# How do you divide  (-4+5i) / (2+i)  in trigonometric form?

$\frac{- 4 + 5 i}{2 + i}$
in trigonometric form is
$\frac{r 1}{r 2} c i s \left(\theta 1 - \theta 2\right)$
where,
$r 1 = \sqrt{\left({\left(- 4\right)}^{2} + {5}^{2}\right)}$
$\theta 1 = {\tan}^{-} 1 \left(\frac{5}{- 4}\right)$
and
$r 2 = \sqrt{\left({2}^{2} + {1}^{2}\right)}$
$\theta 1 = {\tan}^{-} 1 \left(\frac{1}{2}\right)$

#### Explanation:

Divide $\frac{- 4 + 5 i}{2 + i}$
in trigonometric form
Expressing numerator and denominator as
$z = r c i s \theta$
For $z 1 = - 4 + 5 i$
$r 1 = \sqrt{\left({\left(- 4\right)}^{2} + {5}^{2}\right)}$
$\theta 1 = {\tan}^{-} 1 \left(\frac{5}{- 4}\right)$
For $z 2 = 2 + i$
$r 2 = \sqrt{\left({2}^{2} + {1}^{2}\right)}$
$\theta 1 = {\tan}^{-} 1 \left(\frac{1}{2}\right)$
Thus,
$\frac{- 4 + 5 i}{2 + i} = \frac{r 1 c i s \theta 1}{r 2 c i s \theta 2}$
$= \frac{r 1}{r 2} \frac{c i s \theta 1}{c i s \theta 2}$
By De-Moivre's theorem,
$\frac{c i s \theta 1}{c i s \theta 2} = c i s \left(\theta 1 - \theta 2\right)$
Now,
$\frac{- 4 + 5 i}{2 + i} = \frac{r 1}{r 2} c i s \left(\theta 1 - \theta 2\right)$