How do you multiply $(3 + 9 i) (5 + 4 i)$ $(3+9i)(5+4i)$ in trigonometric form?

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How do you multiply $(3 + 9 i) (5 + 4 i)$ $(3+9i)(5+4i)$ in trigonometric form?

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Answer 1 · 2018-06-25T04:54:56

$(3 + 9 i) \cdot (5 + 4 i) = 60.75 \cdot (- 0.3458 + i 0.9383)$ $color(indigo)((3 + 9 i)* (5 + 4i) = 60.75 * (-0.3458 + i 0.9383)$

Explanation:

$z_{1} \cdot z_{2} = (r_{1} \cdot r_{2}) \cdot (({cos θ}_{1} + θ_{2}) + i sin (θ_{1} + θ_{2}))$ $z_1 * z_2 = (r_1 * r_2) * ((cos theta_1 + theta_2) + i sin (theta_1 + theta_2))$

$z_{1} = (3 + 9 i), z_{2} = (5 + 4 i)$ $z_1 = (3 + 9 i), z_2 = (5 + 4i)$

$r_{1} = \sqrt{3^{2} + 9^{2}} = \sqrt{90}$ $r_1 = sqrt (3^2 + 9^2) = sqrt (90)$

$θ_{1} = {tan}^{- 1} (\frac{9}{3}) = {71.57}^{\circ}, I quadrant$ $theta _1 = tan ^-1 (9/3) = 71.57^@, " I quadrant"$

$r_{2} = \sqrt{4^{2} + 5^{2}} = \sqrt{41}$ $r_2 = sqrt (4^2 + 5^2) = sqrt (41)$

$θ_{2} = {tan}^{- 1} (\frac{4}{5}) = {38.66}^{\circ}, I quadrant$ $theta _2 = tan ^-1 (4/5) = 38.66^@, " I quadrant"$

$z_{1} \cdot z_{2} = \sqrt{90} \cdot \sqrt{41}) \cdot (cos (71.57 + 38.66) + i (71.57 + 38.66))$ $z_1 * z_2 = sqrt90 * sqrt 41) * (cos(71.57 + 38.66) + i (71.57 + 38.66))$

$(3 + 9 i) \cdot (5 + 4 i) = 60.75 \cdot (- 0.3458 + i 0.9383)$ $color(indigo)((3 + 9 i)* (5 + 4i) = 60.75 * (-0.3458 + i 0.9383)$

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How do you multiply $(3 + 9 i) (5 + 4 i)$ $(3+9i)(5+4i)$ in trigonometric form?

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How do you multiply (3+9i)(5+4i) (3+9i)(5+4i) in trigonometric form?

1 Answer

Explanation:

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How do you multiply $(3 + 9 i) (5 + 4 i)$ $(3+9i)(5+4i)$ in trigonometric form?