# How do you simplify (1−(sqrt3)*i)^(1/2)?

Dec 11, 2015

Apply the identity
${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$

to find
${\left(1 - \sqrt{3} i\right)}^{\frac{1}{2}} = \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2} i$

#### Explanation:

We will be using the identity

${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$

First, we can make some slight modifications to our original value to make it easier to find $\theta$.

${\left(1 - \sqrt{3} i\right)}^{\frac{1}{2}} = {\left(2 \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right)\right)}^{\frac{1}{2}}$

and now, using

$\cos \left(- \frac{\pi}{3}\right) = \frac{1}{2}$ and $\sin \left(- \frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2}$

together with the above identity, and that ${\left({x}^{a}\right)}^{b} = {x}^{a b}$, we get

${\left(2 \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right)\right)}^{\frac{1}{2}} = {\left(2 \cos \left(- \frac{\pi}{3}\right) + i \sin \left(- \frac{\pi}{3}\right)\right)}^{\frac{1}{2}}$

$= {\left(2 {e}^{i \left(- \frac{\pi}{3}\right)}\right)}^{\frac{1}{2}}$

$= \sqrt{2} {e}^{i \left(- \frac{\pi}{6}\right)}$

$= \sqrt{2} \left(\cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)\right)$

$= \sqrt{2} \left(\frac{\sqrt{3}}{2} - \frac{1}{2} i\right)$

$= \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2} i$