# How do you convert 2 - 2i to polar form?

Aug 4, 2016

$\left(2 , - 2\right) \to \left(2 \sqrt{2} , - \frac{\pi}{4}\right)$

#### 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 = 2 and y = -2

$\Rightarrow r = \sqrt{{2}^{2} + {\left(- 2\right)}^{2}} = \sqrt{8} = 2 \sqrt{2}$

Now 2 - 2i is in the 4th quadrant, hence we must ensure that $\theta$ is in the 4th quadrant.

$\theta = {\tan}^{-} 1 \left(\frac{- 2}{2}\right) = {\tan}^{-} 1 \left(- 1\right) = - \frac{\pi}{4} \text{ in 4th quadrant}$

$\Rightarrow 2 - 2 i = \left(2 , - 2\right) \to \left(2 \sqrt{2} , - \frac{\pi}{4}\right)$