How do you divide $\frac{i + 8}{- 3 i + 2}$ $( i+8) / (-3i +2 )$ in trigonometric form?

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How do you divide $\frac{i + 8}{- 3 i + 2}$ $( i+8) / (-3i +2 )$ in trigonometric form?

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Answer 1 · 2018-07-08T03:45:49

$\Rightarrow 1.06 + i 2.1196$ $color(maroon)(=> 1.06 + i 2.1196)$

Explanation:

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

$z_{1} = 8 + i 3, z_{2} = 2 - i 3$ $z_1 = 8 + i 3, z_2 = 2 - i 3$

$r_{1} = \sqrt{8^{2} + 3^{2}} = \sqrt{73}$ $r_1 = sqrt(8^2 + 3^2) = sqrt 73$

$θ_{1} = {tan}^{- 1} (\frac{1}{8}) = {7.13}^{\circ}, I Quadrant$ $theta_1 = tan ^ (-1) (1/8) = 7.13 ^@, " I Quadrant"$

$r_{2} = \sqrt{2^{2} + {(- 3)}^{2}} = \sqrt{13}$ $r_2 = sqrt(2^2 + (-3)^2) = sqrt 13$

$θ_{2} = {tan}^{- 1} (- \frac{3}{2}) = {303.69}^{\circ}, IV Quadrant$ $theta_2 = tan ^-1 (-3/ 2) = 303.69^@, " IV Quadrant"$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{73}{13}} (cos (7.13 - 303.69) + i sin (7.13 - 303.69))$ $z_1 / z_2 = sqrt(73/13) (cos (7.13- 303.69) + i sin (7.13- 303.69))$

$\Rightarrow 1.06 + i 2.1196$ $color(maroon)(=> 1.06 + i 2.1196)$

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How do you divide $\frac{i + 8}{- 3 i + 2}$ $( i+8) / (-3i +2 )$ in trigonometric form?

1 Answer

Explanation:

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