# How do you find the two square roots of 2i?

Oct 26, 2014

Let $z = r \left(\cos \theta + i \sin \theta\right)$ be the square-roots of $2 i$.

${z}^{2} = {\left[r \left(\cos \theta + i \sin \theta\right)\right]}^{2}$

by De Moivre's Theorem,

$= {r}^{2} \left(\cos 2 \theta + i \sin 2 \theta\right) = 2 i = 2 \left(0 + 1 i\right)$

$R i g h t a r r o w {r}^{2} = 2 R i g h t a r r o w r = \sqrt{2}$

Rightarrow{(cos2theta=0),(sin2theta=1):}}Rightarrow2theta=pi/2+2npi Rightarrow theta=pi/4+npi,

where $n$ is any integer.

So,

$z = \left\{\sqrt{2} \left[\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right] , \sqrt{2} \left[\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right]\right\}$

$= \left\{\sqrt{2} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i\right) , \sqrt{2} \left(- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i\right)\right\}$

$= \left\{1 + i , - 1 - i\right\}$

I hope that this was helpful.