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

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

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Answer 1 · 2018-06-24T05:35:10

$\Rightarrow 0.527 (- 0.9487 + i 0.3112)$ $color(blue)(=> 0.527 ( -0.9487 + i 0.3112)$

Explanation:

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

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

$r_{1} = \sqrt{- 2^{2} + - 1^{2}} = \sqrt{5}$ $r_1 = sqrt(-2^2 + -1^2) = sqrt5$

$θ_{1} = {tan}^{- 1} (- \frac{1}{- 2}) = {tan}^{- 1} (\frac{1}{2}) = {26.57}^{\circ} = {206.57}^{\circ}, III Quadrant$ $theta_1 = tan ^ (-1) (-1/-2) = tan ^-1 (1/2) = 26.57 ^@ = 206.57^@, " III Quadrant"$

$r_{2} = \sqrt{3^{2} + 3^{2}} = \sqrt{18}$ $r_2 = sqrt(3^2 + 3^2) = sqrt 18$

$θ_{2} = {tan}^{\frac{3}{3}} = {tan}^{- 1} (1) = 45^{\circ}$ $theta_2 = tan ^ (3/ 3) = tan^-1 (1) = 45^@$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{5}{8}} (cos (206.57 - 45) + i sin (206.57 - 45))$ $z_1 / z_2 = sqrt(5/8) (cos (206.57- 45) + i sin (206.57 - 45))$

$\Rightarrow 0.527 (- 0.9487 + i 0.3112)$ $color(blue)(=> 0.527 ( -0.9487 + i 0.3112)$

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

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