Gary Liang Notes

Free online notes for Year 11 and 12 HSC, VCE and QCE students

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 $$\cos{\frac{\pi}{5}} \cos{\frac{3\pi}{5}} = -\frac{1}{4}.$$