How do you divide $\frac{i - 3}{7 i + 3}$ $( i-3) / (7i +3)$ in trigonometric form?

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How do you divide $\frac{i - 3}{7 i + 3}$ $( i-3) / (7i +3)$ in trigonometric form?

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Answer 1 · 2018-06-15T04:10:59

$\Rightarrow 0.42 (- 0.7475 + i 0.6643)$ $color(crimson)(=> 0.42 ( -0.7475 + i 0.6643)$

Explanation:

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

$z_{1} = i - 3, z_{2} = 3 + i 7$ $z_1 = i-3, z_2 = 3 + i 7$

$r_{1} = \sqrt{1 + 3^{2}} = \sqrt{10}$ $r_1 = sqrt(1 + 3^2) = sqrt10$

$θ_{1} = {tan}^{- 1} (\frac{1}{- 3}) = - 18.43 = {161.57}^{\circ}, II Quadrant$ $theta_1 = tan ^ (-1) (1/-3) = -18.43 = 161.57^@, " II Quadrant"$

$r_{2} = \sqrt{7^{2} + 3^{2}} = \sqrt{58}$ $r_2 = sqrt(7^2 + 3^2) = sqrt58$

$θ_{2} = {tan}^{- 1} (\frac{3}{7}) = {23.2}^{\circ}$ $theta_2 = tan ^ (-1) (3/7) = 23.2^@$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{10}{58}} (cos (161.57 - 23.2) - i sin (161.57 - 23.2))$ $z_1 / z_2 = sqrt(10/58) (cos (161.57- 23.2) - i sin (161.57 - 23.2))$

$\Rightarrow 0.42 (- 0.7475 + i 0.6643)$ $color(crimson)(=> 0.42 ( -0.7475 + i 0.6643)$

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How do you divide $\frac{i - 3}{7 i + 3}$ $( i-3) / (7i +3)$ in trigonometric form?

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