Let g be defined on a set containing the range of a function f. If f is continuous at z0 and g is continuous at f(z0), then g(f(z)) is continuous at z0.
This is a classic and easy to find on the internet theorem...but let met give you an outline of the proof.
Let and let where are both continuous. (we actually only have to have \mapsto C" alt="g'\mapsto C" /> where but that is neither here nor there). Then is also continuous.
Proof: Let be arbitrary. Since and is continuous there exists some such that . And since is continuous there exists some such that and so
Or much nicer, if you are using the more topological definition (which is equivalent to the regular in metric spaces) that is continuous iff is open in whenever is open in merely note that and is ope in and so is open in . Done.