How do you multiply $(1 + 2 i) (2 - 5 i)$ $(1+2i)(2-5i)$ in trigonometric form?

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How do you multiply $(1 + 2 i) (2 - 5 i)$ $(1+2i)(2-5i)$ in trigonometric form?

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Answer 1 · 2018-06-25T05:34:03

$(1 + 2 i) \cdot (2 - 5 i) = 12.04 \cdot (0.9965 - + i 0.0832)$ $color(orange)((1 + 2 i)* (2 - 5 i) = 12.04 * (0.9965 -+ i 0.0832)$

Explanation:

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

$z_{1} = (1 + 2 i), z_{2} = (2 - 5 i)$ $z_1 = (1 + 2 i), z_2 = (2 - 5 i)$

$r_{1} = \sqrt{1^{2} + 2^{2}} = \sqrt{5}$ $r_1 = sqrt (1^2 + 2^2) = sqrt (5)$

$θ_{1} = {tan}^{- 1} (\frac{2}{1}) = {63.43}^{\circ}, I quadrant$ $theta _1 = tan ^-1 (2/1) = 63.43^@, " I quadrant"$

$r_{2} = \sqrt{2^{2} = - 5^{2}} = \sqrt{29}$ $r_2 = sqrt (2^2 = -5^2) = sqrt (29)$

$θ_{2} = {tan}^{- 1} (- \frac{5}{2}) = {291.8}^{\circ}, IV quadrant$ $theta _2 = tan ^-1 (-5/2) = 291.8^@ , " IV quadrant"$

$z_{1} \cdot z_{2} = (\sqrt{5} \cdot \sqrt{29}) \cdot (cos (63.43 + 291.8) + i (63.43 + 291.8))$ $z_1 * z_2 = (sqrt 5 * sqrt 29) * (cos(63.43 + 291.8) + i (63.43 + 291.8))$

$(1 + 2 i) \cdot (2 - 5 i) = 12.04 \cdot (0.9965 - + i 0.0832)$ $color(orange)((1 + 2 i)* (2 - 5 i) = 12.04 * (0.9965 -+ i 0.0832)$

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How do you multiply $(1 + 2 i) (2 - 5 i)$ $(1+2i)(2-5i)$ in trigonometric form?

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