How do you divide $\frac{- i + 2}{3 i - 4}$ $(-i+2) / (3i-4)$ in trigonometric form?

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How do you divide $\frac{- i + 2}{3 i - 4}$ $(-i+2) / (3i-4)$ in trigonometric form?

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Answer 1 · 2018-06-25T06:08:35

$\frac{2 - i}{- 4 + 3 i} = 2.236 (0.9839 - i 0.1788)$ $color(purple)((2 -i) / (-4 + 3i) = 2.236 ( 0.9839 - i 0.1788)$

Explanation:

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

$z_{1} = 2 - i, z_{2} = - 4 + 3 i$ $z_1 = 2 - i , z_2 = -4 + 3 i$

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

$θ_{1} = {tan}^{- 1} (- \frac{1}{2}) = {333.43}^{\circ} IV Quadrant$ $theta_1 = tan ^ (-1) (-1/2) = 333.43 ^@ " IV Quadrant"$

$| r_{2} | = \sqrt{- 4^{2} + {(3)}^{2}} = 5$ $|r_2| = sqrt(-4^2 + (3)^2) = 5$

$θ_{2} = {tan}^{- 1} (\frac{3}{- 4}) = {143.13}^{\circ}, II Quadrant$ $theta_2 = tan ^-1 (3/ -4) = 143.13^@ , " II Quadrant"$

$\frac{z_{1}}{z_{2}} = ∣ ∣ ∣ (\frac{5}{\sqrt{5}}) ∣ ∣ ∣ \cdot (cos (333.43 - 143.13) + i sin (333.43 - 143.13))$ $z_1 / z_2 = |(5/sqrt5)| * (cos (333.43 - 143.13) + i sin (333.43 - 143.13))$

$\frac{2 - i}{- 4 + 3 i} = 2.236 (0.9839 - i 0.1788)$ $color(purple)((2 -i) / (-4 + 3i) = 2.236 ( 0.9839 - i 0.1788)$

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How do you divide $\frac{- i + 2}{3 i - 4}$ $(-i+2) / (3i-4)$ in trigonometric form?

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