How do you divide $(i+12) / (9i-10)$ in trigonometric form?

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Answer 1 · 2017-04-03T17:08:19

Use $(r_1(cos(theta_1)+isin(theta_1)))/(r_2(cos(theta_2)+isin(theta_2))) = r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))$

Compute $r_1$ :

$r_1 = sqrt(12^2+1^2)$

$r_1 = sqrt(145)$

Compute $theta_1$

$theta_1 = tan^-1(1/12)$

Compute $r_2$ :

$r_2 = sqrt(-10^2+9^2)$

$r_2 = sqrt(181)$

Compute $theta_2$ (second quadrant)

$theta_2 = tan^-1(9/-10)+pi$

The resulting division is:

$sqrt(145/181)(cos(tan^-1(1/12)-tan^-1(9/-10)-pi)+isin(tan^-1(1/12)-tan^-1(9/-10)-pi))$

How do you divide (i+12) / (9i-10)(i+12) / (9i-10) in trigonometric form?