How do you divide $\frac{8 + 7 i}{1 - 5 i}$ $(8+7i)/(1-5i)$ in trigonometric form?

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How do you divide $\frac{8 + 7 i}{1 - 5 i}$ $(8+7i)/(1-5i)$ in trigonometric form?

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Answer 1 · 2018-06-15T04:49:37

$\Rightarrow 2.08 (- 0.5014 + i 0.787)$ $color(brown)(=> 2.08 (-0.5014 + i 0.787)$

Explanation:

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

$z_{1} = 8 + i 7, z - 2 = 1 - i 5$ $z_1 = 8 + i 7, z-2 = 1 - i 5$

$r_{1} = \sqrt{8^{2} + 7^{2}} = \sqrt{113}$ $r_1 = sqrt(8^2 + 7^2) = sqrt113$

$θ_{1} = arctan (\frac{7}{6}) = {49.4}^{\circ}$ $theta_1 = arctan (7/6) = 49.4^@$

$r_{2} = \sqrt{1^{2} + 5^{2}} = \sqrt{26}$ $r_2 = sqrt(1^2 + 5^2) = sqrt26$

$θ_{2} = arctan (- \frac{5}{1}) = {281.31}^{\circ}, IV Quadrant$ $theta_2 = arctan (-5/1) = 281.31^@, " IV Quadrant"$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{113}{26}} (cos (49.4 - 281.31) + i sin (49.4 - 281.31)$ $z_1 / z_2 = sqrt(113/26) (cos (49.4 - 281.31) + i sin(49.4 - 281.31)$

$\Rightarrow 2.08 (- 0.5014 + i 0.787)$ $color(brown)(=> 2.08 (-0.5014 + i 0.787)$

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How do you divide $\frac{8 + 7 i}{1 - 5 i}$ $(8+7i)/(1-5i)$ in trigonometric form?

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How do you divide (8+7i)/(1-5i) 8+7i1−5i (8+7i)/(1-5i) in trigonometric form?

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How do you divide $\frac{8 + 7 i}{1 - 5 i}$ $(8+7i)/(1-5i)$ in trigonometric form?