# Homework Help: Complex Analysis-Analytic Functions

1. Jun 14, 2010

### WannaBe22

1. The problem statement, all variables and given/known data
Prove or find a proper counterexample:
There exists an analytic function f(z) in a pierced neighborhood of z=0 (i.e a neighborhood of z=0 , which doesn't contain z=0) which satisfies: $$f^3 (z) = z^2$$.

2. Relevant equations
3. The attempt at a solution
Got no clue...

Hope you'll be able to help me

2. Jun 14, 2010

### HallsofIvy

If, by $f^3(z)$, you mean $(f(x))^3$ and not $f(f(f(x)))$, then all you need to do is show that $f(x)= z^{2/3}$ is analytic everywhere except at z= 0.

3. Jun 14, 2010

### WannaBe22

Actually, it's excatly what I mean, but why the function you gave isn't analytic at 0?
It's actually an entire function...isn't it?

Thanks

4. Jun 14, 2010

### HallsofIvy

I didn't say it wasn't- but your problem was to determine if "There exists an analytic function f(z) in a pierced neighborhood of z=0 (i.e a neighborhood of z=0 , which doesn't contain z=0)".

If such a function is analytic everywhere it is certainly analytic in that neighborhood!

But what make you so certain $z^{2/3}$ is analytic at z= 0? It's derivative does not exist there!

5. Jun 14, 2010

### Count Iblis

I think the problem here is to show that you can't have such a function, the best you can do is cut away a branch cut line. Just omitting the point z = 0 is not enough.

I think the proof has to use the integral of the derivative of the function and then show that this integral around the point z = 0 is not zero.

6. Jun 15, 2010

### WannaBe22

Thanks ... I'll try this...