Requirements] Here is a proof that I know and like. You need to know theminimum-modulos theorem. The maximum modulos theorem says (let us just say for disks for simplicity) that if is a continous complex function on a closed disk that is analytic on the interior of the disk then assumes it maximum on the boundary of the disk. The minimum-modulos theorem is similar, it says (under the same assumptions) assumes it minimum value on the boundaryprovidedthat is non-zero at any point inside the disk. You also need to know theidentity-theorem, if (under the same assumptions) has a sequence of distinct points inside the disk that are zeros of (i.e. ) and converges to some point inside the disk then must be the zero function on the disk.

Basic Idea] Let be an analytic function on an open set . Remember we want to show that is open, this means for any point in we can find an open disk containing that point that lies wholly in (that is just the definition of being 'open'). Let , be some point. Therefore, for some . We need to show there is some disk containing that lies in . By using translations we can safely assume that (why?). What remains now is that we have to show there is an open disk containing that lies wholly in the image of (i.e. ).

Special Circle] By the identity-theorem there must be a circle around (which is contained in ) so that does not vanish on this circle. To see why this must be true, assume to the contrary that for any circle you draw around (that lies in ) there is always some point for which vanishes. We know for some we can draw a circle around that is contained in . On this circle (as we are assuming) there is a point so that . Now draw a smaller circle around that point of radius and again by assumption there is a point on this circle so that . Now draw an even smaller circle around that point of radius and again there is a on the circle so that . In general we can create by drawing a circle of radius around . Thus, we have a sequence of point . The limit of these points is (because they shrink down to in this limiting process). Now look at the identity theorem. We found a convergent sequence of distinct zeros that converge to a point ( ) which is inside and so it forces to be the zero function on (since is an open connected set, I realized I stated the identity theorem for disks above but it works for open connected sets as well). But is not constant. Therefore, must have a circle, call it , on which it does not vanish.

Minimum] We know that attains a mimimum on (it is a compact set and continous functions on compact sets attain minimums and maximums). Let . We will show that the disk around of radius lies in the image of on the disk bounded by around (and so definitely an open disk around is contained in ). Let and if then . However, , so . Therefore, on the closed disk around with boundary we see that does not attain a minimum on the boundary, and so by minimum-modulos theorem it means for some inside this disk, and so . Thus, we have proven that any point in is the image of some point in that disk bounded by and centered at . Thus, . Thus, is open.