# How do you divide (i+12) / (9i-10) in trigonometric form?

Apr 3, 2017

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

#### Explanation:

Compute ${r}_{1}$:

${r}_{1} = \sqrt{{12}^{2} + {1}^{2}}$

${r}_{1} = \sqrt{145}$

Compute ${\theta}_{1}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{1}{12}\right)$

Compute ${r}_{2}$:

${r}_{2} = \sqrt{- {10}^{2} + {9}^{2}}$

${r}_{2} = \sqrt{181}$

Compute ${\theta}_{2}$ (second quadrant)

${\theta}_{2} = {\tan}^{-} 1 \left(\frac{9}{-} 10\right) + \pi$

The resulting division is:

$\sqrt{\frac{145}{181}} \left(\cos \left({\tan}^{-} 1 \left(\frac{1}{12}\right) - {\tan}^{-} 1 \left(\frac{9}{-} 10\right) - \pi\right) + i \sin \left({\tan}^{-} 1 \left(\frac{1}{12}\right) - {\tan}^{-} 1 \left(\frac{9}{-} 10\right) - \pi\right)\right)$