# How do you divide ( 7i+1) / ( -3i +1 ) in trigonometric form?

Apr 13, 2016

$\frac{7 i + 1}{- 3 i + 1} = \sqrt{5} \left[\cos \arctan \left(- \frac{1}{2}\right) + i \sin \arctan \left(- \frac{1}{2}\right)\right]$

#### Explanation:

We first covert them into trigonometric form. In this form $a + b i = r \left(\cos \theta + i \sin \theta\right)$ or $a + b i = r \cdot {e}^{i \theta}$, where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\theta = \arctan \left(\frac{b}{a}\right)$

Hence, $7 i + 1 = 1 + 7 i \sqrt{50} {e}^{i \alpha}$, where $\alpha = \arctan 7$

and $- 3 i + 1 = 1 - 3 i = \sqrt{10} {e}^{i \beta}$, where $\beta = \arctan \left(- 3\right)$

Hence $\frac{7 i + 1}{- 3 i + 1} = \sqrt{5} {e}^{i \left(\alpha - \beta\right)}$

As $\tan \left(\alpha - \beta\right) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \cdot \tan \beta}$ or

$\tan \left(\alpha + \beta\right) = \frac{7 - \left(- 3\right)}{1 + 7 \cdot \left(- 3\right)} = \frac{10}{- 20} = - \frac{1}{2}$

Hence $\frac{7 i + 1}{- 3 i + 1} = \sqrt{5} {e}^{i \left(\arctan \left(- \frac{1}{2}\right)\right)}$

or $\frac{7 i + 1}{- 3 i + 1} = \sqrt{5} \left[\cos \arctan \left(- \frac{1}{2}\right) + i \sin \arctan \left(- \frac{1}{2}\right)\right]$