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Let $f$ be an entire function without zeroes. I have to show that there exists an entire function $g$ such that $f = g^n$ for some $n \in \mathbb{N}$. I tried proving this by constructing explicitly such a $g$:

If $f = g^n$, we have $f g^{-n} \equiv 1 =: h(z)$. For this to hold, we must have $h'(z) \equiv 0$. We compute $h'(z) = f' g^{-n} + f \cdot (-n) g^{-n-1} g' = f' g^{-n} - n f g^{-n-1} g' = 0$ which is equivalent to $\frac{g'}{g} = \frac{f'}{n f}$. Now I don't really know how to solve this differential equation for $g$, but I think $g(z) := f(z/n)$ should be a solution to it, as $g'(z) = f'(z/n)/n$. But is this really a solution? Because on the left hand side, the argument is $z$ whereas on the right hand side it is $z/n$. This doesn't seem correct to me.

In case my solution is incorrect, how would I actually solve this differential equation? Or is there an even easier approach to prove this claim?

Thanks in advance for any help.

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    I am sorry, what I is actually wanted is what Berci wrote. I don't know why I wrote it the other way, I actually understood the problem correctly.2012-11-08

2 Answers 2

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Hint: Look for a branch of $\log f$, that is a function $h$ so that $\exp(h(z))=f(z)$. Derive a differential equation for $h$.

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$f(z) \neq 0$, for all $z \in \mathbb{C}$,

exists $h$ such that $f = \exp(h)$,

then $g = \exp(\frac{h}{n})$