# How do you divide  (-3+5i)/(2-i)  in trigonometric form?

$\frac{- 3 + 5 i}{2 - i} = \frac{1}{5} \sqrt{170} \left(\cos 2.57 + i \sin 2.57\right)$
$\frac{- 3 + 5 i}{2 - i} = \frac{\left(- 3 + 5 i\right) \left(2 + i\right)}{\left(2 - i\right) \left(2 + i\right)} = \frac{- 11 + 7 i}{5} = - \frac{11}{5} + \frac{7}{5} i$
Use mod-arg form, $a + b i = r \left(\cos \vartheta + i \sin \vartheta\right)$ where $r = \sqrt{- \frac{11}{5} ^ 2 + \frac{7}{5} ^ 2} = \sqrt{\frac{34}{5}} = \frac{1}{5} \sqrt{170}$ and $\vartheta = \arctan \left(\frac{\frac{7}{5}}{- \frac{11}{5}}\right) + \pi = {2.57}^{\text{c}}$