I am looking to find the trig/ polar form of 3 -the Sq root of 3i any help would be appreciated!
So you are asked to convert $\displaystyle z:= 3-\sqrt{3}i$ to the form $\displaystyle |z|\cdot e^{i\phi}$.
You have $\displaystyle |z|=\sqrt{3^2+\sqrt{3}^2}=\sqrt{12}=2\sqrt{3}$, and $\displaystyle \phi = \tan^{-1}\frac{-\sqrt{3}}{3}=-\frac{\pi}{6}$.
To sum up: $\displaystyle 3-\sqrt{3}i=2\sqrt{3}\cdot e^{-i\frac{\pi}{6}}$