How do you divide $\frac{7 i + 1}{i - 6}$ $( 7i+1) / (i -6 )$ in trigonometric form?

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How do you divide $\frac{7 i + 1}{i - 6}$ $( 7i+1) / (i -6 )$ in trigonometric form?

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Answer 1 · 2018-07-26T15:47:44

$\Rightarrow 0.027 - 1.1622 i,$ $color(crimson )(=> 0.027 - 1.1622 i),$ IV QUADRANT

Explanation:

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

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

$r_{1} = {\sqrt{1^{2} + 7^{2}}}^{2}) = \sqrt{50}$ $r_1 = sqrt(1^2 + 7^2)^2) = sqrt 50$

$θ_{1} = {tan}^{- 1} (\frac{7}{1}) = {81.8699}^{\circ} =, I Quadrant$ $theta_1 = tan ^ -1 (7 / 1) = 81.8699^@ = , " I Quadrant"$

$r_{2} = \sqrt{- 6^{2} + {(1)}^{2}} = \sqrt{37}$ $r_2 = sqrt(-6^2 + (1)^2) = sqrt 37$

$θ_{2} = {tan}^{- 1} (\frac{1}{- 6}) \approx {170.5377}^{\circ}, II Quadrant$ $theta_2 = tan ^-1 (1/ -6) ~~ 170.5377^@, " II Quadrant"$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{50}{37}} \cdot (cos (81.8699 - 170.5377) + i sin (81.8699 - 170.5377))$ $z_1 / z_2 = sqrt(50 / 37) * (cos (81.8699 - 170.5377) + i sin (81.8699 - 170.5377))$

$\Rightarrow 0.027 - 1.1622 i,$ $color(crimson )(=> 0.027 - 1.1622 i),$ IV QUADRANT

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How do you divide $\frac{7 i + 1}{i - 6}$ $( 7i+1) / (i -6 )$ in trigonometric form?

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