Let f(z) be analytic in a domain D, and suppose that f(z)-f(zo) has a
zero of the order n at zo in D.  Prove that for epsilon>0 sufficiently
samll, there exists a delta>0 such that for all w in |w-f(zo)|<delta
the equation f(z)-w had exactly n roots in |z-zo|<epsilon

hi fmunshi,

I don't have time now to write out a complete solution, but this
should get you (or whatever researcher decides to work on this)
started:

Assume for simplicity that z0=0, f(z0)=0. Since f has a root of order
n, we may write f(z) = z^n * g(z), where g is analytic in a
neighborhood of 0, and g(0) is NOT 0.

Since g(0) is not 0, there is also a neighborhood of 0 where g(z) is
not 0. Therefore in that neighborhood there exists a function log
g(z), i.e. there exists an analytic function h(z) such that
exp(h(z))=g(z).

Now define a function

w(z) = z * exp((1/n)*h(z))

Then

w(z)^n = z^n * exp(h(z)) = z^n * g(z) = f(z).

In other words, if you think of z->w as a change of coordinates, then
in the w coordinate f(z) "looks like" w^n. So in the w coordinate the
answer to your question is trivial, there are obviously exactly n
solutions to any equation w^n = a  (a not 0).

Now it should be simple to show that z->w is a local analytic
isomorphism, and then use the inverse function theorem to deduce that
the same is true (in the small neighborhood of 0 where we are working)
for the equation f(z)=a.

Cheers,
dannidin