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# How do you add (-5+5i)+(7+i) in trigonometric form?

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Jun 25, 2018

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color(cyan)(=> 2 + 6 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(-5^2+ 5^2))=sqrt 50
${r}_{2} = \sqrt{{7}^{2} + {1}^{2}} = \sqrt{50}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{5}{-} 5\right) \approx {135}^{\circ} , \text{ II quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(\frac{1}{7}\right) \approx {8.13}^{\circ} , \text{ I quadrant}$

${z}_{1} + {z}_{2} = \sqrt{50} \cos \left(135\right) + \sqrt{50} \cos \left(8.13\right) + i \left(\sqrt{50} \sin 135 + \sqrt{50} \sin 8.13\right)$

$\implies - 5 + 7 + i \left(5 + 1\right)$

color(cyan)(=> 2 + 6 i

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