Hey, I don't know if this suffices as a proof or something which is sufficiently within the syllabus, but a possible approach uses the fact that reix = r(cosx + isinx) = rcis(x) (which comes from Euler's formula: http://en.wikipedia.org/wiki/Euler's_formula ... If you look up the taylor series stuff it's understandable using what is within the syllabus and the rest follows).
So that it makes more sense, first consider the simple case where the circle is centered around the origin:
Let f(x) and F(x) be functions where f(x) = reix = rcis[x] = z (and 0 <= x <= 2pi), so f(x) is the anticlockwise circle from 0 to 2pi of radius r in the complex plane, and F(x) = z-1.
F(x) = F(x) = z-1 = (1/r)e-ix which is a circle of radius (1/r) in the complex plane going clockwise from 0 to -2pi.
...
If the circle can be anywhere then let f(x) = reix + z2 = z1 + z2 = z (z2 is any complex number x2 + iy2).
F(x) = z-1
= 1/(z1 + z2)
= conj.(z1 + z2)/(|z1 + z2|)
= (conj.z1 + conj.z2)/k ... (k since the modulus of the denominator will take a real value)
= (r/k)e-ix + (1/k)conj.z2
which is a clockwise cirlce with a radius r/k... note that when z2 = 0 then k = r2 which is the first result. Anyhow, there is my BoS cameo... sorry for any mistakes, I'm quite out of practice.