# How do you add (8-9i)+(4+6i) in trigonometric form?

Jun 25, 2018

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#### Explanation:

$z = a + b i = r \left(\cos \theta + i \sin \theta\right)$

$r = \sqrt{{a}^{2} + {b}^{2}}$
$\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$

${r}_{1} \left(\cos \left({\theta}_{1}\right) + i \sin \left({\theta}_{2}\right)\right) + {r}_{2} \left(\cos \left({\theta}_{2}\right) + i \sin \left({\theta}_{2}\right)\right) = {r}_{1} \cos \left({\theta}_{1}\right) + {r}_{2} \cos \left({\theta}_{2}\right) + i \left({r}_{1} \sin \left({\theta}_{1}\right) + {r}_{2} \sin \left({\theta}_{2}\right)\right)$

r_1=sqrt(8^2+-9^2))=sqrt 145
${r}_{2} = \sqrt{{4}^{2} + {6}^{2}} = \sqrt{52}$

${\theta}_{1} = {\tan}^{-} 1 \left(- \frac{9}{8}\right) \approx {311.63}^{\circ} , \text{ IV quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(\frac{6}{4}\right) \approx {56.31}^{\circ} , \text{ I quadrant}$

${z}_{1} + {z}_{2} = \sqrt{145} \cos \left(311.63\right) + \sqrt{52} \cos \left(56.31\right) + i \left(\sqrt{145} \sin \left(311.63\right) + \sqrt{52} \sin \left(56.31\right)\right)$

$\implies 8 + 4 + i \left(- 9 + 6\right)$
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