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

##### 1 Answer
Jun 16, 2017

Division in trigonometric form is:
$\frac{{r}_{1} \left(\cos \left({\theta}_{1}\right) + i \sin \left({\theta}_{1}\right)\right)}{{r}_{2} \left(\cos \left({\theta}_{2}\right) + i \sin \left({\theta}_{2}\right)\right)} = {r}_{1} / {r}_{2} \left(\cos \left({\theta}_{1} - {\theta}_{2}\right) + i \sin \left({\theta}_{1} - {\theta}_{2}\right)\right)$

#### Explanation:

Given: $\frac{2 i - 7}{- 2 i - 8}$

${r}_{1} = \sqrt{{\left(- 7\right)}^{2} + {2}^{2}}$

${r}_{1} = \sqrt{53}$

To find the value of ${\theta}_{1}$, we must observe that the real part is negative and the imaginary part is positive; this places the angle in the 2nd quadrant:

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

${\theta}_{1} = \pi - {\tan}^{-} 1 \left(\frac{2}{7}\right)$

Moving on to ${r}_{2}$:

${r}_{2} = \sqrt{{\left(- 8\right)}^{2} + {\left(- 2\right)}^{2}}$

${r}_{2} = \sqrt{68}$

To find the value of ${\theta}_{2}$, we must observe that the real part is negative and the imaginary part is negative; this places the angle in the 3nd quadrant:

${\theta}_{2} = \pi + {\tan}^{-} 1 \left(\frac{- 2}{-} 8\right)$

${\theta}_{2} = \pi + {\tan}^{-} 1 \left(\frac{1}{4}\right)$

$\frac{2 i - 7}{- 2 i - 8} = \sqrt{\frac{53}{68}} \left(\cos \left(- {\tan}^{-} 1 \left(\frac{2}{7}\right) - {\tan}^{-} 1 \left(\frac{1}{4}\right)\right) + i \sin \left(- {\tan}^{-} 1 \left(\frac{2}{7}\right) - {\tan}^{-} 1 \left(\frac{1}{4}\right)\right)\right)$