# How do you convert  -4 + 8i to polar form?

Jul 15, 2017

In polar coordinates, the point will be $\left(r , \theta\right) = \left(6.92 i , - 1.11 i\right)$

#### Explanation:

For coordinates in polar form, $\left(r , \theta\right)$, we need to find the value of $r$ and the value of $\theta$.

We will use Pythagoras' theorem to find $r$ and the $\tan$ function to find $\theta$.

$r = \sqrt{{\left(- 4\right)}^{2} + 8 {i}^{2}} = \sqrt{16 - 64} = \sqrt{- 48} = 6.92 i$

(since $8 {i}^{2} = 8 i \times 8 i = 8 \sqrt{- 1} \times 8 \sqrt{- 1} = 64 \times - 1 = - 64$)

$\tan \theta = \frac{o p p}{a \mathrm{dj}} = \frac{8 i}{-} 4 = - \frac{8}{4} i = - 2 i$

So $\theta = {\tan}^{-} 1 \left(- 2 i\right) = - 1.11 i$

Jul 15, 2017

$\left(4 \sqrt{5} , 2.03\right)$

#### Explanation:

$\text{to convert from "color(blue)"cartesian to polar form}$

$\text{that is "(x,y)to(r,theta)" where}$

•color(white)(x)r=sqrt(x^2+y^2)

•color(white)(x)theta=tan^-1(y/x)color(white)(x)-pi< theta <=pi

$\text{here " x=-4" and } y = 8$

$\Rightarrow r = \sqrt{{\left(- 4\right)}^{2} + {8}^{2}} = \sqrt{80} = 4 \sqrt{5}$

$- 4 + 8 i \text{ is in the second quadrant }$

$\text{so we must ensure that "theta" is in the second quadrant}$

$\theta = {\tan}^{-} 1 \left(2\right) = 1.11 \leftarrow \textcolor{red}{\text{ related acute angle}}$

$\Rightarrow \theta = \left(\pi - 1.11\right) = 2.03 \leftarrow \textcolor{red}{\text{ in second quadrant}}$

$\Rightarrow - 4 + 8 i \to \left(- 4 , 8\right) \to \left(4 \sqrt{5} , 2.03\right)$