# How do you add (-1+8i)+(4-2i) in trigonometric form?

Jun 25, 2018

color(indigo)(=> 3+ 6 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(-1^2+ 8^2))=sqrt 65
${r}_{2} = \sqrt{- {2}^{2} + {4}^{2}} = \sqrt{20}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{8}{-} 1\right) \approx {97.13}^{\circ} , \text{ II quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(- \frac{2}{4}\right) \approx {333.43}^{\circ} , \text{ IV quadrant}$

${z}_{1} + {z}_{2} = \sqrt{65} \cos \left(97.13\right) + \sqrt{20} \cos \left(333.43\right) + i \left(\sqrt{65} \sin 97.13 + \sqrt{20} \sin 333.43\right)$

$\implies - 1 + 4 + i \left(8 - 2\right)$

color(indigo)(=> 3+ 6 i