How do you divide $\frac{2 i - 7}{- 2 i + 6}$ $(2i-7) / (-2i+6)$ in trigonometric form?

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How do you divide $\frac{2 i - 7}{- 2 i + 6}$ $(2i-7) / (-2i+6)$ in trigonometric form?

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Answer 1 · 2018-07-08T02:43:43

$\frac{- 7 + 2 i}{6 - 2 i} = - 1.15 - 0.05 i$ $color(maroon)((-7 + 2i) / (6 - 2i) = -1.15 - 0.05i)$

Explanation:

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

$z_{1} = - 7 + 2 i, z_{2} = 6 - 2 i$ $z_1 = -7 + 2i , z_2 = 6 - 2i$

$| r_{1} | = \sqrt{- 7^{2} + 2^{2}} = \sqrt{53}$ $|r_1| = sqrt(-7^2 + 2^2) = sqrt 53$

$θ_{1} = {tan}^{- 1} (\frac{2}{- 7}) = {164.05}^{\circ} II Quadrant$ $theta_1 = tan ^ (-1) (2/-7) = 164.05 ^@ " II Quadrant"$

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

$θ_{2} = {tan}^{- 1} (- \frac{2}{6}) = {341.57}^{\circ}, IV Quadrant$ $theta_2 = tan ^-1 (-2/ 6) = 341.57^@ , " IV Quadrant"$

$\frac{z_{1}}{z_{2}} = ∣ ∣ ∣ \sqrt{\frac{53}{40}} ∣ ∣ ∣ \cdot (cos (164.05 - 341.57) + i sin (164.05 - 341.57))$ $z_1 / z_2 = |sqrt(53/40)| * (cos (164.05- 341.57) + i sin (164.05- 341.57))$

$\frac{- 7 + 2 i}{6 - 2 i} = - 1.15 - 0.05 i$ $color(maroon)((-7 + 2i) / (6 - 2i) = -1.15 - 0.05i)$

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How do you divide $\frac{2 i - 7}{- 2 i + 6}$ $(2i-7) / (-2i+6)$ in trigonometric form?

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How do you divide (2i-7) / (-2i+6)2i−7−2i+6(2i-7) / (-2i+6) in trigonometric form?

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How do you divide $\frac{2 i - 7}{- 2 i + 6}$ $(2i-7) / (-2i+6)$ in trigonometric form?