How do you divide $\frac{- 6 i + 5}{- 2 i + 6}$ $( - 6 i + 5) / (-2i+6)$ in trigonometric form?

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How do you divide $\frac{- 6 i + 5}{- 2 i + 6}$ $( - 6 i + 5) / (-2i+6)$ in trigonometric form?

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Answer 1 · 2018-06-15T04:43:05

$\Rightarrow 1.235 (0.8503 - i 0.526)$ $color(violet)(=> 1.235 (0.8503 - i 0.526)$

Explanation:

$\frac{z_{1}}{z_{2}} = (\frac{r_{1}}{r_{2}}) (cos (θ_{1} - θ_{2}) + i sin (θ_{1} - θ_{2}))$ $z_1/ z_2 = (r_1 / r_2 ) (cos(theta_1 - theta_2) + i sin (theta_1 - theta_2))$

$z_{1} = 5 - i 6, z - 2 = 6 - i 2$ $z_1 = 5 - i6, z-2 = 6 - i2$

$r_{1} = \sqrt{5^{2} + 6^{2}} = \sqrt{61}$ $r_1 = sqrt(5^2 + 6^2) = sqrt61$

$θ_{1} = arctan (- \frac{6}{5}) = {309.81}^{\circ}, IV Quadrant$ $theta_1 = arctan (-6/5) = 309.81^@, " IV Quadrant"$

$r_{2} = \sqrt{6^{2} + 2^{2}} = \sqrt{40}$ $r_2 = sqrt(6^2 + 2^2) = sqrt40$

$θ_{2} = arctan (- \frac{2}{6}) = {341.57}^{\circ}, IV Quadrant$ $theta_2 = arctan (-2/6) = 341.57^@, " IV Quadrant"$

$\frac{z_{1}}{z_{2}} = (\sqrt{\frac{61}{40}} (cos (309.81 - 341.57) + i sin (129.81 - 161.57)$ $z_1 / z_2 = (sqrt(61/40) (cos (309.81 - 341.57) + i sin(129.81 - 161.57)$

$\Rightarrow 1.235 (0.8503 - i 0.5264)$ $color(violet)(=> 1.235 (0.8503 - i 0.5264)$

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How do you divide $\frac{- 6 i + 5}{- 2 i + 6}$ $( - 6 i + 5) / (-2i+6)$ in trigonometric form?

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How do you divide −6i+5−2i+6( - 6 i + 5) / (-2i+6) in trigonometric form?

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How do you divide $\frac{- 6 i + 5}{- 2 i + 6}$ $( - 6 i + 5) / (-2i+6)$ in trigonometric form?