How do you divide $\frac{- i + 1}{2 i - 7}$ $( -i+1) / (2i -7 )$ in trigonometric form?

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How do you divide $\frac{- i + 1}{2 i - 7}$ $( -i+1) / (2i -7 )$ in trigonometric form?

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Answer 1 · 2018-06-24T06:03:03

$\Rightarrow 0.1943 (- 0.8742 - i 0.4856)$ $color(magenta)(=> 0.1943 ( -0.8742 - i 0.4856)$

Explanation:

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

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

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

$θ_{1} = {tan}^{- 1} (- \frac{1}{1}) = tan \cdot - 1 (- 1) = - 45^{\circ} = 315^{\circ}, IV Quadrant$ $theta_1 = tan ^ (-1) (-1/1) = tan *-1 (-1) = -45 ^@ = 315^@, " IV Quadrant"$

$r_{2} = \sqrt{{(- 7)}^{2} + {(2)}^{2}} = \sqrt{53}$ $r_2 = sqrt((-7)^2 + (2)^2) = sqrt 53$

$θ_{2} = {tan}^{- 1} (\frac{2}{- 7}) = - {15.95}^{\circ} = {164.05}^{\circ}, II Quadrant$ $theta_2 = tan ^-1 (2/ -7) = -15.95^@ = 164.05^@, " II Quadrant"$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{2}{53}} (cos (315 - 164.05) - i sin (315 - 164.05))$ $z_1 / z_2 = sqrt(2/53) (cos (315- 164.05) - i sin (315 - 164.05))$

$\Rightarrow 0.1943 (- 0.8742 - i 0.4856)$ $color(magenta)(=> 0.1943 (- 0.8742 - i 0.4856)$

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How do you divide $\frac{- i + 1}{2 i - 7}$ $( -i+1) / (2i -7 )$ in trigonometric form?

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Explanation:

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How do you divide $\frac{- i + 1}{2 i - 7}$ $( -i+1) / (2i -7 )$ in trigonometric form?