# How do you divide ( 2i-6) / ( -12i +4 ) in trigonometric form?

May 19, 2016

$\left(2 i - 6\right) \div \left(- 12 i + 4\right) = \frac{1}{2} \left(\cos \rho + i \sin \rho\right)$ where $\rho = {\tan}^{- 1} \left(- \frac{5}{3}\right)$

#### Explanation:

Let us first write $\left(2 i - 6\right)$ and $\left(- 12 i + 4\right)$ in trigonometric form.

$a + i b$ can be written in trigonometric form $r {e}^{i \theta} = r \cos \theta + i r \sin \theta = r \left(\cos \theta + i \sin \theta\right)$,
where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\tan \theta = \frac{b}{a}$ or $\theta = \arctan \left(\frac{b}{a}\right)$

Hence $2 i - 6 = \left(- 6 + 2 i\right) = \sqrt{{\left(- 6\right)}^{2} + {2}^{2}} \left[\cos \alpha + i \sin \alpha\right]$ or

$\sqrt{40} {e}^{i \alpha}$, where $\tan \alpha = \frac{- 1}{3}$ and

$- 12 i + 4 = \left(4 - 12 i\right) = \sqrt{{4}^{2} + {\left(- 12\right)}^{2}} \left[\cos \beta + i \sin \beta\right]$ or

$\sqrt{160} {e}^{i \beta}$, where $\tan \beta = \frac{- 12}{4} = - 3$

Hence $\left(2 i - 6\right) \div \left(- 12 i + 4\right) = \frac{\sqrt{40} {e}^{i \alpha}}{\sqrt{160} {e}^{i \beta}} = \sqrt{\frac{1}{4}} {e}^{i \left(\alpha - \beta\right)} = {e}^{i \left(\alpha - \beta\right)} / 2 = \frac{1}{2} \left(\cos \left(\alpha - \beta\right) + i \sin \left(\alpha - \beta\right)\right)$

Now, $\tan \left(\alpha - \beta\right) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

= $\frac{- \frac{1}{3} + \left(- 3\right)}{1 + \left(- \frac{1}{3}\right) \cdot \left(- 3\right)} = \frac{- \frac{10}{3}}{2} = - \frac{5}{3}$

Hence $\left(2 i - 6\right) \div \left(- 12 i + 4\right) = \frac{1}{2} \left(\cos \rho + i \sin \rho\right)$ where $\rho = {\tan}^{- 1} \left(- \frac{5}{3}\right)$