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))#

Related questions
See all questions in The Trigonometric Form of Complex Numbers
Impact of this question
1508 views around the world
You can reuse this answer
Creative Commons License