# How do you convert sqrt3 - sqrt3i to polar form?

##### 1 Answer
Oct 22, 2016

The polar form is $\sqrt{3} \left(\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right)$

#### Explanation:

Let $z = \sqrt{3} - i \sqrt{3}$

Then the modulus of z is ∣z∣=sqrt((sqrt3)^2+ (sqrt3)^2) =sqrt6
Rewriting z

z=∣z∣(sqrt3-isqrt3)=sqrt6((sqrt3/sqrt6)-i(sqrt3/sqrt6))

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

Comparing this to the standard form $z = r \left(\cos \theta + i \sin \theta\right)$

We see that $\cos \theta = \frac{1}{\sqrt{2}}$

and $\sin \theta = - \frac{1}{\sqrt{2}}$
This can occur in the 4th quadrant and $\theta = - \frac{\pi}{4}$

And finally we have $z = \sqrt{6} \left(\cos \left(- \frac{\pi}{4}\right) + \sin \left(- \frac{\pi}{4}\right)\right)$

You can also write the resultas $z = \sqrt{6} {e}^{- i \frac{\pi}{4}}$