Is there any way to find a angle of a complex number without a calculator? Transforming the complex number $z=-\sqrt{3}+3i$ into polar form will bring me to the problem to solve this two equations to find the angle $\phi$: $\cos{\phi}=\frac{\Re z}{|z|}$ and $\sin{\phi}=\frac{\Im z}{|z|}$. 
For $z$ the solutions are $\cos{\phi}=-0,5$ and $\sin{\phi}=-0,5*\sqrt{3}$.
Using Wolfram Alpha or my calculator I can get $\phi=\frac{2\pi}{3}$ as solution. But using a calculator is forbidden in my examination. 
Do you know any (cool) ways to get the angle without any other help?
 A: memorize sin/cos for angles $0,{\pi \over 6},{\pi \over 4},{\pi \over 3},{\pi \over 2}$ and in your examination look at the unit circle to figure out what is going on
A: You could start with known angles (s.a. multiples of 45 or 30 degrees) and work your way from there using the formulas for trigonometric half-angles, and sums of angles. If you don't remember them, use: http://www.sosmath.com/trig/Trig5/trig5/trig5.html
For instance, in degrees, if you want cos(41), you can use the sequence:
45+120=165
165/2=82.5 
82.5/2=41.25
and use the trigonometric identities to fall back from cos(45) and cos(120) to the approximation cos(41.25) 
A: The direct calculation is 
$$\arg(-\sqrt 3+ 3i)=\arctan\frac{3}{-\sqrt{3}}=\arctan (-\sqrt 3)=\arctan \frac{\sqrt 3/2}{-1/2}$$
As the other two answers remark, you must learn by heart the values of at least the sine and cosine at the main angle values between zero and $\,\pi/2\,$ and then, understanding the trigonometric circle, deduce the functions' values anywhere on that circle.
The solution you said you got is incorrectly deduced as you wrote 
$$\,cos\phi=-0,5\,,\,\sin\phi=-0,5\cdot \sqrt 3\,$$ 
which would give you both values of $\,\sin\,,\,\cos\,$ negative, thus putting you in the third quadrant of the trigonometric circle, $\,\{(x,y)\;:\;x,y<0\}\,$, which is wrong as the value indeed is $\,2\pi/3\,$ (sine is positive!), but who knows how did you get to it.
In the argument's calculation above please do note the minus sign is at $\,1/2\,$ in the denominator, since that's what corresponds to the $\,cos\,$ in the polar form, but the sine is positive thus putting us on the second quadrant $\,\{(x,y,)\;:\;x<0<y\}\,$ .
So knowing that 
$$\sin x = \sin(\pi - x)\,,\,\cos x=-\cos(\pi -x)\,,\,0\leq x\leq \pi/2$$
and knowing the basic values for the basic angles, gives you now
$$-\frac{1}{2}=-\cos\frac{\pi}{3}\stackrel{\text{go to 2nd quad.}}=\cos\left(\pi-\frac{\pi}{3}\right)=\cos\frac{2\pi}{3}$$
$$\frac{\sqrt 3}{2}=\sin\frac{\pi}{3}\stackrel{\text{go to 2nd quad.}}=\sin\left(\pi-\frac{\pi}{3}\right)=\sin\frac{2\pi}{3}$$
