# Roots of Complex Numbers

## Key Questions

• To evaluate the $n t h$ root of a complex number I would first convert it into trigonometric form:
$z = r \left[\cos \left(\theta\right) + i \sin \left(\theta\right)\right]$
and then use the fact that:
${z}^{n} = {r}^{n} \left[\cos \left(n \cdot \theta\right) + i \sin \left(n \cdot \theta\right)\right]$
and:
$n \sqrt{z} = {z}^{\frac{1}{n}} = {r}^{\frac{1}{n}} \cdot \left[\cos \left(\frac{\theta + 2 k \pi}{n}\right) + i \sin \left(\frac{\theta + 2 k \pi}{n}\right)\right]$
Where $k = 0. . n - 1$

For example: consider $z = 2 + 3.46 i$ and let us try $\sqrt{z}$;
$z$ can be written as:
$z = 4 \left[\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right]$
So:
$k = 0$
$\sqrt{z} = {z}^{\frac{1}{2}} = {4}^{\frac{1}{2}} \left[\cos \left(\frac{\frac{\pi}{3} + 0}{2}\right) + i \sin \left(\frac{\frac{\pi}{3} + 0}{2}\right)\right] =$
=2[cos(pi/6)+isin(pi/6))]
And:
$k = n - 1 = 2 - 1 = 1$
$\sqrt{z} = {z}^{\frac{1}{2}} = {4}^{\frac{1}{2}} \left[\cos \left(\frac{\frac{\pi}{3} + 2 \pi}{2}\right) + i \sin \left(\frac{\frac{\pi}{3} + 2 \pi}{2}\right)\right] =$
=2[cos(7pi/6)+isin(7pi/6))]
Which gives, in total, two solutions.

If you express your complex number in polar form as $r \left(\cos \theta + i \sin \theta\right)$, then it has fourth roots:

$\alpha = \sqrt{r} \left(\cos \left(\frac{\theta}{4}\right) + i \sin \left(\frac{\theta}{4}\right)\right)$, $i \alpha$, $- \alpha$ and $- i \alpha$

#### Explanation:

Given $a + i b$, let $r = \sqrt{{a}^{2} + {b}^{2}}$, $\theta = \text{atan2} \left(b , a\right)$

Then $a + i b = r \left(\cos \theta + i \sin \theta\right)$

This has one $4 t h$ root $\alpha = \sqrt{r} \left(\cos \left(\frac{\theta}{4}\right) + i \sin \left(\frac{\theta}{4}\right)\right)$

There are three other $4 t h$ roots: $i \alpha$, $- \alpha$ and $- i \alpha$

• A root of unity is a complex number that when raised to some positive integer will return 1.

It is any complex number $z$ which satisfies the following equation:

${z}^{n} = 1$

where $n \in \mathbb{N}$, which is to say that n is a natural number. A natural number is any positive integer: (n = 1, 2, 3, ...). This is sometimes referred to as a counting number and the notation for it is $\mathbb{N}$.

For any $n$, there may be multiple $z$ values that satisfy that equation, and those values comprise the roots of unity for that n.

When $n = 1$
Roots of unity: $1$

When $n = 2$
Roots of unity: $- 1 , 1$

When $n = 3$
Roots of unity = $1 , \frac{1 + \sqrt{3} i}{2} , \frac{1 - \sqrt{3} i}{2}$

When $n = 4$
Roots of unity = $- 1 , i , 1 , - i$

• To evaluate the square root (and in general any root) of a complex number I would first convert it into trigonometric form:
$z = r \left[\cos \left(\theta\right) + i \sin \left(\theta\right)\right]$
and then use the fact that:
${z}^{n} = {r}^{n} \left[\cos \left(n \cdot \theta\right) + i \sin \left(n \cdot \theta\right)\right]$

Where, in our case, $n = \frac{1}{2}$ (remembering that $\sqrt{x} = {x}^{\frac{1}{2}}$).
To evaluate the $n t h$ root of a complex number I would write:

$n \sqrt{z} = {z}^{\frac{1}{n}} = {r}^{\frac{1}{n}} \cdot \left[\cos \left(\frac{\theta + 2 k \pi}{n}\right) + i \sin \left(\frac{\theta + 2 k \pi}{n}\right)\right]$
Where $k = 0. . n - 1$

For example: consider $z = 2 + 3.46 i$ and let us try $\sqrt{z}$;
$z$ can be written as:
$z = 4 \left[\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right]$
So:
$k = 0$
$\sqrt{z} = {z}^{\frac{1}{2}} = {4}^{\frac{1}{2}} \left[\cos \left(\frac{\frac{\pi}{3} + 0}{2}\right) + i \sin \left(\frac{\frac{\pi}{3} + 0}{2}\right)\right] =$
=2[cos(pi/6)+isin(pi/6))]
And:
$k = n - 1 = 2 - 1 = 1$
$\sqrt{z} = {z}^{\frac{1}{2}} = {4}^{\frac{1}{2}} \left[\cos \left(\frac{\frac{\pi}{3} + 2 \pi}{2}\right) + i \sin \left(\frac{\frac{\pi}{3} + 2 \pi}{2}\right)\right] =$
=2[cos(7pi/6)+isin(7pi/6))]
Which gives, in total, two solutions.