# How do you divide ( 3i+8) / (i +4 ) in trigonometric form?

May 21, 2016

$\frac{6 i - 4}{- 3 i - 5} = \sqrt{\frac{73}{17}} \left(\cos \rho + i \sin \rho\right)$ where $\rho = {\tan}^{- 1} \left(\frac{4}{35}\right)$

#### Explanation:

Let us first write $\left(3 i + 8\right)$ and $\left(i + 4\right)$ 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 $3 i + 8 = \left(8 + 3 i\right) = \sqrt{{8}^{2} + {3}^{2}} \left[\cos \alpha + i \sin \alpha\right]$ or

$\sqrt{73} {e}^{i \alpha}$, where $\tan \alpha = \frac{3}{8}$ and

$i + 4 = \left(4 + i\right) = \sqrt{{4}^{2} + {1}^{2}} \left[\cos \beta + i \sin \beta\right]$ or

$\sqrt{17} {e}^{i \beta}$, where $\tan \beta = \frac{1}{4}$

Hence $\frac{3 i + 8}{i + 4} = \frac{\sqrt{73} {e}^{i \alpha}}{\sqrt{17} {e}^{i \beta}} = \sqrt{\frac{73}{17}} {e}^{i \left(\alpha - \beta\right)} = \sqrt{\frac{73}{17}} \left(\cos \left(\alpha - \beta\right) + i \sin \left(\alpha - \beta\right)\right)$

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

= $\frac{\frac{3}{8} - \frac{1}{4}}{1 + \frac{3}{8} \cdot \frac{1}{4}} = \frac{\frac{1}{8}}{1 + \frac{3}{32}} = \frac{1}{8} \cdot \frac{32}{35} = \frac{4}{35}$

Hence $\frac{6 i - 4}{- 3 i - 5} = \sqrt{\frac{73}{17}} \left(\cos \rho + i \sin \rho\right)$ where $\rho = {\tan}^{- 1} \left(\frac{4}{35}\right)$