How do you divide $\frac{2 - 9 i}{6 + 2 i}$ $(2-9i)/(6+2i)$ in trigonometric form?

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How do you divide $\frac{2 - 9 i}{6 + 2 i}$ $(2-9i)/(6+2i)$ in trigonometric form?

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Answer 1 · 2018-06-24T03:41:28

$\Rightarrow 1.33 (- 0.1028 - i 0.9947)$ $color(crimson)(=> 1.33 ( -0.1028 - i 0.9947)$

Explanation:

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

$z_{1} = 2 - i 9, z_{2} = 6 + i 2$ $z_1 = 2 - i 9, z_2 = 6 + i 2$

$r_{1} = \sqrt{2^{2} + 9^{2}} = \sqrt{85}$ $r_1 = sqrt(2^2 + 9^2) = sqrt85$

$θ_{1} = {tan}^{- 1} (- \frac{9}{2}) = - tan \cdot - 1 - 4.5 = - 77.47 = {282.53}^{\circ}, IV Quadrant$ $theta_1 = tan ^ (-1) (-9/2) = -tan *-1 -4.5 = -77.47 = 282.53^@, " IV Quadrant"$

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

$θ_{2} = {tan}^{- 1} (\frac{2}{6}) = {18.43}^{\circ}$ $theta_2 = tan ^ (-1) (2/6) = 18.43^@$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{85}{40}} (cos (282.53 - 18.43) - i sin (282.53 - 18.43))$ $z_1 / z_2 = sqrt(85/40) (cos (282.53- 18.43) - i sin (282.53 - 18.43))$

$\Rightarrow 1.33 (- 0.1028 - i 0.9947)$ $color(crimson)(=> 1.33 ( -0.1028 - i 0.9947)$

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How do you divide $\frac{2 - 9 i}{6 + 2 i}$ $(2-9i)/(6+2i)$ in trigonometric form?

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Explanation:

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