# How do you add (-9+2i)+(2+4i) in trigonometric form?

Jul 27, 2018

color(magenta)(=> -7 + 6 i, " II Quadrant"

#### 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+ 2^2))=sqrt 85
${r}_{2} = \sqrt{{2}^{2} + {4}^{2}} = \sqrt{20}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{2}{-} 9\right) \approx {167.4712}^{\circ} , \text{ II quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(\frac{4}{2}\right) \approx {63.4349}^{\circ} , \text{ I quadrant}$

${z}_{1} + {z}_{2} = \sqrt{85} \cos \left(167.4712\right) + \sqrt{20} \cos \left(63.4349\right) + i \left(\sqrt{85} \sin 167.4712 + \sqrt{20} \sin 63.4349\right)$

$\implies - 9 + 2 + i \left(2 + 4\right)$

color(magenta)(=> -7 + 6 i, " II Quadrant"