# Complex Inverse Function Theorem

1. Feb 23, 2009

### NineIsPrime

Hi, the statement of Inverse Function Theorem for Complex functions I learned is this:

"let f:A->C, g:B->C be continuous s.t. f(A) \subset B. A,B are both open subsets of C, the complex plane. If g(f(z)) = z for all z in A, and g is analytic on f(z) where g'(f(z)) != 0, then f is analytic on z and f'(z) = 1/(g'(f(z))"

My question is this: suppose the condition that "g is analytic on f(z)" is not true. Could f(z) still be analytic at that point?

In another word, does there exist function f s.t. g(f(z)) = z, g is not analytic at f(z) but f is analytic at z ?

I've been playing around with z^(1/n) and z^n where it is possible that g(f(z)) = z but f(g(z)) != z, but I still can't find any examples.

Thanks!

2. Feb 23, 2009

### HallsofIvy

If there were, you could reverse the rolls of f and g in the theorem to say:
"let f:A->C, g:B->C be continuous s.t. g(B) \subset A. A,B are both open subsets of C, the complex plane. If f(g(z)) = z for all z in A, and f is analytic on g(z) where f'(g(z)) != 0, then g is analytic on z and g'(z) = 1/(g'(f(z))"
In order that g NOT be analytic, we would have to have f'(g(z))= 0.