How do you divide #(2i-7) / (-2i-8)# in trigonometric form?

1 Answer
Douglas K.
Jun 16, 2017

Division in trigonometric form is:
#(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))#

Explanation:

Given: #(2i-7)/(-2i-8)#

#r_1=sqrt((-7)^2+2^2)#

#r_1=sqrt53#

To find the value of #theta_1#, we must observe that the real part is negative and the imaginary part is positive; this places the angle in the 2nd quadrant:

#theta_1=pi+tan^-1(2/-7)#

#theta_1=pi-tan^-1(2/7)#

Moving on to #r_2#:

#r_2 = sqrt((-8)^2+(-2)^2)#

#r_2=sqrt(68)#

To find the value of #theta_2#, we must observe that the real part is negative and the imaginary part is negative; this places the angle in the 3nd quadrant:

#theta_2= pi+tan^-1((-2)/-8)#

#theta_2= pi+tan^-1(1/4)#

#(2i-7)/(-2i-8)=sqrt(53/68)(cos(-tan^-1(2/7)-tan^-1(1/4))+isin(-tan^-1(2/7)-tan^-1(1/4)))#

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