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

Apr 13, 2016

$\left(- 1 + 2 i\right) \cdot \left(3 - i\right) = \sqrt{50} \left[\cos \arctan \left(- 7\right) + i \sin \arctan \left(- 7\right)\right]$

#### 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, $- 1 + 2 i = \sqrt{5} {e}^{i \alpha}$, where $\alpha = \arctan \left(- 2\right)$

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

Hence $\left(- 1 + 2 i\right) \cdot \left(3 - i\right) = \sqrt{50} {e}^{i \left(\alpha + \beta\right)}$

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

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

Hence $\left(- 1 + 2 i\right) \cdot \left(3 - i\right) = \sqrt{50} {e}^{i \left(\arctan \left(- 7\right)\right)}$

or $\left(- 1 + 2 i\right) \cdot \left(3 - i\right) = \sqrt{50} \left[\cos \arctan \left(- 7\right) + i \sin \arctan \left(- 7\right)\right]$