How do you add (2-5i) and (-2-2i) in trigonometric form?

Jul 9, 2018

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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(2^2+ -5^2))=sqrt 29
${r}_{2} = \sqrt{- {2}^{2} + - {2}^{2}} = \sqrt{8}$

${\theta}_{1} = {\tan}^{-} 1 \left(- \frac{5}{2}\right) \approx {291.8}^{\circ} , \text{ IV quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(- \frac{2}{-} 2\right) \approx {225}^{\circ} , \text{ III quadrant}$

${z}_{1} + {z}_{2} = \sqrt{29} \cos \left(291.8\right) + \sqrt{8} \cos \left(225\right) + i \left(\sqrt{29} \sin 291.8 + \sqrt{8} \sin 225\right)$

$\implies 2 - 2 + i \left(- 5 - 2\right)$

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