How do you divide ( 2i -7) / (- 5 i -8 ) in trigonometric form?

Apr 25, 2018

$0.51 - 0.58 i$

Explanation:

We have $z = \frac{- 7 + 2 i}{- 8 - 5 i} = \frac{7 - 2 i}{8 + 5 i}$

For $z = a + b i$, $z = r \left(\cos \theta + i \sin \theta\right)$, where:

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

For $7 - 2 i$:

$r = \sqrt{{7}^{2} + {2}^{2}} = \sqrt{53}$
$\theta = {\tan}^{-} 1 \left(- \frac{2}{7}\right) \approx - {0.28}^{c}$, however $7 - 2 i$ is in quadrant 4 and so must add $2 \pi$ to it to make it positive, also $2 \pi$ would be going around a circle back.

$\theta = {\tan}^{-} 1 \left(- \frac{2}{7}\right) + 2 \pi \approx {6}^{c}$

For $8 + 5 i$:
$r = \sqrt{{8}^{2} + {5}^{2}} = \sqrt{89}$
$\theta = {\tan}^{-} 1 \left(\frac{5}{8}\right) \approx {0.56}^{c}$

When we have ${z}_{1} / {z}_{1}$ in trig form, we do r_1/r_1(cos(theta_1-theta_2)+isin(theta_1-theta_2)

z_1/z_2=sqrt53/sqrt89(cos(6-0.56)+isin(6-0.56))=sqrt4717/89(cos( 5.44)+isin(5.44))=0.51-0.58i

Proof:
$\frac{7 - 2 i}{8 + 5 i} \cdot \frac{8 - 5 i}{8 - 5 i} = \frac{56 - 51 i - 10}{64 + 25} = \frac{46 - 51 i}{89} = 0.52 - 0.57$