# How do I find the square root of a complex number?

##### 1 Answer
Jan 16, 2015

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.