# How do I find the trigonometric form of the complex number -1-isqrt3?

To represent it in trigonometric form I first need to calculate the modulus $r$ using Pitagora's theorem:
$r = \sqrt{{\left(- 1\right)}^{2} + {\left(- \sqrt{3}\right)}^{2}} = \sqrt{1 + 3} = 2$
Then I need to find the angle $\theta$;
$\theta = \pi + \arctan \left(\sqrt{3}\right) = \frac{4}{3} \pi$
$z = - 1 - i \sqrt{3} = 2 \left[\cos \left(\frac{4}{3} \pi\right) + i \sin \left(\frac{4}{3} \pi\right)\right]$