How do you divide ( - 6 i + 5) / (i+ 2 ) in trigonometric form?

1 Answer
Shiva Prakash M V
Aug 9, 2018

(-6i+5)/(i+2)=sqrt(61/5)cis(-tan^-1(17/-9))

Explanation:

(-6i+5)/(i+2)=f(r,theta)
z_1 = -6i+5
r_1 = sqrt((-6)^2+5^2) =sqrt(36+25)=sqrt61
theta_1=tan^-1 (5/-6)=2pi-tan^-1(5/6)
z_2=i+2
r_2 = sqrt(1^2+2^2) =sqrt(1+4)=sqrt5
theta_2=tan^-1 (2/1)=2pi+tan^-1 2
Thus,
(-6i+5)/(i+2)=(sqrt61, (2pi-tan^-1(5/6)))/(sqrt5, (2pi+tan^-1 2))
By De-Moivre's Theorem,

(sqrt61, (2pi-tan^-1(5/6)))/(sqrt5, (2pi+tan^-1 2))
=sqrt 61 /(sqrt 5) xxcis(2pi-tan^-1(5/6)-(2pi+tan^-1 2))

=sqrt(61/5)cis(-tan^-1(5/6)-tan^-1 2)
=sqrt(61/5)cis(-(tan^-1(5/6)+tan^-1 2))

tan^-1 (5/6) +tan^-1 2 = tan^-1 ((5/6+2)/(1-5/6xx2))

=tan^-1((5xx1+2xx6)/(1xx6-5xx3))=tan^-1 ((5+12)/(6-15))

=tan^-1(17/-9)
Thus,
(-6i+5)/(i+2)=sqrt(61/5)cis(-tan^-1(17/-9))

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