# Find 4 values of following in exponential forms ?

## Find 4 values of following in exponential forms ? ${\left(\sqrt{3} - i\right)}^{\frac{1}{4}}$

Jan 17, 2018

See below

#### Explanation:

let
$\sqrt{3} - i = {x}^{4}$

${x}^{4} = 2 \left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right)$

${x}^{4} = 2 \left[\cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)\right]$

${x}^{4} = 2 \left[\cos \left(2 m \pi - \frac{\pi}{6}\right) + i \sin \left(2 m \pi - \frac{\pi}{6}\right)\right]$

(This small addition will not the change the value of the equation. Check out yourself.)

$x = {2}^{\frac{1}{4}} {\left[c i s \left(2 m \pi - \frac{\pi}{6}\right)\right]}^{\frac{1}{4}}$

Short form of writing $\cos \theta + i \sin \theta = c i s \theta$

$x = {2}^{\frac{1}{4}} \left[c i s \left(\frac{2 m \pi}{4} - \left(\frac{\pi}{6}\right) \cdot \frac{1}{4}\right)\right]$ (applying Demoveries theorm)

$x = {2}^{\frac{1}{4}} \left[c i s \left(\frac{m \pi}{2} - \frac{\pi}{24}\right)\right]$

Since the equation was of degree 4, there will be only four roots.

1st root
putting m =0

$x = {2}^{\frac{1}{4}} \left[c i s \left(- \frac{\pi}{24}\right)\right] = {2}^{\frac{1}{4}} \cdot {e}^{- i \frac{\pi}{24}}$

2nd root
putting m =1

$x = {2}^{\frac{1}{4}} \left[c i s \left(\frac{11 \pi}{24}\right)\right] = {2}^{\frac{1}{4}} \cdot {e}^{i \frac{11 \pi}{24}}$

3rd root
putting m =2

x = 2^(1/4)[cis((23pi)/24)] = 2^(1/4)*e^(i(23pi)/(24)

4th root
putting m =3

$x = {2}^{\frac{1}{4}} \left[c i s \left(\frac{35 \pi}{24}\right)\right] = {2}^{\frac{1}{4}} \cdot {e}^{i \frac{35 \pi}{24}}$