Question

How do you divide `( 2i-1) / ( 3i +5 )` in trigonometric form?

Answer 1

`(2i-1)/(3i+5)=sqrt(5/34)(cosrho+isinrho)` where `rho=tan^(-1)13`

Explanation:

Let us first write `(2i-1)` 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 `2i-1=(-1+2i)=sqrt((-1)^2+2^2)[cosalpha+isinalpha]` or

`sqrt5e^(ialpha)`, where `tanalpha=6/(-4)=2/(-1)=-2` and

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

`sqrt34e^(ibeta]`, where `tanbeta=3/5`

Hence `(2i-1)/(3i+5)=(sqrt5e^(ialpha))/(sqrt34e^(ibeta])=sqrt(5/34)e^(i(alpha-beta))=sqrt(5/34)(cos(alpha-beta)+isin(alpha-beta))`

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

= `(-2-3/5)/(1+(-2)*(3/5))` = `(-13/5)/(-1/5)=-13/5*(-5)/1=13`

Hence `(2i-1)/(3i+5)=sqrt(5/34)(cosrho+isinrho)` where `rho=tan^(-1)13`