# How do you divide ( 2i-1) / ( 3i +5 ) in trigonometric form?

May 21, 2016

$\frac{2 i - 1}{3 i + 5} = \sqrt{\frac{5}{34}} \left(\cos \rho + i \sin \rho\right)$ where $\rho = {\tan}^{- 1} 13$

#### Explanation:

Let us first write $\left(2 i - 1\right)$ 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 $2 i - 1 = \left(- 1 + 2 i\right) = \sqrt{{\left(- 1\right)}^{2} + {2}^{2}} \left[\cos \alpha + i \sin \alpha\right]$ or

$\sqrt{5} {e}^{i \alpha}$, where $\tan \alpha = \frac{6}{- 4} = \frac{2}{- 1} = - 2$ and

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

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

Hence $\frac{2 i - 1}{3 i + 5} = \frac{\sqrt{5} {e}^{i \alpha}}{\sqrt{34} {e}^{i \beta}} = \sqrt{\frac{5}{34}} {e}^{i \left(\alpha - \beta\right)} = \sqrt{\frac{5}{34}} \left(\cos \left(\alpha - \beta\right) + i \sin \left(\alpha - \beta\right)\right)$

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

= $\frac{- 2 - \frac{3}{5}}{1 + \left(- 2\right) \cdot \left(\frac{3}{5}\right)}$ = $\frac{- \frac{13}{5}}{- \frac{1}{5}} = - \frac{13}{5} \cdot \frac{- 5}{1} = 13$

Hence $\frac{2 i - 1}{3 i + 5} = \sqrt{\frac{5}{34}} \left(\cos \rho + i \sin \rho\right)$ where $\rho = {\tan}^{- 1} 13$