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!
 A: The problem is that points can be expressed in polar form in more than one way.  Take a look at the diagram below:

There's your point (in blue).  As you can see, it's in the second quadrant, and the angle $\theta = \frac{2\pi}{3}$ passes through it.  So one could accurately say that the point's polar representation is $r = 1, \theta = \frac{2\pi}{3}$, i.e. $$z = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)$$
You'll notice, though, that $\theta = -\frac{\pi}{3}$ represents the same angle, but in the opposite direction.  With this in mind, one could just as accurately say that the point's polar representation is $r = -1, \theta = -\frac{\pi}{3}$, i.e. $$z = -\cos\left(-\frac{\pi}{3}\right) - i\sin\left(-\frac{\pi}{3}\right)$$
Because $\cos\left(\frac{2\pi}{3}\right) = -\cos\left(-\frac{\pi}{3}\right)$ and $\sin\left(\frac{2\pi}{3}\right) = -\sin\left(-\frac{\pi}{3}\right)$, these two polar representations are equivalent.  In other words, there's nothing wrong with concluding, as you did, that $\theta = -\frac{\pi}{3}$  You and the so-called "required answer" can reasonably disagree here and both be correct.
But you still did something wrong.  If you choose to use $\theta = -\frac{\pi}{3}$, then you must also choose $r = -1$, yet you choose $r = 1$, based on your modulus computation.  How can you avoid making this mistake in the future?
Simple: look at the quadrant in which the point lies.  The point $-\frac{1}{2} + \frac{\sqrt{3}}{2}i$ has negative real part and positive imaginary part.  Therefore it must be in the second quadrant.  The angle $\theta = -\frac{\pi}{3}$ extends into the fourth quadrant, so if you intend to use that angle, then you must also make $r$ negative.
A: If you draw the picture, it often points you in the right direction.  You can see in the picture how the 30-60-90 triangle is oriented and you will get it right every time. 
A: The problem is in how you calculated the argument using the arctangent. The range of the arctangent is only $(-\pi/2,\pi/2)$, so you can't get the full range of arguments in this way. Your number is in the second quadrant, which yields the same tangent values as the fourth quadrant, and the argument you calculated is in the fourth quadrant. You need to add $\pi$ to get the right argument.
There's another, unrelated error: The cosine is even and the sine is odd, so the negative sign of the argument you calculated should have appeared in front of the sine but not the cosine; i.e., the polar form corresponding to the argument you calculated is
$$\cos\frac{\pi}{3}-\mathrm i\sin\frac{\pi}{3}\;.$$
A: 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 
$$
\text{arg}(-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i) = \tan^{-1} (\dfrac{\sqrt 3}{2} \times \dfrac{-2}{1})+\pi = -\dfrac{\pi}{3}+\pi=\frac{2\pi}{3}.
$$
