Question

How do you convert `((2sqrt3) 2i)/(1-sqrt3i)` to polar form?

Answer 1

`((2sqrt3)2i)/(1-sqrt3i)=2sqrt3(cos(-pi/6)+isin(-pi/6))`

Explanation:

`((2sqrt3)2i)/(1-sqrt3i)`

Let us first rationalize it by multiplying numerator and denominator by complex conjugate of denominator. Then above becomes

`(((2sqrt3)2i)(1+sqrt3i))/((1-sqrt3i)(1+sqrt3i))`

= `(4sqrt3i+12i^2)/(1^2-(sqrt3i)^2)`

= `(4sqrt3i+12i^2)/(1^2-(3i^2))=(4sqrt3i+12i^2)/(1+3)`

= `(-12+4sqrt3i)/4=-3+sqrt3i`

Now when a complex number `a+bi` is written in polar form as `r(costheta+isintheta)`, `r=sqrt(a^2+b^2)` and `theta=arctan(b/a)`

Hence, here `r=sqrt((-3)^2+(sqrt3)^2)=sqrt(9+3)=sqrt12=2sqrt3`

and `theta=arctan(sqrt3/(-3))=arctan(-1/sqrt3)=-pi/6`

Hence, `((2sqrt3)2i)/(1-sqrt3i)=2sqrt3(cos(-pi/6)+isin(-pi/6))`