Where do Conformal maps map regions to? 
Question: How do you generally work out where a certain region $U$ gets mapped to under a given conformal map $f(z)$?


Context:
I have a region, $U$ which is a unit circle with a smaller circle removed, this circle being $|z-\frac25|\leq\frac25$. The conformal map is $f(z)=\frac{z-2}{2z-1}$. I need to find out where $U$ gets mapped to under $f$.

Thoughts + attempt:
I don't know how to work out where this region gets mapped to. I let $z=\cos\theta+i\sin\theta$ and determined that any point on the unit circle gets mapped to another point on the unit circle. Also $f^{-1}(z)=f(z)$. I think this is enough to conclude that the unit circle gets mapped to itself. However this method was quite long-winded so I feel there should be a more efficient way to work this out. I am not sure how to proceed with finding where $f$ maps the other circle, I plugged in $f(z)$ in place of $z$ in the equation of the circle, and got $|z-8|=4|z-\frac12|$, but I don't know what sort of line this is. Is this method the correct way to go about it?
PS: I don't know if I may be missing any tags, so if you feel like some tag which I haven't mentioned fits, please feel free to add it in.

Your $f$ is a Möbius transformation.  These have a number of important properties which make it easy to identify $f(U)$:


*

*They are one-to-one from the Riemann sphere to itself.

*They map straight lines or circles to straight lines or circles.


Thus all you have to do is find the straight line or circle to which $f$ maps the boundary circles of $U$, and $f(U)$ is the region between these.  Note also that a straight line or circle can be determined by three points on it.
Thus if you can show that $f(1)$, $f(-1)$ and $f(i)$ are on the unit circle, you can conclude that $f$ maps the unit circle to itself.
