# How do you convert z= i -1/cos (pi/3) + isin(pi/3) in polar form?

Jun 4, 2017

Given: $z = i - \frac{1}{\cos} \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)$

We know that $- \frac{1}{\cos} \left(\frac{\pi}{3}\right) = - 2$

$z = - 2 + i + i \sin \left(\frac{\pi}{3}\right)$

We know that $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

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

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

$r \approx 2.7$

The angle is in the second quadrant:

$\theta = \pi + {\tan}^{-} 1 \left(- \frac{1}{2} - \frac{\sqrt{3}}{4}\right)$

$\theta \approx 2.39$

$z = 2.7 {e}^{i 2.39}$