How do you write the following in trigonometric form and perform the operation given $- 2 i (1 + i)$ $-2i(1+i)$ ?

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How do you write the following in trigonometric form and perform the operation given $- 2 i (1 + i)$ $-2i(1+i)$ ?

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Answer 1 · 2018-07-09T11:18:25

$- 2 i \cdot (1 + i) = 2 - 2 i$ $color(purple)(-2i * (1 + i) = 2 - 2 i)$

Explanation:

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

$z_{1} = 0 - 2 i, z_{2} = 1 + i$ $z_1 = 0 - 2 i, z_2 = 1 + i$

$r_{1} = \sqrt{0^{2} + - 2^{2}} = 2$ $r_1 = sqrt(0^2 + -2^2) = 2$

$θ_{1} = {tan}^{- 1} (- \frac{2}{0}) = 270^{\circ}, IV Quadrant$ $theta_1 = tan ^ (-1) (-2/0) = 270 ^@, " IV Quadrant"$

$r_{2} = \sqrt{1^{2} + {(1)}^{2}} = \sqrt{2}$ $r_2 = sqrt(1^2 + (1)^2) = sqrt 2$

$θ_{2} = {tan}^{- 1} (\frac{1}{1}) = 45^{\circ}, I Quadrant$ $theta_2 = tan ^-1 (1/ 1) = 45^@, " I Quadrant"$

$z_{1} \cdot z_{2} = (2 \cdot \sqrt{2}) (cos (270 + 45) + i sin (270 + 45))$ $z_1 * z_2 = (2*sqrt(2)) (cos (270 + 45) + i sin (270 + 45))$

$- 2 i \cdot (1 + i) = 2 - 2 i$ $color(purple)(-2i * (1 + i) = 2 - 2 i)$

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How do you write the following in trigonometric form and perform the operation given $- 2 i (1 + i)$ $-2i(1+i)$ ?

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Explanation:

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How do you write the following in trigonometric form and perform the operation given -2i(1+i)−2i(1+i)-2i(1+i)?

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How do you write the following in trigonometric form and perform the operation given $- 2 i (1 + i)$ $-2i(1+i)$ ?