# What is the trigonometric form of  (5-2i)(2+3i) ?

Jan 6, 2016

$\sqrt{377} \left\{\cos \left({\tan}^{-} 1 \left(\frac{11}{16}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{11}{16}\right)\right)\right\}$

#### Explanation:

$\left(5 - 2 i\right) \left(2 + 3 i\right)$

Multiply using FOIL

$\left(5\right) \left(2\right) + 5 \left(3 i\right) + \left(- 2 i\right) \left(2\right) + \left(- 2 i\right) \left(3 i\right)$
$10 + 15 i - 4 i - 6 {i}^{2}$
$= 10 + 11 i - 6 \left(- 1\right)$ since ${i}^{2} = - 1$
$= 10 + 6 + 11 i$
$= 16 + 11 i$

Trigonometric form is $r \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)$
if $z = x + i y$

$r = \sqrt{{x}^{2} + {y}^{2}}$ and $\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)$

$r = \sqrt{{16}^{2} + {11}^{2}}$ and $\theta = {\tan}^{-} 1 \left(\frac{11}{16}\right)$
$r = \sqrt{256 + 121}$
$r = \sqrt{377}$
Trigonometric form

$\sqrt{377} \left\{\cos \left({\tan}^{-} 1 \left(\frac{11}{16}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{11}{16}\right)\right)\right\}$