# How do you divide  (-3-4i)/(5+2i)  in trigonometric form?

Apr 26, 2018

$\frac{5}{\sqrt{29}} \left(\cos \left(0.540\right) + i \sin \left(0.540\right)\right) \approx 0.79 + 0.48 i$

#### Explanation:

$\frac{- 3 - 4 i}{5 + 2 i} = - \frac{3 + 4 i}{5 + 2 i}$

$z = a + b i$ can be written as $z = r \left(\cos \theta + i \sin \theta\right)$, where

• $r = \sqrt{{a}^{2} + {b}^{2}}$
• $\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$

For ${z}_{1} = 3 + 4 i$:
$r = \sqrt{{3}^{2} + {4}^{2}} = 5$
$\theta = {\tan}^{-} 1 \left(\frac{4}{3}\right) = \approx 0 , 927$

For ${z}_{2} = 5 + 2 i$:
$r = \sqrt{{5}^{2} + {2}^{2}} = \sqrt{29}$
$\theta = {\tan}^{-} 1 \left(\frac{2}{5}\right) = \approx 0.381$

For ${z}_{1} / {z}_{2}$:
${z}_{1} / {z}_{2} = {r}_{1} / {r}_{2} \left(\cos \left({\theta}_{1} - {\theta}_{2}\right) + i \sin \left({\theta}_{1} - {\theta}_{2}\right)\right)$

${z}_{1} / {z}_{2} = \frac{5}{\sqrt{29}} \left(\cos \left(0.921 - 0.381\right) + i \sin \left(0.921 - 0.381\right)\right)$

${z}_{1} / {z}_{2} = \frac{5}{\sqrt{29}} \left(\cos \left(0.540\right) + i \sin \left(0.540\right)\right) = 0.79 + 0.48 i$

Proof:
$- \frac{3 + 4 i}{5 + 2 i} \cdot \frac{5 - 2 i}{5 - 2 i} = - \frac{15 + 20 i - 6 i + 8}{25 + 4} = \frac{23 + 14 i}{29} = 0.79 + 0.48 i$