# How do you divide ( 2i-9) / ( -7 i + 7 ) in trigonometric form?

Apr 27, 2018

$z = a + b i$ = $r \left[z\right] {e}^{i \cdot \arctan \left(\frac{b}{a}\right)}$ by definition,
for $a , b , r \left[z\right] \in \mathbb{R}$, $i = \sqrt{- 1}$ , and $r \left[z\right] = \sqrt{{a}^{2} + {b}^{2}}$
so for complex numbers $u = v + i w \mathmr{and} z = a + b i$,
$\left(\frac{u}{z}\right)$ =$\frac{r \left[u\right]}{r \left[z\right]}$ e^(i(arctan(w/v)-arctan(b/a))

In this case:
$u = 2 i - 9$ = $\sqrt{{2}^{2} + {9}^{2}} \cdot {e}^{i \cdot \arctan \left(- \frac{2}{9}\right)} = 9.2195 {e}^{- .2187 i}$
$z = 7 - 7 i = \sqrt{2 \cdot \left({7}^{2}\right)} \cdot {e}^{i \cdot \arctan \left(- 1\right)} = 9.899 {e}^{- .25 i \pi}$

So:
$\left(\frac{2 i - 9}{- 7 i + 7}\right)$ = $\left(\frac{9.2195}{9.899}\right)$ e^(i*(-.2187-.25*pi)
= $0.9314 \cdot {e}^{- 1.0041 i}$