# How do you divide  (6+9i)/(1+i)  in trigonometric form?

Apr 5, 2016

Sorry for the delay. Below is my answer.

#### Explanation:

First step, get rid of complex denominator by multiplying top and bottom by its complex conjugate
$r = \left(6 + 9 i\right) \frac{1 - i}{\left(1 + i\right) \left(1 - i\right)} = \frac{6 + 9 i - 6 i + 9}{2} = \frac{15}{2} + 3 \frac{i}{2}$

Next you factor out the modulus or norm of the complex number. and write

$r = \frac{\sqrt{225 + 9}}{2} \left(\cos \alpha + i \sin \alpha\right)$

where $\alpha$ is such that $\tan \alpha = \frac{1}{5}$