How do you divide #( 6i-4) / ( -3 i -5 )# in trigonometric form?

1 Answer
Shwetank Mauria
May 20, 2016

#(6i-4)/(-3i-5)=sqrt(26/17)(cosrho+isinrho)# where #rho=tan^(-1)21#

Explanation:

Let us first write #(6i-4)# and #(-3i-5)# in trigonometric form.

#a+ib# can be written in trigonometric form #re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)#,
where #r=sqrt(a^2+b^2)# and #tantheta=b/a# or #theta=arctan(b/a)#

Hence #6i-4=(-4+6i)=sqrt((-4)^2+6^2)[cosalpha+isinalpha]# or

#sqrt52e^(ialpha)#, where #tanalpha=6/(-4)=-3/2# and

#-3i-5=(-5-3i)=sqrt((-5)^2+(-3)^2)[cosbeta+isinbeta]# or

#sqrt34e^(ibeta]#, where #tanbeta=(-3)/(-5)=3/5#

Hence #(6i-4)/(-3i-5)=(sqrt52e^(ialpha))/(sqrt34e^(ibeta])=sqrt(26/17)e^(i(alpha-beta))=sqrt(28/17)(cos(alpha-beta)+isin(alpha-beta))#

Now, #tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)#

= #(-3/2-3/5)/(1+(-3/2)*(3/5))=(-21/10)/(1-9/10)=-21/10*10/1=-21#

Hence #(6i-4)/(-3i-5)=sqrt(26/17)(cosrho+isinrho)# where #rho=tan^(-1)21#

Related questions
See all questions in The Trigonometric Form of Complex Numbers
Impact of this question
1146 views around the world
You can reuse this answer
Creative Commons License