# How do you write the complex number 6-8i in polar form?

Dec 12, 2016

$\left(10 , - 0.927\right)$

#### Explanation:

To express a $\textcolor{b l u e}{\text{complex number in polar form}}$

$\text{that is } \left(x , y\right) \to \left(r , \theta\right)$

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

and $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{ where } - \pi < \theta \le \pi$

$\text{here " x=6" and } y = - 8$

$\Rightarrow r = \sqrt{{6}^{2} + {\left(- 8\right)}^{2}} = \sqrt{36 + 64} = 10$

Now, 6 - 8i is in the 4th quadrant so we must ensure that theta" is in the 4th quadrant.

$\theta = {\tan}^{-} 1 \left(- \frac{8}{6}\right) = - 0.927 \leftarrow \text{ in 4th quadrant}$

$\Rightarrow \left(6 , - 8\right) \to \left(10 , - 0.927\right) \to \left(10 , - {53.13}^{\circ}\right)$