# How do you write the complex number in trigonometric form -1+sqrt3i?

Aug 25, 2016

$2 \left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right)$

#### Explanation:

To convert from $\textcolor{b l u e}{\text{complex to trigonometric form}}$

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

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

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

here x = - 1 and y $= \sqrt{3}$

$\Rightarrow r = \sqrt{{\left(- 1\right)}^{2} + {\left(\sqrt{3}\right)}^{2}} = \sqrt{4} = 2$

Now $- 1 + \sqrt{3} i$ is in the 2nd quadrant, so we must ensure that $\theta$ is in the 2nd quadrant.

$\theta = {\tan}^{-} 1 \left(- \sqrt{3}\right) = - \frac{\pi}{3} \text{ in 4th quadrant}$

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

$\Rightarrow - 1 + \sqrt{3} i = 2 \left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right)$