How do you divide $\frac{- 4 + 5 i}{6 + i}$ $(-4+5i)/(6+i)$ in trigonometric form?

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How do you divide $\frac{- 4 + 5 i}{6 + i}$ $(-4+5i)/(6+i)$ in trigonometric form?

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Answer 1 · 2018-06-15T04:00:43

$\Rightarrow 1.05 (- 0.482 + i 0.876)$ $color(purple)(=> 1.05 (-0.482 + i 0.876)$

Explanation:

$\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} (cos (θ_{1} + t h e t_{2}) + i sin (θ_{1} + θ_{2})$ $z_1 / z_2 = r_1/ r_2 (cos (theta_1 + thet_2) + i sin (theta_1 + theta_2)$

$z_{1} = (- 4 + i 5), z_{2} = (6 + i)$ $z_1 = (-4 + i5), z_2 = (6 + i)$

$\Rightarrow r_{1} = \sqrt{4^{2} + 5^{2}} = \sqrt{41} = 6.4$ $=> r_1 = sqrt(4^2 + 5^2) = sqrt41 = 6.4$

$θ_{1} = {tan}^{- 1} (\frac{5}{- 4}) = - {51.34}^{\circ} = {128.26}^{\circ}, II Quadrant$ $theta_1 = tan^(-1) (5/-4)= -51.34^@ = 128.26^@, " II Quadrant"$

$\Rightarrow r_{2} = \sqrt{1^{2} + 6^{2}} = \sqrt{37} = 6.08$ $=> r_2 = sqrt(1^2 + 6^2) = sqrt37 = 6.08$

$θ_{2} = {tan}^{- 1} (\frac{1}{6}) = {9.46}^{\circ}$ $theta_2 = tan^(-1) (1/6)= 9.46^@$

$\frac{z_{1}}{z_{2}} = \frac{\sqrt{41}}{\sqrt{37}} (cos (128.26 - 9.46) + i sin (128.26 - 9.46) 0$ $z_1 / z_2 = sqrt41/sqrt37 (cos (128.26 - 9.46) + i sin(128.26 - 9.46)0$

$\Rightarrow 1.05 (- 0.482 + i 0.876)$ $color(purple)(=> 1.05 (-0.482 + i 0.876)$

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How do you divide $\frac{- 4 + 5 i}{6 + i}$ $(-4+5i)/(6+i)$ in trigonometric form?

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