Complex Analysis?

Question

Two questions: Please keep in mind that this is an introductory course that mixes computation and a little bit of theory. Your help will be greatly appreciated.

1.) Show that f(z)=|z| is continous.

2.) Use the fact that a function is continuous if and only if the inverse image of every open set is open to show that a composition of two continuous functions is continuous.

Please include an explanation. Thanks...

Answer 1

1) Let z0 be a complex number. For every z, we have, according to one of the variations of the triangle inequality, that | |z| - |z0)| | <= |z - z0|. Given eps >0, if we put d = eps, then for ever complex z satisfying |z - z0| < d, we have
| |z| - |z0)| | < d = eps. Since eps is arbitrary, it follows f is continuous on each zo of the complex plane. So, it's continuous.

2) Suppose g is defined on a set Dg and has values in a set A and f is defined on a subset Dg of A and has values in a set B. Let O be any open subset of B and let h = f og. We know the inverse image of O under h is given by g^(-1)[f^(-1)(O)]. Since f is continuous and O is open in B, f^(-1)(O) is a subset of Df, open in A. Since g is continuous and f^(-1)(O) is open in A, it follows g^(-1)[f^(-1)(O)] = h(-1)(O) is open in Dg. Since this holds for every open set of B, it follows h = f o g is continuous on it's domain, which is a subset of Dg.

Actually, this conclusion is valid in every topological space, not only in the complexes.