# How do you convert -3-3sqrt(3)i to polar form?

Aug 12, 2016

$\left(- 3 , - 3 \sqrt{3}\right) \to \left(6 , - \frac{2 \pi}{3}\right)$

#### Explanation:

To convert from $\textcolor{b l u e}{\text{cartesian to polar form}}$

That is $\left(x , y\right) \to \left(r , \theta\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{r = \sqrt{{x}^{2} + {y}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

here x = - 3 and $y = - 3 \sqrt{3}$

$\Rightarrow r = \sqrt{{\left(- 3\right)}^{2} + {\left(- 3 \sqrt{3}\right)}^{2}} = \sqrt{9 + 27} = 6$

Now $- 3 - 3 \sqrt{3}$ is in the 3rd quadrant hence we must ensure that $\theta$ is in the 3rd quadrant.

$\Rightarrow \theta = {\tan}^{-} 1 \left(\frac{- 3 \sqrt{3}}{- 3}\right) = {\tan}^{-} 1 \left(\sqrt{3}\right) = \frac{\pi}{3}$

However, $\frac{\pi}{3}$ is in the 1st quadrant and we require the 3rd.

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{- \pi < \theta \le \pi} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow \theta = - \pi + \frac{\pi}{3} = - \frac{2 \pi}{3}$

$\Rightarrow - 3 - 3 \sqrt{3} i = \left(- 3 , - 3 \sqrt{3}\right) \to \left(6 , - \frac{2 \pi}{3}\right)$