(\pi/6+2k\pi/3)}$, so the roots are $\sqrt3+\mathrm{i}$, $-\sqrt3+\mathrm{i}$ and $-2\mathrm{i}$.

If $w$ is given as $x+\mathrm{i}y$, first convert to $r\mathrm{e}^{\mathrm{i}\theta}$: $r=\sqrt{x^2+y^2}$ and $\theta$ from the quadrant of $w$ (sketch it first). Use $-\pi<\theta\le\pi$ for the principal argument. Worked example: $w=-4+4\sqrt3\,\mathrm{i}$ has $r=8$ and lies in the second quadrant with $\tan^{-1}\sqrt3=\frac{\pi}{3}$, so $\theta=\frac{2\pi}{3}$. Its cube roots are $2\mathrm{e}^{\mathrm{i}(2\pi/9+2k\pi/3)}$, with arguments $\frac{2\pi}{9},\frac{8\pi}{9},\frac{14\pi}{9}$. Give the last as $-\frac{4\pi}{9}$ if the question asks for $-\pi<\theta\le\pi$. To give a root as $x+\mathrm{i}y$, use $x=r\cos\theta$ and $y=r\sin\theta$, keeping exact values such as $\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$.

The solutions of $z^n=1$ are the $n$th roots of unity. Since $1=\mathrm{e}^{2k\pi\mathrm{i}}$, $z=\mathrm{e}^{2k\pi\mathrm{i}/n},\qquad k=0,1,\dots,n-1.$ Write $\omega=\mathrm{e}^{2\pi\mathrm{i}/n}$; then the roots are $1,\omega,\omega^2,\dots,\omega^{n-1}$, each a power of $\omega$, and $\omega^n=1$. Key facts:

• The roots have modulus 1, so they lie on the unit circle.
• Their sum is zero: $1+\omega+\dots+\omega^{n-1}=\frac{1-\omega^n}{1-\omega}=0$ for $n\ge2$.
• They come in conjugate pairs, since $\omega^{n-k}=\overline{\omega^k}$; for example $\omega+\omega^{n-1}=2\cos\frac{2\pi}{n}$.
• For $z^n=a$ with $a$ real and positive, the roots are $a^{1/n}$ times the $n$th roots of unity.

All $n$ roots of $z^n=w$ lie on the circle $|z|=r^{1/n}$, centre the origin, and each is rotated by $\frac{2\pi}{n}$ from the previous one. So they are the vertices of a regular $n$-gon in the Argand diagram (a regular hexagon for $n=6$, an equilateral triangle for $n=3$). The side length is the distance between neighbouring roots. With radius $R$ and angle $\frac{2\pi}{n}$, the cosine rule gives $\text{side}=2R\sin\frac{\pi}{n}$. For $z^6=64$ the radius is 2 and the side is $2\times2\sin\frac{\pi}{6}=2$. Because multiplying by $\omega$ rotates by $\frac{2\pi}{n}$ about the origin, you can find one root and rotate it to generate the rest.

Once the roots are plotted, geometric questions become algebra with complex numbers.

• Lengths: the distance between roots $z_1,z_2$ is $|z_1-z_2|$.
• Sub-shapes: taking every $m$th root gives a smaller regular polygon, e.g. alternate roots of $z^6=64$ ($A_0,A_2,A_4$) form an equilateral triangle.
• Area: split the polygon into $n$ isosceles triangles at the origin, each of area $\frac12R^2\sin\frac{2\pi}{n}$, so the area of the polygon is $\frac{n}{2}R^2\sin\frac{2\pi}{n}$. Example: the triangle $2,\ -1+\sqrt3\mathrm{i},\ -1-\sqrt3\mathrm{i}$ has side $2\sqrt3$ and area $\frac{\sqrt3}{4}(2\sqrt3)^2=3\sqrt3$.