1.2.4 Roots of Complex Numbers

Overview

Finding the roots of complex numbers involves expressing complex numbers in their modulus-argument (polar) form and then applying the properties of complex numbers.
This process is based on de Moivre's Theorem, which makes it easier to calculate the roots of complex numbers.

If $z = r(\cos \theta + i \sin \theta)$ is a complex number in polar form, then its nth roots are given by:

z^{\frac{1}{n}} = r^{\frac{1}{n}} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right)

for $k = 0, 1, 2, \dots, n-1$ , where:

$r$ is the modulus,
$\theta$ is the argument of the complex number,
$k$ represents the different values that provide all possible roots. Each different value of $k$ gives a different nth root of the complex number, and there will be $n$ distinct roots.

infoNote

Question: Let's find the square roots of $z = 4 + 4i$

Step 1: Express $z$ in modulus-argument form.

Modulus $r$ :

r = |z| = \sqrt{4^2 + 4^2}

= \sqrt{16 + 16} = \sqrt{32}

= 4\sqrt{2}

Argument $\theta$ :

\theta = \tan^{-1}\left( \frac{4}{4} \right) = \frac{\pi}{4} \text{ radians}

So:

z = 4\sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)

Step 2: Apply the formula for square roots $n=2$

The two square roots are given by:

z^{\frac{1}{2}} = \left( 4\sqrt{2} \right)^{\frac{1}{2}} \left( \cos \frac{\frac{\pi}{4} + 2k\pi}{2} + i \sin \frac{\frac{\pi}{4} + 2k\pi}{2} \right)

For $k=0$

z_0 = \sqrt{4\sqrt{2}} \left( \cos \frac{\pi}{8} + i \sin \frac{\pi}{8} \right)

For $k=1$

z_1 = \sqrt{4\sqrt{2}} \left( \cos \frac{9\pi}{8} + i \sin \frac{9\pi}{8} \right)

Step 3: Simplify the modulus and trigonometric expressions.

The square roots of $z = 4 + 4i$ are the two complex numbers:

z_0 = 2 \left( \cos \frac{\pi}{8} + i \sin \frac{\pi}{8} \right)

z_1 = 2 \left( \cos \frac{9\pi}{8} + i \sin \frac{9\pi}{8} \right)

infoNote

Question: Find the cube roots of $z = 8$

Step 1: Express $z=8$ in polar form.

Since $z$ is real, it can be written as:

z = 8 \left( \cos 0 + i \sin 0 \right)

Step 2: Apply the formula for cube roots $n=3$

The three cube roots are:

z^{\frac{1}{3}} = 8^{\frac{1}{3}} \left( \cos \frac{0 + 2k\pi}{3} + i \sin \frac{0 + 2k\pi}{3} \right)

For $k = 0, 1, 2$

z_0 = 2 (\cos 0 + i \sin 0) = 2

z_1 = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) = -1 + i\sqrt{3}

z_2 = 2 \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) = -1 - i\sqrt{3}

Thus, the cube roots of $8$ are $2, -1 + i\sqrt{3}, -1 - i\sqrt{3}$

infoNote

z^{\frac{1}{n}} = r^{\frac{1}{n}} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right)

where $r$ is the modulus and $\theta$ is the argument.

Each value of $k = 0, 1, 2, \dots, n-1$ gives a distinct root.
These roots are evenly spaced on the Argand diagram, forming a regular polygon around the origin.