# How do you divide  (7-4i)/(3+2i)  in trigonometric form?

Mar 2, 2016

sqrt5xxe^(i(alpha-beta), where $\alpha = {\tan}^{-} 1 \left(- \frac{4}{7}\right)$ and $\beta = {\tan}^{-} 1 \left(\frac{2}{3}\right)$

#### Explanation:

$\left(a + b i\right)$ an be written in as sqrt(a^2+b^2)e^(i(tan^-1(b/a))

Hence, $\left(7 - 4 i\right) = \sqrt{{7}^{2} + {4}^{2}} \left({e}^{i {\tan}^{-} 1 \left(- \frac{4}{7}\right)}\right) = \sqrt{65} {e}^{i {\tan}^{-} 1 \left(- \frac{4}{7}\right)}$

Similarly, $\left(3 + 2 i\right) = \sqrt{{3}^{2} + {2}^{2}} \left({e}^{i {\tan}^{-} 1 \left(\frac{2}{3}\right)}\right) = \sqrt{13} {e}^{i {\tan}^{-} 1 \left(\frac{2}{3}\right)}$

Hence $\frac{7 - 4 i}{3 + 2 i} = \frac{\sqrt{65} {e}^{i {\tan}^{-} 1 \left(- \frac{4}{7}\right)}}{\sqrt{13} {e}^{i {\tan}^{-} 1 \left(\frac{2}{3}\right)}}$

or $\sqrt{5} \times {e}^{i \left({\tan}^{-} 1 \left(- \frac{4}{7}\right)\right) - \left({\tan}^{-} 1 \left(\frac{2}{3}\right)\right)}$ or

sqrt5xxe^(i(alpha-beta), where $\alpha = {\tan}^{-} 1 \left(- \frac{4}{7}\right)$ and $\beta = {\tan}^{-} 1 \left(\frac{2}{3}\right)$