# How do you convert 3 sqrt 2 - 3 sqrt2i to polar form?

Oct 24, 2016

In polar form $3 \sqrt{2} - 3 \sqrt{2} i$ is $6 \left(\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right)$

#### Explanation:

A complex number $a + b i$ can be written in polar form as $r \left(\cos \theta + i \sin \theta\right)$ i.e. $a = r \cos \theta$ and $b = r \sin \theta$.

Squaring and adding the last two, we get ${a}^{2} + {b}^{2} = {r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta$

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

and $\frac{b}{a} = \frac{r \sin \theta}{r \cos \theta} = \tan \theta$

Here in the complex number $3 \sqrt{2} - 3 \sqrt{2} i$, we have $a = 3 \sqrt{2}$ and $b = - 3 \sqrt{2}$ and hence $r = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{{\left(3 \sqrt{2}\right)}^{2} + {\left(- 3 \sqrt{2}\right)}^{2}}$

= $\sqrt{18 + 18} = \sqrt{36} = 6$ and

$\cos \theta = \frac{3 \sqrt{2}}{6} = \frac{1}{\sqrt{2}}$ and $\cos \theta = \frac{- 3 \sqrt{2}}{6} = - \frac{1}{\sqrt{2}}$

i.e. $\theta = \left(- \frac{\pi}{4}\right)$

Hence in polar form $3 \sqrt{2} - 3 \sqrt{2} i$ is $6 \left(\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right)$