# How do you convert i(6 + 6i)  to polar form?

Aug 19, 2016

$\left(6 \sqrt{2} , \frac{3 \pi}{4}\right)$

#### Explanation:

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

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

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

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

The first step, however, is to distribute the bracket.

$\Rightarrow i \left(6 + 6 i\right) = 6 i + 6 {i}^{2} \text{ and } \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{i}^{2} = {\left(\sqrt{- 1}\right)}^{2} = - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow 6 i + 6 {i}^{2} = - 6 + 6 i$

here x = -6 and y = 6

$\Rightarrow r = \sqrt{{\left(- 6\right)}^{2} + {6}^{2}} = \sqrt{72} = 6 \sqrt{2}$

Now - 6 + 6i is in the 2nd quadrant so we must ensure that $\theta$ is in the 2nd quadrant.

$\theta = {\tan}^{-} 1 \left(\frac{6}{-} 6\right) = {\tan}^{-} 1 \left(- 1\right) = - \frac{\pi}{4} \text{ in 4th quadrant}$

$\Rightarrow \theta = \left(\pi - \frac{\pi}{4}\right) = \frac{3 \pi}{4} \text{ in 2nd quadrant}$

$\Rightarrow - 6 + 6 i = \left(- 6 , 6\right) \to \left(6 \sqrt{2} , \frac{3 \pi}{4}\right) \text{ in polar form}$