How do you divide $\frac{3 i - 7}{- 3 i - 9}$ $( 3i-7) / ( -3 i -9 )$ in trigonometric form?

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How do you divide $\frac{3 i - 7}{- 3 i - 9}$ $( 3i-7) / ( -3 i -9 )$ in trigonometric form?

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Answer 1 · 2018-06-24T03:47:06

$\Rightarrow 0.8028 (0.7475 - i 0.6643)$ $color(blue)(=> 0.8028 (0.7475 - i 0.6643)$

Explanation:

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

$z_{1} = - 7 + i 3, z_{2} = - 9 - i 3$ $z_1 = -7 + i 3, z_2 = -9 - i 3$

$r_{1} = \sqrt{7^{2} + 3^{2}} = \sqrt{58}$ $r_1 = sqrt(7^2 + 3^2) = sqrt58$

$θ_{1} = {tan}^{- 1} (\frac{3}{- 7}) = - tan \cdot - 1 (- \frac{3}{7}) = - {23.2}^{\circ} = {156.8}^{\circ}, II Quadrant$ $theta_1 = tan ^ (-1) (3/-7) = -tan *-1 (-3/7) = -23.2 ^@ = 156.8^@, " II Quadrant"$

$r_{2} = \sqrt{9^{2} + 3^{2}} = \sqrt{90}$ $r_2 = sqrt(9^2 + 3^2) = sqrt90$

$θ_{2} = {tan}^{- \frac{3}{- 9}} = {tan}^{- 1} (\frac{1}{3}) = {18.43}^{\circ} {198.43}^{\circ}, III Quadrant$ $theta_2 = tan ^ (-3/ -9) = tan^-1 (1/3) = 18.43^@ 198.43^@, " III Quadrant"$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{58}{90}} (cos (156.8 - 198.43) + i sin (156.8 - 198.43))$ $z_1 / z_2 = sqrt(58/90) (cos (156.8- 198.43) + i sin (156.8 - 198.43))$

$\Rightarrow 0.8028 (0.7475 - i 0.6643)$ $color(blue)(=> 0.8028 ( 0.7475 - i 0.6643)$

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How do you divide $\frac{3 i - 7}{- 3 i - 9}$ $( 3i-7) / ( -3 i -9 )$ in trigonometric form?

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How do you divide 3i−7−3i−9( 3i-7) / ( -3 i -9 ) in trigonometric form?

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How do you divide $\frac{3 i - 7}{- 3 i - 9}$ $( 3i-7) / ( -3 i -9 )$ in trigonometric form?