How do you multiply  (9-i)(7-3i)  in trigonometric form?

1 Answer
Feb 12, 2016

$2 \sqrt{1189} \left(\cos \left(\alpha + \beta\right) + i \sin \left(\alpha + \beta\right)\right)$, where $\alpha = {\tan}^{- 1} \left(- \frac{1}{9}\right)$ and $\beta = {\tan}^{- 1} \left(- \frac{3}{7}\right)$

Explanation:

A complex number of form $a + b i$ can be written as $r \left(\cos \theta + i \sin \theta\right)$, where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\theta = {\tan}^{- 1} \left(\frac{b}{a}\right)$.

Using this $\left(9 - i\right) = \sqrt{82} \left(\cos \alpha + i \sin \alpha\right)$, and $\alpha = {\tan}^{- 1} \left(- \frac{1}{9}\right)$

Similarly, $\left(7 - 3 i\right) = \sqrt{58} \left(\cos \beta + i \sin \beta\right)$, and $\beta = {\tan}^{- 1} \left(- \frac{3}{7}\right)$

Multiplication of (sqrt82(cos alpha+i sin alpha) and $\sqrt{58} \left(\cos \beta + i \sin \beta\right)$ is given by

$2 \sqrt{1189} \left(\cos \left(\alpha + \beta\right) + i \sin \left(\alpha + \beta\right)\right)$, where $\alpha = {\tan}^{- 1} \left(- \frac{1}{9}\right)$ and $\beta = {\tan}^{- 1} \left(- \frac{3}{7}\right)$, as $\sqrt{82 \cdot 58} = 2 \sqrt{1189}$.