How do you divide $\frac{- 3 + 5 i}{2 - i}$ $(-3+5i)/(2-i)$ in trigonometric form?

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How do you divide $\frac{- 3 + 5 i}{2 - i}$ $(-3+5i)/(2-i)$ in trigonometric form?

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Answer 1 · 2017-08-15T14:39:33

$\frac{- 3 + 5 i}{2 - i} = \frac{1}{5} \sqrt{170} (cos 2.57 + i sin 2.57)$ $(-3+5i)/(2-i) = 1/5sqrt (170)(cos2.57+isin2.57)$

Explanation:

$\frac{- 3 + 5 i}{2 - i} = \frac{(- 3 + 5 i) (2 + i)}{(2 - i) (2 + i)} = \frac{- 11 + 7 i}{5} = - \frac{11}{5} + \frac{7}{5} i$ $(-3+5i)/(2-i)=((-3+5i)(2+i))/((2-i)(2+i))= (-11+7i)/5 = -11/5+7/5i$

Use mod-arg form, $a + b i = r (cos ϑ + i sin ϑ)$ $a+bi=r(cosvartheta+isinvartheta)$ where $r = \sqrt{- \frac{11}{5^{2}} + \frac{7}{5^{2}}} = \sqrt{\frac{34}{5}} = \frac{1}{5} \sqrt{170}$ $r= sqrt (-11/5^2+7/5^2) = sqrt (34/5) = 1/5sqrt170$ and $ϑ = arctan (\frac{\frac{7}{5}}{- \frac{11}{5}}) + π = {2.57}^{c}$ $vartheta = arctan ((7/5)/(-11/5))+pi = 2.57^"c"$

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How do you divide $\frac{- 3 + 5 i}{2 - i}$ $(-3+5i)/(2-i)$ in trigonometric form?

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