Convert to polar form

Question

How do I proceed to convert $$\frac{14\sqrt{3} +30i}{3\sqrt{3}i - 2}$$ to polar form. I've taken the conjugate twice to get: $$\frac{(30i+14\sqrt{3})(3\sqrt{3}i+2)(-4-27i)}{745}$$

Can't seem to get further than this, any pointers would be great!

Answer 1

Hint: multiply numerator and denominator by the conjugate of $3\sqrt{3}i - 2$. Then you should obtain something like this: $$z:=\frac{14\sqrt{3} +30i}{3\sqrt{3}i - 2}=\frac{(14\sqrt{3} +30i)(-3\sqrt{3}i - 2)}{(3\sqrt{3})^2+4}\\=\frac{62(-3i+\sqrt{3})}{31}=2(-3i+\sqrt{3})=4\sqrt{3}\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)=4\sqrt{3}e^{-i\pi/3}.$$ So $|z|=4\sqrt{3}$ and $\mbox{arg}(z)=-\pi/3$ (or $2\pi-\pi/3=5\pi/3$).

Answer 2

Hint:

Why don't we convert the numerator to polar form $r_1\exp (i \theta_1)$

and the denominator to polar from $r_2 \exp(i \theta_2)$.

then the magnitude would be $\frac{r_1}{r_2}$ and the corresponding angle is $\theta_1-\theta_2$