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

Sep 30, 2016

$- 6 + 6 i = 6 \sqrt{2} \left(\cos \left(\frac{3}{4} \pi\right) + i \sin \left(\frac{3}{4} \pi\right)\right)$

The general $\theta \in {Q}_{2}$ is $2 k \pi + \frac{3}{4} \pi , k = 0 , \pm 1 , | = 2 , \pm 3 , \ldots$

#### Explanation:

The conversion formula for z=x+iy=(x, y) is

$\left(x , y\right) = r \left(\cos \theta , \sin \theta\right)$, giving

$r = \sqrt{{x}^{2} + {y}^{2}}$ (the principal square root) $\ge 0$,

$\cos \theta = \frac{x}{r} \mathmr{and} \sin \theta = \frac{y}{r}$

Here, $x = - 6 \mathmr{and} y = 6$. So,

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

$\cos \theta = - \frac{6}{6 \sqrt{2}} = - \frac{1}{\sqrt{2}} \mathmr{and} \sin \theta = \frac{6}{6 \sqrt{2}} = \frac{1}{\sqrt{2}}$.

Note that there is no common principal value for both.

In $\left(0 , 2 \pi\right)$, the value is $\frac{3}{4} \pi \in {Q}_{2}$., wherein

sine is positive and cosine is negative..

And so,

$- 6 + 6 i = 6 \sqrt{2} \left(\cos \left(\frac{3}{4} \pi\right) + i \sin \left(\frac{3}{4} \pi\right)\right)$

The general $\theta \in {Q}_{2}$ is $2 k \pi + \frac{3}{4} \pi , k = 0 , \pm 1 , | = 2 , \pm 3 , \ldots$