How do you divide $\frac{3 i + 8}{6 i + 4}$ $( 3i+8) / (6i +4 )$ in trigonometric form?

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How do you divide $\frac{3 i + 8}{6 i + 4}$ $( 3i+8) / (6i +4 )$ in trigonometric form?

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Answer 1 · 2018-06-25T06:02:46

$\frac{8 + 3 i}{4 + 6 i} = 1.1007 (0.9731 - i 0.2306)$ $color(purple)((8 + 3i) / (4 + 6i) = 1.1007 ( 0.9731 - i 0.2306)$

Explanation:

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

$z_{1} = 8 + 3 i, z_{2} = 4 + 6 i$ $z_1 = 8 +3 i , z_2 = 4 + 6 i$

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

$θ_{1} = {tan}^{- 1} (\frac{3}{8}) = {20.56}^{\circ} I Quadrant$ $theta_1 = tan ^ (-1) (3/8) = 20.56 ^@ " I Quadrant"$

$| r_{2} | = \sqrt{4^{2} + {(6)}^{2}} = \sqrt{52}$ $|r_2| = sqrt(4^2 + (6)^2) = sqrt 52$

$θ_{2} = {tan}^{- 1} (\frac{4}{6}) = {33.69}^{\circ}, I Quadrant$ $theta_2 = tan ^-1 (4/ 6) = 33.69^@ , " I Quadrant"$

$\frac{z_{1}}{z_{2}} = ∣ ∣ ∣ \sqrt{\frac{63}{52}} ∣ ∣ ∣ \cdot (cos (20.56 - 33.69) + i sin (20.56 - 33.69))$ $z_1 / z_2 = |sqrt(63/52)| * (cos (20.56 - 33.69) + i sin (20.56 - 33.69))$

$\frac{8 + 3 i}{4 + 6 i} = 1.1007 (0.9731 - i 0.2306)$ $color(purple)((8 + 3i) / (4 + 6i) = 1.1007 ( 0.9731 - i 0.2306)$

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How do you divide $\frac{3 i + 8}{6 i + 4}$ $( 3i+8) / (6i +4 )$ in trigonometric form?

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