Question

How do you divide `( 2i-6) / ( -12i +4 )` in trigonometric form?

Answer 1

`(2i-6)-:(-12i+4)=1/2(cosrho+isinrho)` where `rho=tan^(-1)(-5/3)`

Explanation:

Let us first write `(2i-6)` and `(-12i+4)` 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 `2i-6=(-6+2i)=sqrt((-6)^2+2^2)[cosalpha+isinalpha]` or

`sqrt40e^(ialpha)`, where `tanalpha=(-1)/3` and

`-12i+4=(4-12i)=sqrt(4^2+(-12)^2)[cosbeta+isinbeta]` or

`sqrt160e^(ibeta]`, where `tanbeta=(-12)/4=-3`

Hence `(2i-6)-:(-12i+4)=(sqrt40e^(ialpha))/(sqrt160e^(ibeta])=sqrt(1/4)e^(i(alpha-beta))=e^(i(alpha-beta))/2=1/2(cos(alpha-beta)+isin(alpha-beta))`

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

= `(-1/3+(-3))/(1+(-1/3)*(-3))=(-10/3)/2=-5/3`

Hence `(2i-6)-:(-12i+4)=1/2(cosrho+isinrho)` where `rho=tan^(-1)(-5/3)`