# How do you divide ( 6i-4) / ( -3 i -5 ) in trigonometric form?

May 20, 2016

$\frac{6 i - 4}{- 3 i - 5} = \sqrt{\frac{26}{17}} \left(\cos \rho + i \sin \rho\right)$ where $\rho = {\tan}^{- 1} 21$

#### Explanation:

Let us first write $\left(6 i - 4\right)$ and $\left(- 3 i - 5\right)$ 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 $6 i - 4 = \left(- 4 + 6 i\right) = \sqrt{{\left(- 4\right)}^{2} + {6}^{2}} \left[\cos \alpha + i \sin \alpha\right]$ or

$\sqrt{52} {e}^{i \alpha}$, where $\tan \alpha = \frac{6}{- 4} = - \frac{3}{2}$ and

$- 3 i - 5 = \left(- 5 - 3 i\right) = \sqrt{{\left(- 5\right)}^{2} + {\left(- 3\right)}^{2}} \left[\cos \beta + i \sin \beta\right]$ or

$\sqrt{34} {e}^{i \beta}$, where $\tan \beta = \frac{- 3}{- 5} = \frac{3}{5}$

Hence $\frac{6 i - 4}{- 3 i - 5} = \frac{\sqrt{52} {e}^{i \alpha}}{\sqrt{34} {e}^{i \beta}} = \sqrt{\frac{26}{17}} {e}^{i \left(\alpha - \beta\right)} = \sqrt{\frac{28}{17}} \left(\cos \left(\alpha - \beta\right) + i \sin \left(\alpha - \beta\right)\right)$

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

= $\frac{- \frac{3}{2} - \frac{3}{5}}{1 + \left(- \frac{3}{2}\right) \cdot \left(\frac{3}{5}\right)} = \frac{- \frac{21}{10}}{1 - \frac{9}{10}} = - \frac{21}{10} \cdot \frac{10}{1} = - 21$

Hence $\frac{6 i - 4}{- 3 i - 5} = \sqrt{\frac{26}{17}} \left(\cos \rho + i \sin \rho\right)$ where $\rho = {\tan}^{- 1} 21$