Complex Analysis-Analytic Functions

[WannaBe22]

Homework Statement

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: $$f^3 (z) = z^2$$.

Homework Equations

The Attempt at a Solution

Got no clue...

Hope you'll be able to help me


Thanks in advance

 

[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.

 

[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

 

[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!

 

[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.

 

[WannaBe22]

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