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))#