How do you divide $\frac{9 + 2 i}{5 - 6 i}$ $(9+2i) / (5-6i)$ in trigonometric form?

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How do you divide $\frac{9 + 2 i}{5 - 6 i}$ $(9+2i) / (5-6i)$ in trigonometric form?

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Answer 1 · 2018-06-25T01:46:28

$\frac{9 + 2 i}{5 - 6 i} = 1.18 (0.611 + i 0.7917)$ $color(maroon)((9 + 2i) / (5-6i) = 1.18 ( 0.611 + i 0.7917)$

Explanation:

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

$z_{1} = 9 + 2 i, z_{2} = 5 - 6 i$ $z_1 = 9 + 2 i , z_2 = 5 - 6 i$

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

$θ_{1} = {tan}^{- 1} (\frac{2}{9}) = {12.53}^{\circ} I Quadrant$ $theta_1 = tan ^ (-1) (2/9) = 12.53 ^@ " I Quadrant"$

$r_{2} = \sqrt{5^{2} + {(- 67)}^{2}} = \sqrt{61}$ $r_2 = sqrt(5^2 + (-67)^2) = sqrt 61$

$θ_{2} = {tan}^{- 1} (- \frac{6}{5}) = - {39.81}^{\circ} = 320.19, IV Quadrant$ $theta_2 = tan ^-1 (-6/ 5) = -39.81^@ = 320.19, " IV Quadrant"$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{85}{61}} (cos (12.53 - 320.19) + i sin (12.53 - 320.19))$ $z_1 / z_2 = sqrt(85/61) (cos (12.53- 320.19) + i sin (12.53 - 320.19))$

$\frac{9 + 2 i}{5 - 6 i} = 1.18 (0.611 + i 0.7917)$ $color(maroon)((9 + 2i) / (5-6i) = 1.18 ( 0.611 + i 0.7917)$

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How do you divide $\frac{9 + 2 i}{5 - 6 i}$ $(9+2i) / (5-6i)$ in trigonometric form?

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How do you divide 9+2i5−6i (9+2i) / (5-6i) in trigonometric form?

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Explanation:

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How do you divide $\frac{9 + 2 i}{5 - 6 i}$ $(9+2i) / (5-6i)$ in trigonometric form?