# How do I find the trigonometric form of the complex number 3-3sqrt3 i?

Apr 29, 2018

In trigonometric form: $6 \left(\cos 5.236 + i \sin 5.236\right)$

#### Explanation:

Let Z=a+ib ; Z=3- 3 sqrt 3i ; a=3 ,b = - 3sqrt3 ;

$Z$ is in $4$ th quadrant. Modulus $| Z | = \sqrt{{a}^{2} + {b}^{2}}$

$= \left(\sqrt{{3}^{2} + {\left(- 3 \sqrt{3}\right)}^{2}}\right) = \sqrt{36} = 6$

$\tan \alpha = | \frac{b}{a} | = \frac{3 \sqrt{3}}{3} \mathmr{and} \tan \alpha = \sqrt{3}$

$\alpha = {\tan}^{-} 1 \left(\sqrt{3}\right) \approx 1.0472$

$\theta$ is on $4$th quadrant $\therefore \theta = 2 \pi - 1.0472 \approx 5.236 \therefore$

Argument , $\theta = 5.236 \therefore$ In trigonometric form expressed as

$| Z | \left(\cos \theta + i \sin \theta\right) = 6 \left(\cos 5.236 + i \sin 5.236\right)$ [Ans]