# How do you find the square root of 8i?

Mar 25, 2018

$2 \sqrt{2 i}$

#### Explanation:

$\sqrt{8 i}$

$\sqrt{4 \cdot 2 \cdot i} \rightarrow$ 4 is a perfect square; it can be taken out of the radical

$2 \sqrt{2 i}$

Mar 25, 2018

$\sqrt{8 i} = 2 + 2 i$

#### Explanation:

Note that:

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

Hence by de Moivre's theorem, one square root is:

$\sqrt{8} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right) = 2 \sqrt{2} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i\right) = 2 + 2 i$

This is the principal square root. The other square root is $- 2 - 2 i$