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

Oct 21, 2014

Let $z = \sqrt{3} - i$.

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

By factoring out $2$,

$z = 2 \left(\frac{\sqrt{3}}{2} - \frac{1}{2} i\right) = r \left(\cos \theta + i \sin \theta\right)$

by matching the real part and the imaginary part,

$R i g h t a r r o w \left\{\begin{matrix}r = 2 \\ \cos \theta = \frac{\sqrt{3}}{2} \\ \sin \theta = - \frac{1}{2}\end{matrix}\right.$

$R i g h t a r r o w \theta = - \frac{\pi}{6}$

Hence,

$z = 2 \left[\cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)\right]$

since cosine is even and sine is odd, we can also write

$z = 2 \left[\cos \left(\frac{\pi}{6}\right) - i \sin \left(\frac{\pi}{6}\right)\right]$

I hope that this was helpful.