# How do you write (2)-(2isqrt3) in trigonometric form?

Aug 26, 2016

$4 \left(\cos \left(\frac{\pi}{3}\right) - i \sin \left(\frac{\pi}{3}\right)\right)$

#### Explanation:

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

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

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

here x = 2 and $y = - 2 \sqrt{3}$

$\Rightarrow r = \sqrt{{2}^{2} + {\left(2 \sqrt{3}\right)}^{2}} = \sqrt{4 + 12} = \sqrt{16} = 4$

Now $2 - 2 \sqrt{3} i$ is in the 4th quadrant so we must ensure that $\theta$ is in the 4th quadrant.

$\theta = {\tan}^{-} 1 \left(\frac{- 2 \sqrt{3}}{2}\right) = {\tan}^{-} 1 \left(- \sqrt{3}\right)$

$= - \frac{\pi}{3} \text{ in 4th quadrant}$

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

which can also be expressed as $4 \left(\cos \left(\frac{\pi}{3}\right) - i \sin \left(\frac{\pi}{3}\right)\right)$

Since $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\cos \left(- \theta\right) = \cos \theta \text{ and } \sin \left(- \theta\right) = - \sin \theta} \textcolor{w h i t e}{\frac{a}{a}} |}}}$