# How do you write the complex number in trigonometric form 4-4sqrt3i?

Mar 18, 2017

$z = 8 \left[\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right]$

#### Explanation:

$\left(x , y\right) \to r \left(\cos \theta + i \sin \theta\right)$

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

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

$\text{here "x=4" and } y = - 4 \sqrt{3}$

$\Rightarrow r = \sqrt{{\left(4\right)}^{2} + {\left(- 4 \sqrt{3}\right)}^{2}} = \sqrt{64} = 8$

$\text{now } 4 - 4 \sqrt{3}$ is in the 2nd quadrant so we must ensure that
$\theta$ is in the 2nd quadrant.

$\Rightarrow {\tan}^{-} 1 \left(\frac{4 \sqrt{3}}{4}\right) = {\tan}^{-} 1 \left(\sqrt{3}\right) = \frac{\pi}{3}$

$\Rightarrow \theta = \left(\pi - \frac{\pi}{3}\right) = \frac{2 \pi}{3} \leftarrow \textcolor{red}{\text{ in 2nd quadrant}}$

$\Rightarrow 4 - 4 \sqrt{3} i = 8 \left[\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right]$