# How do you divide ( 3i-8) / (-i +7 ) in trigonometric form?

Mar 31, 2018

$\frac{- 59}{50} + \frac{29}{50} i$

#### Explanation:

Let's convert each to trigonometric form:

Numerator: $3 i - 8 = - 8 + 3 i$
Magnitude: ${r}_{N} = \sqrt{{\left(- 8\right)}^{2} + {3}^{2}} = \sqrt{73}$
Angle: ${\theta}_{N} = \arctan \left(\frac{3}{-} 8\right) = - \arctan \left(\frac{3}{8}\right)$

Denominator: $- i + 7 = 7 - i$
Magnitude: ${r}_{D} = \sqrt{{7}^{2} + {\left(- 1\right)}^{2}} = \sqrt{50} = 5 \sqrt{2}$
Angle: ${\theta}_{D} = \arctan \left(- \frac{1}{7}\right) = - \arctan \left(\frac{1}{7}\right)$

Therefore, the first equation can be written as
$\frac{- 8 + 3 i}{7 - i} = \frac{{r}_{N} {e}^{i {\theta}_{N}}}{{r}_{D} {e}^{i {\theta}_{D}}} = {r}_{N} / {r}_{D} {e}^{i \left({\theta}_{N} - {\theta}_{D}\right)}$

By Euler's formula, we get
$= \sqrt{\frac{73}{50}} \left[\cos \left({\theta}_{N} - {\theta}_{D}\right) + i \sin \left({\theta}_{N} - {\theta}_{D}\right)\right]$
$= \sqrt{\frac{73}{50}} \left[\cos \left({\theta}_{N}\right) \cos \left({\theta}_{D}\right) + \sin \left({\theta}_{N}\right) \sin \left({\theta}_{D}\right)\right] + i \sqrt{\frac{73}{50}} \left[\sin \left({\theta}_{N}\right) \cos \left({\theta}_{D}\right) - \cos \left({\theta}_{N}\right) \sin \left({\theta}_{D}\right)\right]$

We know the cosines and sines of these angles based on the original numbers:

$\cos \left({\theta}_{N}\right) = \frac{- 8}{\sqrt{73}} \setminus \setminus \setminus \setminus \setminus \setminus \sin \left({\theta}_{N}\right) = \frac{3}{\sqrt{73}}$
$\cos \left({\theta}_{D}\right) = \frac{7}{\sqrt{50}} \setminus \setminus \setminus \setminus \setminus \setminus \sin \left({\theta}_{D}\right) = \frac{- 1}{\sqrt{50}}$

Hence,

$= \sqrt{\frac{73}{50}} \left[\frac{- 8}{\sqrt{73}} \cdot \frac{7}{\sqrt{50}} + \frac{3}{\sqrt{73}} \cdot \frac{- 1}{\sqrt{50}}\right] + i \sqrt{\frac{73}{50}} \left[\frac{3}{\sqrt{73}} \cdot \frac{7}{\sqrt{50}} + \frac{- 8}{\sqrt{73}} \cdot \frac{- 1}{\sqrt{50}}\right]$
$= \frac{1}{50} \left[- 56 - 3\right] + \frac{i}{50} \left[21 + 8\right] = \frac{- 59}{50} + \frac{29}{50} i$

We can check this with a Cartesian calculation:
$\frac{- 8 + 3 i}{7 - i} \cdot \frac{7 + i}{7 + i} = \frac{- 56 + 21 i - 3 + 8 i}{50} = \frac{- 59}{50} + \frac{29}{50} i$
as expected.