# How do you divide  (-4+5i) / (-1+6i)  in trigonometric form?

$\frac{1}{37} \left(34 + 19 i\right)$

#### Explanation:

We have

$\setminus \frac{- 4 + 5 i}{- 1 + 6 i}$

$= \setminus \frac{\setminus \sqrt{41} {e}^{i \left(\setminus \pi - \setminus {\tan}^{- 1} \left(\frac{5}{4}\right)\right)}}{\setminus \sqrt{37} {e}^{i \left(\setminus \pi - \setminus {\tan}^{- 1} \left(6\right)\right)}}$

$= \setminus \sqrt{\setminus \frac{41}{37}} \left({e}^{i \left(\setminus \pi - \setminus {\tan}^{- 1} \left(\frac{5}{4}\right)\right)} {e}^{- i \left(\setminus \pi - \setminus {\tan}^{- 1} \left(6\right)\right)}\right)$

$= \setminus \sqrt{\setminus \frac{41}{37}} \left({e}^{i \left(\setminus {\tan}^{- 1} \left(6\right) - \setminus {\tan}^{- 1} \left(\frac{5}{4}\right)\right)}\right)$

$= \setminus \sqrt{\setminus \frac{41}{37}} {e}^{i \setminus {\tan}^{- 1} \left(\frac{19}{34}\right)}$

$= \setminus \sqrt{\setminus \frac{41}{37}} \left(\setminus \cos \left(\setminus {\tan}^{- 1} \left(\frac{19}{34}\right)\right) + i \setminus \sin \left(\setminus {\tan}^{- 1} \left(\frac{19}{34}\right)\right)\right)$

$= \setminus \sqrt{\setminus \frac{41}{37}} \left(\frac{34}{\setminus} \sqrt{1517} + i \frac{19}{\setminus} \sqrt{1517}\right)$

$= \setminus \sqrt{\setminus \frac{41}{37 \setminus \times 1517}} \left(34 + 19 i\right)$

$= \frac{1}{37} \left(34 + 19 i\right)$