# How do you add (-9+i)+(-1+3i) in trigonometric form?

Jul 9, 2018

color(crimson)(=> -10+ 4i

#### 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(-9^2+ 1^2))=sqrt 82
${r}_{2} = \sqrt{- {1}^{2} + {3}^{2}} = \sqrt{10}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{1}{-} 9\right) \approx {173.66}^{\circ} , \text{ II quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(\frac{3}{-} 1\right) \approx {108.43}^{\circ} , \text{ II quadrant}$

${z}_{1} + {z}_{2} = \sqrt{82} \cos \left(173.66\right) + \sqrt{10} \cos \left(108.43\right) + i \left(\sqrt{82} \sin 173.66 + \sqrt{10} \sin 108.43\right)$

$\implies - 9 - 1 + i \left(1 + 3\right)$

color(crimson)(=> -10+ 4i