Finding argument of complex number without calculator I am solving some exercises in the book I am reading. In this particular exercise I should find real and imaginary part of 
$$ \left ( {1 + i \sqrt{3}\over 2}\right )^n$$
My idea was to calculate the argument and the absolute value and then use polar representation. 
But I think the idea is not to use a calculator. So I am stuck on 
$$ \arctan \sqrt{3}$$
If I use a calculator to find this value I can easily solve the exercise. 

How to calculate $ \arctan \sqrt{3}$ without using a calculator? Is it
  possible?

Edit
If it's possible any general method is most appreciated since I am already stuck at the next exercise where I am trying to find the argument of $-3+i$. 
 A: There's an even easier way. You can verify by direct computation that
$$z^6=\left({1+i\sqrt 3\over 2}\right)^6=1$$
and that no smaller power works. This along with the fact that both real and imaginary parts are positive, i.e. in the first quadrant tells you that $z=e^{i\pi\over 3}$, since this is the only $6^{th}$ root of $1$ in the first quadrant. (all others aside from $1$ have an argument at least $120^\circ={2\pi\over 3}$ which is outside the first quadrant.
Then you know that if $n=6k+r$ with $0\le r\le 5$ that the argument is $r$ times that of $z$, i.e. $\theta={\pi r\over 3}$.
A: $$(\frac{1+i\sqrt{3}}{2})^n$$
$$(\frac{1}{2} + \frac{i\sqrt{3}}{2})^n$$
$$Let \ \ z=\frac{1}{2} + \frac{i\sqrt{3}}{2}$$
$$r = |z| = \sqrt{ (\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = 1$$
$$\sin\theta = \frac{y}{r} \ \ , \ \ \cos\theta = \frac{x}{r}$$
$$\sin\theta = \frac{\sqrt{3}}{2} \ \ , \ \ \cos\theta = \frac{1}{2}$$
Here both $\sin\theta$ and $\cos\theta$ are in 1st quadrant so
$$\theta = \frac{\pi}{3}$$ 
$$\operatorname{Arg} z = \frac{\pi}{3}$$
