How do you divide $\frac{i - 3}{2 i + 4}$ $( i-3) / (2i+4)$ in trigonometric form?

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How do you divide $\frac{i - 3}{2 i + 4}$ $( i-3) / (2i+4)$ in trigonometric form?

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Answer 1 · 2018-06-25T04:18:07

$\frac{3 - i}{4 + 2 i} = \frac{1}{2} \cdot (- 1 + i)$ $color(maroon)((3 - i)/ (4 + 2 i) = 1/2 * (-1 + i)$

Explanation:

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

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

$r_{1} = \sqrt{- 3^{2} + 1^{2}} = \sqrt{10}$ $r_1 = sqrt(-3^2 + 1^2) = sqrt10$

$θ_{1} = {tan}^{- 1} (\frac{1}{- 3}) = - 18.43 = {161.57}^{\circ}, II quadrant$ $theta _ 1 = tan ^ -1 (1/-3) = -18.43 = 161.57^@, " II quadrant"$

$r_{2} = \sqrt{4^{2} + 2^{2}} = \sqrt{20}$ $r_2 = sqrt(4^2 + 2^2) = sqrt20$

$θ_{2} = {tan}^{- 1} (\frac{2}{4}) = {26.57}^{\circ}, I quadrant$ $theta _ 2 = tan ^ -1 (2/4) = 26.57^@, " I quadrant"$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{10}{20}} \cdot (cos (161.57 - 26.57) + i (161.57 - 26.57))$ $z_1 / z_2 = sqrt(10/20) * (cos (161.57 - 26.57) + i (161.57 - 26.57))$

$\frac{3 - i}{4 + 2 i} = \frac{1}{\sqrt{2}} \cdot (- \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) = \frac{1}{2} \cdot (- 1 + i)$ $(3 - i)/ (4 + 2 i) = 1/sqrt2 * (-1/sqrt2 + i 1/sqrt2) = 1/2 * (-1 + i)$

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How do you divide $\frac{i - 3}{2 i + 4}$ $( i-3) / (2i+4)$ in trigonometric form?

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