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

Question

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

1 Answer

Impact of this question

1610 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-25T05:13:43

$(9 - 2 i) \cdot (2 + i) = 18.44 \cdot (0.9701 + i 0.2426)$ $color(crimson)((9 - 2 i)* (2 + i) = 18.44 * (0.9701 + i 0.2426)$

Explanation:

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

$z_{1} = (9 - 2 i), z_{2} = (2 + i)$ $z_1 = (9 - 2 i), z_2 = (2 + i)$

$r_{1} = \sqrt{9^{2} + - 23^{2}} = \sqrt{85}$ $r_1 = sqrt (9^2 + -23^2) = sqrt (85)$

$θ_{1} = {tan}^{- 1} (- \frac{2}{9}) = {347.47}^{\circ}, IV quadrant$ $theta _1 = tan ^-1 (-2/9) = 347.47^@, " IV quadrant"$

$r_{2} = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$ $r_2 = sqrt (2^2 + 1^2) = sqrt (5)$

$θ_{2} = {tan}^{- 1} (\frac{1}{2}) = {26.57}^{\circ}, I quadrant$ $theta _2 = tan ^-1 (1/2) = 26.57^@ , " I quadrant"$

$z_{1} \cdot z_{2} = (\sqrt{85} \cdot \sqrt{5}) \cdot (cos (347.47 + 26.57) + i (347.47 + 26.57))$ $z_1 * z_2 = (sqrt 85 * sqrt 5) * (cos(347.47 + 26.57) + i (347.47 + 26.57))$

$(9 - 2 i) \cdot (2 + i) = 18.44 \cdot (0.9701 + i 0.2426)$ $color(crimson)((9 - 2 i)* (2 + i) = 18.44 * (0.9701 + i 0.2426)$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you multiply (9−2i)(2+i) (9-2i)(2+i) in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

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