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

Jul 9, 2018

color(chocolate)(=> 10 - 6i

#### 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+ -7^2))=sqrt 130
${r}_{2} = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$

${\theta}_{1} = {\tan}^{-} 1 \left(- \frac{7}{9}\right) \approx {322.13}^{\circ} , \text{ IV quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(\frac{1}{1}\right) \approx {45}^{\circ} , \text{ I quadrant}$

${z}_{1} + {z}_{2} = \sqrt{130} \cos \left(322.13\right) + \sqrt{2} \cos \left(45\right) + i \left(\sqrt{130} \sin 322.13 + \sqrt{2} \sin 45\right)$

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

color(chocolate)(=> 10 - 6i