How do you divide $\frac{- i - 9}{i - 2}$ $( -i-9) / (i-2)$ in trigonometric form?

Question

How do you divide $\frac{- i - 9}{i - 2}$ $( -i-9) / (i-2)$ in trigonometric form?

1 Answer

Impact of this question

1602 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-27T02:08:28

$\Rightarrow 3.4 + 2.2 i, I Quadrant$ $color(indigo)(=> 3.4 + 2.2 i, " I Quadrant"$

Explanation:

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

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

$r_{1} = {\sqrt{- 9^{2} + - 1^{2}}}^{2}) = \sqrt{82}$ $r_1 = sqrt(-9^2 + -1^2)^2) = sqrt 82$

$θ_{1} = {tan}^{- 1} (- \frac{1}{- 9}) \approx {186.3402}^{\circ} =, III Quadrant$ $theta_1 = tan ^-1 (-1/ -9) ~~ 186.3402^@ = , " III Quadrant"$

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

$θ_{2} = {tan}^{- 1} (\frac{1}{- 2}) \approx {153.4349}^{\circ}, II Quadrant$ $theta_2 = tan ^-1 (1/ -2) ~~ 153.4349^@, " II Quadrant"$

$\frac{z_{1}}{z_{2}} = \sqrt{\frac{82}{5}} (cos (186.3402 - 153.4349) + i sin (186.3402 - 153.4349))$ $z_1 / z_2 = sqrt(82 / 5) (cos (186.3402 - 153.4349) + i sin (186.3402 - 153.4349))$

$\Rightarrow 3.4 + 2.2 i, I Quadrant$ $color(indigo)(=> 3.4 + 2.2 i, " I Quadrant"$

Science

Math

Humanities

... and beyond

How do you divide $\frac{- i - 9}{i - 2}$ $( -i-9) / (i-2)$ in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you divide ( -i-9) / (i-2)−i−9i−2( -i-9) / (i-2) in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

How do you divide $\frac{- i - 9}{i - 2}$ $( -i-9) / (i-2)$ in trigonometric form?