# How do you add (1+6i)+(-3+i) in trigonometric form?

Jun 25, 2018

color(blue)(=> -2 + 7 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+ 6^2))=sqrt 37
${r}_{2} = \sqrt{- {3}^{2} + {1}^{2}} = \sqrt{10}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{6}{1}\right) \approx {80.54}^{\circ} , \text{ I quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(\frac{1}{-} 3\right) \approx {161.57}^{\circ} , \text{ II quadrant}$

${z}_{1} + {z}_{2} = \sqrt{37} \cos \left(80.54\right) + \sqrt{10} \cos \left(161.57\right) + i \left(\sqrt{37} \sin 80.54 + \sqrt{10} \sin 161.57\right)$

$\implies 1 - 3 + i \left(6 + 1\right)$

color(blue)(=> -2 + 7 i