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Prove that $\not \exists$ a function $f: \mathbb{C} \to \mathbb{C}$ such that $f$ is continuous on $0< |z| < 1$ and satisfies the property: $(f(z))^n = z \ $ for $n > 1$.

Comment: I am assuming this requires contradiction. But I am failing to find the contradictory point.

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See also: math.stackexchange.com/q/113876 and math.stackexchange.com/q/71775 –  user44795 Oct 15 '12 at 22:56
    
Your question is a bit ambiguous. Is $f$ supposed to be continuous on $\mathbb{C}$, or just on the open unit disc? Not that it really matters - neither exist - but still... –  fgp Oct 16 '12 at 0:06
    
I clarified it now. Thanks! –  user44069 Oct 16 '12 at 0:15
    
@OrestXherija You might want to either post your solution as an answer, or delete the question (which you can do since it hasn't been answer yet, I think) –  fgp Oct 16 '12 at 0:43
    
I did not solve it! I just edited the question so that it is not ambiguous anymore. –  user44069 Oct 16 '12 at 1:22

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