# What is the trigonometric form of  (2-4i)*(3-2i) ?

Jul 3, 2017

See the explanation below.

#### Explanation:

First, expand the expression.

$\left(2 - 4 i\right) \cdot \left(3 - 2 i\right)$
= $6 - 4 i - 12 i + 8 {i}^{2}$
= $6 - 4 i - 12 i - 8$
= $- 2 - 16 i$

To convert this to trigonometric form, you need to know the values of $r$ and $\theta$.

You can use the following equations:
${r}^{2} = {x}^{2} + {y}^{2}$ and $\tan \theta = \frac{y}{x}$

${r}^{2} = {x}^{2} + {y}^{2}$
$r = \sqrt{{x}^{2} + {y}^{2}}$
$r = \sqrt{{\left(- 2\right)}^{2} + {\left(- 16\right)}^{2}}$
$r = 2 \sqrt{65}$

$\tan \theta = \frac{y}{x}$
$\tan \theta = \frac{- 16}{- 2}$
$\tan \theta = 8$
$\theta = {\tan}^{-} 1 \left(8\right)$
$\theta \approx 1.45$

So, the answer is $2 \sqrt{65}$ $c i s$ $1.45$ or $2 \sqrt{65} \left(\cos 1.45 + i \sin 1.45\right)$.