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Let $f$ be an entire function and $\xi=e^{\frac{2\pi i}{n}}$ for some $n\in \mathbb{N}$. Suppose that $f\left(\xi z\right)=f(z)$ for all $z\in \mathbb{C}$. Show that there is a entire function $g$ such that $f(z)=g\left(z^{n}\right)$ for all $z \in \mathbb{C}$.

Remark: I have tried to express $g$ in terms of $f\left(\xi z\right)$ but my attempts have been unsuccessful.

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  • $\begingroup$ For each $z \in \mathbb C - \{0\}$, the set of all $n$ $n$th roots of $z$ varies continuously as a function of $z$. Call this set $A(z)$. Consider $g(z) = \frac{1}{n} \sum_{w \in A(z)} f(w)$. (You'll need to define $g(0)$ as well, and show continuity there, etc.) $\endgroup$ – John Hughes Feb 6 '16 at 22:06
  • $\begingroup$ @JohnHughes Not know if the right is $g(z) = \frac{1}{n} \sum_{w \in A(z)} f(\xi w)$. For other hand, What is the importance of $\xi=e^{\frac{2\pi i}{n}}$? $\endgroup$ – Diego Fonseca Feb 6 '16 at 22:28
  • $\begingroup$ $\xi$ is a primative $n$th root of unity. That is, $\xi^n = 1$ and for $0 < k < n, \xi^k \ne 1$. All of the $n$th roots of unity are $\xi^k, k = 0, ..., n-1$. The expression that John gives for $g$ can be written $$g(z) = \frac1n \sum_{k = 0}^{n-1} f(\xi^k\sqrt[n]z)$$, where $\sqrt[n]z$ maybe taken as the root with argument between $0$ and $2\pi/n$. $\endgroup$ – Paul Sinclair Feb 7 '16 at 0:53
  • $\begingroup$ @PaulSinclair Then I have to show that $g$ is an entire function, so, I think the problem is reduced to show that the function $\sqrt[n]{z}$ with the argument between $0$ and $2\pi / n$ is an entire function. It does not know how to show it. $\endgroup$ – Diego Fonseca Feb 7 '16 at 3:17
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    $\begingroup$ $\sqrt[n]z = \exp(\log(z)/n)$. if $z \ne 0$, you can choose a branch of the logarithm function that has $z$ in its interior. As a composition of holomorphic functions, $\sqrt[n]z$ is holomorphic on a neighborhood of $z$. Use the invariance of $f$ under interior multiplication by $\xi$ to show that it doesn't matter which branch of the logarithm you pick. You always get the same value for $g(z)$. This shows that $g$ is holomorphic everywhere but $0$. Clearly $g(0)$ should be $f(0)$. Continuity at 0 is easy. Play with the Cauchy-Riemann conditions to show it is holomorphich there as well. $\endgroup$ – Paul Sinclair Feb 7 '16 at 20:02
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Using power series is cheating because it makes it too easy. Say $$f(z)=\sum_{k=1}^\infty a_k z^k.$$

Now $f(\xi z)=f(z)$, with uniqueness of the power series coefficients, shows that $$\xi^ka_k=a_k$$for all $k$. If $k$ is not a multiple of $n$ then $\xi^k\ne1$, so $a_k=0$. Hence $$f(z)=\sum_{j=0}^\infty a_{jn}z^{jn}=g(z^n)$$if $$g(z)=\sum_{j=0}^\infty a_{jn}z^j.$$

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  • $\begingroup$ Not really "too easy", entire functions are ones which have power series as stated. That is one of their most fundamental properties. $\endgroup$ – vonbrand Feb 9 '16 at 1:42
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    $\begingroup$ @vonbrand Puts me in the position of denigrating my own solution. A manlier solution would deal with branches of $z^{1/n}$ and analytic continuation and so on. That's better because a solution like that would work for $f\in H(V)$ where $V$ is any open set invariant under multiplication by $\xi$. Power series, or more generally Laurent series, give a solution for a much smaller class of domains. But for the problem as stated what the hell may as well wimp out... $\endgroup$ – David C. Ullrich Feb 9 '16 at 1:48
  • $\begingroup$ @DavidC.Ullrich $g(z)$ is a power serie, What is the reason for which the serie converge?, furthermore, What is the reason for which $g$ is entire? $\endgroup$ – Diego Fonseca Feb 9 '16 at 14:09
  • $\begingroup$ @DiegoFonseca Have you thought about this? It's completely trivial. Since $f$ is entire the power series for $f$ converges in the entire plane, which says $\sum|a_j|r^j<\infty$ for every $r>0$. Restricting to a subsequence, this shows that $\sum_j|a_{jn}|r^{jn}<\infty$ for every $r>0$. Setting $R=r^n$, that says that $\sum_j|a_{jn}|R^j<\infty$ for every $R>0$, which says that the series for $g$ converges in the entire plane. $\endgroup$ – David C. Ullrich Feb 9 '16 at 14:36
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I should look back at the original problem more. Since $f(\xi z) = f(z)$, it holds that $f(\xi^k \sqrt[n]z) = f(\sqrt[n]z)$, and hence the formula I gave in the comment for $g(z)$ reduces to just $$g(z) = f(\sqrt[n]z)$$

You need the invariance condition $f(\xi z) = f(z)$ to show that this definition is independent of which branch of the n-th root you use. As already mentioned, for $z \ne 0$, $g$ is holomorphic at $z$ because $f$ and $\sqrt[n]z$ both are, for some branch of n-th root. At $0$, you will need to pull out the Cauchy-Riemann conditions.

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For any $a\neq 0$, we can define a local branch of $\log$ in a small ball $U$ around $a$ and define $$g(z)=f\left(e^{\frac{\log z}{n}}\right)$$ for $z\in U$. Furthermore, the function $g$ is actually independent of the branch of $\log z$ chosen, since any different branch just multiplies $e^{\frac{\log z}{n}}$ by some power of $\xi$. Thus these local definitions of $g$ actually give a single well-defined function on all of $\mathbb{C}\setminus\{0\}$. Since any of the local definitions are clearly holomorphic, $g$ is holomorphic on all of $\mathbb{C}\setminus\{0\}$. It is also clear that $g(z^n)=f(z)$ for any $z\neq 0$. Finally, the singularity at $0$ is removable, since $g$ is bounded in any ball around $0$ (if $0<|z|<r$, then $|g(z)|\leq\sup \{|f(w)|:|w|\leq \sqrt[n]{r}\}$). Thus $g$ extends to a holomorphic function on all of $\mathbb{C}$, and the identity $g(z^n)=f(z)$ also holds at $z=0$ by continuity.

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