How do you multiply $(- 10 - 2 i) (3 - i)$ $(-10-2i)(3-i)$ in trigonometric form?

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How do you multiply $(- 10 - 2 i) (3 - i)$ $(-10-2i)(3-i)$ in trigonometric form?

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Answer 1 · 2018-06-25T05:02:27

$(10 - 2 i) \cdot (3 - i) = 32.25 \cdot (- 0.9923 + i 0.1239)$ $color(magenta)((10 - 2 i)* (3 - i) = 32.25 * (-0.9923 + i 0.1239)$

Explanation:

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

$z_{1} = (- 10 - 2 i), z_{2} = (3 - i)$ $z_1 = (-10 - 2 i), z_2 = (3 - i)$

$r_{1} = \sqrt{- 10^{2} + - 2^{2}} = \sqrt{104}$ $r_1 = sqrt (-10^2 + -2^2) = sqrt (104)$

$θ_{1} = {tan}^{- 1} (- \frac{2}{- 10}) + {191.31}^{\circ}, III quadrant$ $theta _1 = tan ^-1 (-2/-10) + 191.31^@, " III quadrant"$

$r_{2} = \sqrt{3^{2} = - 1^{2}} = \sqrt{10}$ $r_2 = sqrt (3^2 = -1^2) = sqrt (10)$

$θ_{2} = {tan}^{- 1} (- \frac{1}{3}) = - {18.43}^{\circ} = {341.57}^{\circ}, IV quadrant$ $theta _2 = tan ^-1 (-1/3) = -18.43^@ = 341.57^@, " IV quadrant"$

$z_{1} \cdot z_{2} = (\sqrt{104} \cdot \sqrt{10}) \cdot (cos (191.31 + 341.57) + i (191.31 + 341.57))$ $z_1 * z_2 = (sqrt104 * sqrt 10) * (cos(191.31 + 341.57) + i (191.31 + 341.57))$

$(10 - 2 i) \cdot (3 - i) = 32.25 \cdot (- 0.9923 + i 0.1239)$ $color(magenta)((10 - 2 i)* (3 - i) = 32.25 * (-0.9923 + i 0.1239)$

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How do you multiply $(- 10 - 2 i) (3 - i)$ $(-10-2i)(3-i)$ in trigonometric form?

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