We should have learnt how to find square roots of complex numbers, but we can also find \(n\)th roots of complex numbers, using a form of de Moivre’s theorem.

Recall that de Moivre’s theorem says that for any real number \(\theta\) and integer \(n\), we have $$(\cos\theta + i \sin\theta)^n = \cos{n\theta} + i \sin{n\theta}.$$ We noted that de Moivre’s theorem also works for rational fractions, but it only gives one of many possible values.

**Method to find \(n\)th root of a complex number \(w\). **

- Write down the equation corresponding to the roots i.e. \(z^n=w.\)
- Put \(w\) in polar or exponential form.
- Add \(2k\pi\) to the argument, where \(k\) is any integer.
- Take both sides to the power of \(\frac{1}{n}\) and use de Moivre’s theorem.
- Take \(n\) consecutive value of \(k\) to find the \(n\) roots.

A **root of unity** is any \(n\)th root of 1.

**Worked exercise. **Find the cube roots of unity in the form \(x+yi\) and plot them on the complex plane. What is the geometrical relationship between the cube roots of unity?**Solution. **The cube roots of unity satisfy the equation $$z^3=1 = e^{2k\pi i},$$ where \(k\) is an integer. Taking both sides to the power of \(\frac{1}{3}\), we have $$z= e^{\frac{2k\pi i}{3}},$$ and substituting \(k=-1,0,1\) gives us $$z=1,-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i.$$

The cube roots of unity form an **equilateral triangle**.

In general, the \(n\)th roots of any complex number will be spaced evenly around the origin and form a regular \(n\)-gon.

## Problems

- Find:
- The 4th roots of \(16\)
- The cube roots of \(-8\)
- The 5th roots of \(i\)
- The 6th roots of \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i\)

- Find the seventh roots of unity in polar form, and plot them on the complex plane.
- Suppose that \(\omega\) is a root of the equation \(z^3=1\), where \(\omega\ne1\). Simplify:
- \(\omega^7\)
- \(1+\omega+\omega^2\)
- \(\frac{1}{\omega+\omega^2}\)
- \(\omega^8+\omega^9+\omega^{10}\)
- \(1+\omega^4+\omega^8\)
- \((1+\omega^2)^6\)

- Suppose that \(w^5=1\), where \(w\ne1\).
- Show that \(1+w+w^2+w^3+w^4=0\).
- Reduce the equation to a quadratic equation by dividing by \(w^2\) and letting \(z=w+\frac{1}{w}\).
- By considering the product of roots, show that $$\cos{\frac{\pi}{5}} \cos{\frac{3\pi}{5}} = -\frac{1}{4}.$$