Rotation around complex number 
The function
  $$f(z) = \frac{(-1 + i \sqrt{3}) z + (-2 \sqrt{3} - 18i)}{2}$$
  represents a rotation around some complex number $c$. Find $c$.

How would I start this? 
Thanks.
 A: So far I can see two solutions. The first one is the one suggested by hardmath, which is probably the simplest one. I think it is quite instructive to write it down. We must solve the equation $f(c)=c$, which gives:
$$(-1+\sqrt{3}i)c+(-2\sqrt{3}-18i) = 2c $$
$$(-1+\sqrt{3}i)c-2c+(-2\sqrt{3}-18i) = 0 $$
$$ (-3+\sqrt{3}i)c = 2\sqrt{3}+18i $$
$$ c = \frac{2\sqrt{3}+18i}{-3+\sqrt{3}i} = \sqrt{3}-5i $$
My second solution would be to compute $f(0) = -\sqrt{3}-9i$, and then compute $f(f(0)) = f(-\sqrt{3}-9i) = 4\sqrt{3}-6i$. These three complex numbers (i.e. $0$, $-\sqrt{3}-9i$, and $4\sqrt{3}-6i$) form the vertices of a triangle, and the center of its circumcircle would be the center of rotation $c$ that we are looking for. Now we can find $c$ as the intersection of two perpendicular bisectors. 
Nevertheless, if we still have the curiosity to compute $f(f(f(0))) = f(4\sqrt{3}-6i)$, we find that it yields $0$. That means that our triangle is equilateral, and its circumcenter $c$ now becomes easier to find, since it coincides with the incenter and the orthocenter. Additionally, we see that the angle of  rotation is $120º$. 
