# How do you multiply  (5+3i)(3+2i)  in trigonometric form?

Nov 13, 2017

$\left(5 + 3 i\right) \left(3 + 2 i\right) \approx 21.02 \left(\cos \left(64.65\right) + i \sin \left(64.65\right)\right)$

#### Explanation:

$a + b i$ in trig form is r(cos\theta+isin\theta), where:

• $r = \sqrt{{a}^{2} + {b}^{2}}$
• $\setminus \theta = \left\mid \arctan \left(\frac{b}{a}\right) \right\mid$

$\left(5 + 3 i\right) \left(3 + 2 i\right) \cong \left({r}_{1} \left(\cos \setminus {\theta}_{1} + i \sin \setminus {\theta}_{1}\right)\right) \left({r}_{2} \left(\cos \setminus {\theta}_{2} + i \sin \setminus {\theta}_{2}\right)\right)$
$= {r}_{1} {r}_{2} \left(\cos \left(\setminus {\theta}_{1} + \setminus {\theta}_{2}\right) + i \sin \left(\setminus {\theta}_{1} + \setminus {\theta}_{2}\right)\right)$
$= \sqrt{{5}^{2} + {3}^{2}} \sqrt{{3}^{2} + {2}^{2}} \left(\cos \left(\left\mid \arctan \left(\frac{3}{5}\right) \right\mid + \left\mid \arctan \left(\frac{2}{3}\right) \right\mid\right) + i \sin \left(\left\mid \arctan \left(\frac{3}{5}\right) \right\mid + \left\mid \arctan \left(\frac{2}{3}\right) \right\mid\right)\right)$
$\approx \sqrt{34} \sqrt{13} \left(\cos \left(64.65\right) + i \sin \left(64.65\right)\right)$
$= \sqrt{442} \left(\cos \left(64.65\right) + i \sin \left(64.65\right)\right)$
$\approx 21.02 \left(\cos \left(64.65\right) + i \sin \left(64.65\right)\right)$

Given $\tan {\theta}_{1} = \frac{3}{5}$ and $\tan {\theta}_{1} = \frac{2}{3}$

$\tan \theta = \frac{\frac{3}{5} + \frac{2}{3}}{1 - \frac{3}{5} \times \frac{2}{3}} = \frac{\frac{19}{15}}{\frac{3}{5}} = \frac{19}{9}$

and in exact form we can write product as

sqrt442(cos(arctan(19/9))+isin((arctan(19/9)))