# Roots of complex numbers

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 $$w$$.

1. Write down the equation corresponding to the roots i.e. $$z^n=w.$$
2. Put $$w$$ in polar or exponential form.
3. Add $$2k\pi$$ to the argument, where $$k$$ is any integer.
4. Take both sides to the power of $$\frac{1}{n}$$ and use de Moivre’s theorem.
5. 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

1. Find:
1. The 4th roots of $$16$$
2. The cube roots of $$-8$$
3. The 5th roots of $$i$$
4. The 6th roots of $$\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i$$
2. Find the seventh roots of unity in polar form, and plot them on the complex plane.
3. Suppose that $$\omega$$ is a root of the equation $$z^3=1$$, where $$\omega\ne1$$. Simplify:
1. $$\omega^7$$
2. $$1+\omega+\omega^2$$
3. $$\frac{1}{\omega+\omega^2}$$
4. $$\omega^8+\omega^9+\omega^{10}$$
5. $$1+\omega^4+\omega^8$$
6. $$(1+\omega^2)^6$$
4. Suppose that $$w^5=1$$, where $$w\ne1$$.
1. Show that $$1+w+w^2+w^3+w^4=0$$.
2. Reduce the equation to a quadratic equation by dividing by $$w^2$$ and letting $$z=w+\frac{1}{w}$$.
3. By considering the product of roots, show that $$\cos{\frac{\pi}{5}} \cos{\frac{3\pi}{5}} = -\frac{1}{4}.$$