How do you add (-7+9i) and (-1+6i) in trigonometric form?

Jul 9, 2018

color(indigo)(=> -8+ 15i

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(-7^2+ 9^2))=sqrt 130
${r}_{2} = \sqrt{- {1}^{2} + {6}^{2}} = \sqrt{37}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{9}{-} 7\right) \approx {127.87}^{\circ} , \text{ II quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(\frac{6}{-} 1\right) \approx {333.43}^{\circ} , \text{ II quadrant}$

${z}_{1} + {z}_{2} = \sqrt{130} \cos \left(127.87\right) + \sqrt{37} \cos \left(99.46\right) + i \left(\sqrt{130} \sin 127.87 + \sqrt{37} \sin 99.46\right)$

$\implies - 7 - 1 + i \left(9 + 6\right)$

color(indigo)(=> -8+ 15i