# How do you convert 4+4i  to polar form?

Jul 13, 2016

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

#### Explanation:

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

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

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

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

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

Now 4 + 4i is in the 1st quadrant ,hence we must ensure that $\theta$ is an angle in the 1st quadrant.

here x = 4 and y = 4

$\Rightarrow r = \sqrt{{4}^{2} + {4}^{2}} = \sqrt{32} = 4 \sqrt{2}$

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

$\Rightarrow \left(4 , 4\right) = \left(4 \sqrt{2} , \frac{\pi}{4}\right) \text{ in polar form}$