# What is the square root of 2i?

Sep 28, 2014

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

Let us look at some details.

Let $z = \sqrt{2 i}$.
(Note that $z$ are complex numbers.)

by squaring,

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

by using the exponential form $z = r {e}^{i \theta}$,

$R i g h t a r r o w {r}^{2} {e}^{i \left(2 \theta\right)} = 2 i = 2 {e}^{i \left(\frac{\pi}{2} + 2 n \pi\right)}$

Rightarrow {(r^2=2 Rightarrow r=sqrt{2}), (2theta=pi/2+2npi Rightarrow theta=pi/4+npi):}

So, $z = \sqrt{2} {e}^{i \left(\frac{\pi}{4} + n \pi\right)}$

by Eular's Formula: ${e}^{i \theta} = \cos \theta + i \sin \theta$

$R i g h t a r r o w z = \sqrt{2} \left[\cos \left(\frac{\pi}{4} + n \pi\right) + i \sin \left(\frac{\pi}{4} + n \pi\right)\right]$

$= \sqrt{2} \left(\pm \frac{1}{\sqrt{2}} \pm \frac{1}{\sqrt{2}} i\right) = \pm 1 \pm i$

I kept the following original post just in case someone needs it.
${\left(2 i\right)}^{\frac{1}{2}}$ = ${\left(2\right)}^{\frac{1}{2}}$ ${\left(i\right)}^{\frac{1}{2}}$,

${\left(i\right)}^{\frac{1}{2}}$ = -1

${\left(2 i\right)}^{\frac{1}{2}}$ = ${\left(2\right)}^{\frac{1}{2}}$ x -1

${\left(2\right)}^{\frac{1}{2}}$ = 1.41

${\left(2 i\right)}^{\frac{1}{2}}$ = 1.41 x -1 = -1.41