Question

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

Answer 1

`(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`