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

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

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Answer 1 · 2018-07-17T07:37:43

$\Rightarrow 0.064 - 0.048 i$ $color(blue)(=> 0.064 - 0.048 i)$

Explanation:

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

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

$r_{1} = {\sqrt{2^{2} +_{- 2}}}^{2}) = \sqrt{8}$ $r_1 = sqrt(2^2 + _-2)^2) = sqrt8$

$θ_{1} = \frac{{tan}^{- 2}}{2} = - 45^{\circ} = 315^{\circ}, IV Quadrant$ $theta_1 = tan ^ (-2)/ (2) = -45^@ = 315^@, " IV Quadrant"$

$r_{2} = \sqrt{7^{2} + {(- 1)}^{2}} = \sqrt{50}$ $r_2 = sqrt(7^2 + (-1)^2) = sqrt 50$

$θ_{2} = {tan}^{- 1} (- \frac{1}{7}) \approx {351.87}^{\circ}, IV Quadrant$ $theta_2 = tan ^-1 (-1/ 7) ~~ 351.87^@, " IV Quadrant"$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{8}{50}} (cos (315 - 351.87) + i sin (315 - 351.87))$ $z_1 / z_2 = sqrt(8/50) (cos (315- 351.87) + i sin (315- 351.87))$

$\Rightarrow 0.064 - 0.048 i$ $color(blue)(=> 0.064 - 0.048 i)$

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

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