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

Jun 25, 2018

color(cyan)(=> -2 - 10 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
${r}_{2} = \sqrt{- {1}^{2} \pm {2}^{2}} = \sqrt{5}$

${\theta}_{1} = {\tan}^{-} 1 \left(- \frac{8}{-} 1\right) \approx {262.87}^{\circ} , \text{ III quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(- \frac{2}{-} 1\right) \approx {243.43}^{\circ} , \text{ III quadrant}$

${z}_{1} + {z}_{2} = \sqrt{65} \cos \left(262.87\right) + \sqrt{5} \cos \left(243.43\right) + i \left(\sqrt{65} \sin \left(262.87\right) + \sqrt{5} \sin \left(243.43\right)\right)$

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

color(cyan)(=> -2 - 10 i