How do you multiply $(- 2 - 9 i) (- 1 - 6 i)$ $(-2-9i)(-1-6i)$ in trigonometric form?

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How do you multiply $(- 2 - 9 i) (- 1 - 6 i)$ $(-2-9i)(-1-6i)$ in trigonometric form?

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Answer 1 · 2018-07-08T02:54:45

$\frac{- 2 - 9 i}{- 1 - 6 i} = - 50.66 - 24.06$ $color(chocolate)((-2 - 9i) / (-1 - 6i) = -50.66 - 24.06)$

Explanation:

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

$z_{1} = - 2 - 9 i, z_{2} = - 1 - 6 i$ $z_1 = -2 - 9i , z_2 = -1 - 6i$

$| r_{1} | = \sqrt{- 2^{2} + - 9^{2}} = \sqrt{85}$ $|r_1| = sqrt(-2^2 + -9^2) = sqrt 85$

$θ_{1} = {tan}^{- 1} (- \frac{9}{- 2}) = {254.05}^{\circ} III Quadrant$ $theta_1 = tan ^ (-1) (-9/-2) = 254.05 ^@ " III Quadrant"$

$| r_{2} | = \sqrt{- 1^{2} + {(- 6)}^{2}} = \sqrt{37}$ $|r_2| = sqrt(-1^2 + (-6)^2) = sqrt 37$

$θ_{2} = {tan}^{- 1} (- \frac{6}{- 1}) = {260.54}^{\circ}, III Quadrant$ $theta_2 = tan ^-1 (-6/ -1) = 260.54^@ , " III Quadrant"$

$\frac{z_{1}}{z_{2}} = ∣ ∣ \sqrt{85 \cdot 37} ∣ ∣ \cdot (cos (254.05 + 260.54) + i sin (254.05 + 260.54))$ $z_1 / z_2 = |sqrt(85*37)| * (cos (254.05 + 260.54) + i sin (254.05 + 260.54))$

$\frac{- 2 - 9 i}{- 1 - 6 i} = - 50.66 - 24.06$ $color(chocolate)((-2 - 9i) / (-1 - 6i) = -50.66 - 24.06)$

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How do you multiply $(- 2 - 9 i) (- 1 - 6 i)$ $(-2-9i)(-1-6i)$ in trigonometric form?

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Explanation:

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