Complex number to polar form I need to take a raincheck with this problem. I want to make sure I haven't messed up some fundamental idea.

Convert the complex number $$-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i$$ to polar form.

I took the modulus as below,
$$\lvert-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i \rvert = \sqrt{(\dfrac{-1}{2})^2 + (\dfrac{\sqrt 3}{2})^2} = 1$$
And the argument as below,
$$arg(-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i) = \tan^{-1} (\dfrac{\sqrt 3}{2} \times \dfrac{-2}{1}) = -60 = -\dfrac{\pi}{3}$$
Hence the complex number in polar form is,
$$-\cos \dfrac{\pi}{3} + i \sin \dfrac{\pi}{3}$$


But, The required answer is $$\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}$$


I thought of converting the -60 to positive, as 360 - 60 = 300, ie:- $$\dfrac{5\pi}{3}$$. I have a feeling I am missing something important. Can you guys tell me where I am going wrong? Thanks for all your help!

The problem is you forgot to add $\pi$ in calculating the argument. See the wiki page which shows all the formulas for computing the argument depending on $x$ and $y$.  
Taking $z=x+yi$ to be your complex number, here $x$ is negative and $y$ is positive, so 
$$
\text{arg}(-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i) = \tan^{-1} (\dfrac{\sqrt 3}{2} \times \dfrac{-2}{1})+\pi = -\dfrac{\pi}{3}+\pi=\frac{2\pi}{3}.
$$
