# How do you add (8+4i) and (3-3i) in trigonometric form?

Jun 25, 2018

color(crimson)(=> 11+ i

#### Explanation:

$z = a + b i = r \left(\cos \theta + i \sin \theta\right)$

$r = \sqrt{{a}^{2} + {b}^{2}} , \text{ } \theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$

${r}_{1} \left(\cos \left({\theta}_{1}\right) + i \sin \left({\theta}_{2}\right)\right) + {r}_{2} \left(\cos \left({\theta}_{2}\right) + i \sin \left({\theta}_{2}\right)\right) = {r}_{1} \cos \left({\theta}_{1}\right) + {r}_{2} \cos \left({\theta}_{2}\right) + i \left({r}_{1} \sin \left({\theta}_{1}\right) + {r}_{2} \sin \left({\theta}_{2}\right)\right)$

r_1=sqrt(4^2+ 8^2))=sqrt 80
${r}_{2} = \sqrt{{3}^{2} + - {3}^{2}} = \sqrt{18}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{4}{8}\right) \approx {26.57}^{\circ} , \text{ I quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(- \frac{3}{3}\right) \approx {315}^{\circ} , \text{ IV quadrant}$

${z}_{1} + {z}_{2} = \sqrt{80} \cos \left(26.57\right) + \sqrt{18} \cos \left(315\right) + i \left(\sqrt{80} \sin 26.57 + \sqrt{18} \sin 315\right)$

$\implies 8 + 3 + i \left(4 - 3\right)$

color(crimson)(=> 11+ i