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Using power series, I have to find all holomorphic functions $f$ such that $f(0) = 0$ and $f(f(z)) = z$ near $0$. If I'm not mistaken, $f(0)=0$ restricts the power series to a form $\sum_n a_n z^n$ but now I have no idea how to proceed. If I just plug in, I get $$\sum_n a_n \left( \sum_k a_k z^k \right)^n = z$$ for $z$ near $0$, but what next? Thank you very much for any hints.

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    @Teddy I think that the OP asked for holomorphic functions with the same asymptotic behavior of $z$ near $0$. Because if $f(z)=z$ in a whole open set then $f(z)=z$ on the whole complex plane. Look for example: http://en.wikipedia.org/wiki/Power_series at the paragraph "Analytic functions". Did I get your question correctly @studeth?2012-10-11

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If your assumption is just that $f$ is holomorphic in a neighborhood of $0$, i.e., that it has a convergent power series about $0$ (and that seems to be the only thing you can assume if you have to "use power series"), then there are lots of solutions. In particular, for every convergent power series $g(z) = z + b_1 z + b_2 z^2 + \ldots$, the function $f(z) = g^{-1}(-g(z))$ (i.e., a local conjugate of $z \mapsto -z$ satisfies $f(f(z)) = z$. So you get infinitely many solutions, and I don't know of an easy way to classify all of them via power series coefficients.

Just as an example, it is easy to check that $f(z) = \frac{z}{z-1} = -z - z^2 - z^3 - \ldots$ is a convergent power series in $|z|<1$ with $f(f(z))=z$.

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    Can the downvoter please explain what is wrong with my answer?2012-10-15
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    Dear Lukas, clearly it was me who downvoted your answer to call back your attention on an issue which was not clear enough. I'm glad you did explain your position! So I'm going to remove the downvote! All the best!2012-10-16
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    Alright, thanks Giovanni! I hope the example I added clarified the point I was trying to make.2012-10-16
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    It did so! The point is that we started from two different "holomorphicity" requests on $f(z)$ (namely global and local), and while your example show that the local request allows the form of $f(z)$ to be "not so rigid", I still don't see exactly how the request of global holomorphicity would affect the shape of $f(z)$.2012-10-16
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Important Note This answer is wrong, in specific it is so the argument I use to compare power series as Lukas kindly pointed out. I'm keeping it here only as a reference for the amount of comments made on it.

Answer So we are looking for $f(z)$ holomorphic such that there exists an open set $0\in U \subset \mathbb{C}$ with $f(f(z))=z \: \:\forall \;z\in U$.

Let $f(z) = \sum_{n=0}^\infty a_nz^n$ be the power series of $f(z)$ around zero. From the condition $f(0)=0$ we deduce $a_0=0$.

Then $z=f(f(z))= \sum_{k=1}^\infty a_k \left(\sum_{n=1}^\infty a_n z^n\right)^k = a_1^2z +O(z^2)$. In specific $a_1=\pm 1$ and $a_i=0$ for $i>1$. This implies $f(z)= \pm z \:\: \forall z\in U$.

By the paragraph "Analytic function" of this Wikipedia page. We deduce that $f(z)=\pm z \:\: \forall z\in U$ implies that $f(z)=\pm z \:\: \forall \; z\in \mathbb{C}$, because $f(z)$ and $\pm z$ are holomorphic functions. This implies $f(z)= \pm z$.

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    The OP wants $f(f(z))=z$ exactly, not just $f(f(z))=z+O(z^2)$2012-10-11
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    Ok, evidently I read it wrong! Thank you for pointing it out! I just changed the answer to be relative to the correct question!2012-10-11
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    Why are the only holomorphic functions which are their own inverses $\pm z$? I can't see an obvious proof of this fact. There are certainly other non-holomoprhic functions which are their own inverse, i.e. $\bar z$.2012-10-11
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    @AlexBecker Indeed you are right, I had mixed up together different arguments, now I've put them in order. I hope that now everything is clear and correct.2012-10-11
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    So basically you compare coefficients which leads to $f(z) = \pm z$? That result is somewhat too trivial, I thought the restriction to an open set around $0$ would give more interesting functions. Say I want $f(0) = 0$ and $f(z) = f(2z)$ near $0$ (another exercise of the problem set). By comparison of coefficients, $a_n = 0$ for all $n \in \mathbb{N}$. Maybe I misunderstand the exercise, but these two solutions seem way too trivial to me and the restriction to an open set around $0$ seems negligible.2012-10-11
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    The key is that *an analytic function is globally determined by its local behavior*. So if you make a constraint around zero then you will have it on the whole complex plane! (In this last sentence I'm being a bit heuristical...)2012-10-11
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    How do you get $a_i=0$ for $i>1$?2012-10-11
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    Since $f(z)$ is holomorphic also $f(f(z))$ is, so it coincides with its Taylor series in a neighborhood of $z=0$. The same holds for the holomorphic function $g(z)=z$. In specific the two Taylor series $z +\sum_{n=2}^\infty 0 z^n$ and $\sum_{n=0}^\infty c_n z^n$ (the TS of f(f(z))) are coincident and then equal term by term. The difference between my answer and yours is that I'm classifying $f(z)$ such that it is *globally* holomorphic and not just in a neighborhood of $0$. If you disagree or have further questions please let me know! :)2012-10-11
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    Giovanni, your answer is still not convincing as a power series argument. You compare coefficients of $f(f(z))$ and $z$, but this still does not show how the coefficients of $f(z)$ compare to those of $-z$. And if you just use recursive relations for coefficients, you can't get any further, since there are convergent examples of power series $f(z) = -z + a_2 z^2+ \ldots$ with $a_2 \ne 0$ such that $f(f(z)) = z$. In these cases, the power series for $f(z)$ does not converge in the whole plane, but the one for $f(f(z))$ does.2012-10-15
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    @LukasGeyer, Ok! I'm glad you came back on the question because now I see your point! What I don't see are counterexamples to my statement (which I didn't prove though!). In the sense that the counterexamples you cite refuse to be holomorphic somehow. Or do you have an example of such a function? In any case I'm going to edit my answer according to your observation! :)2012-10-16
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    If you assume that $f$ is holomorphic in the whole plane, then $f(f(z))=z$, $f(0)=0$, only has solutions $f(z)=z$ and $f(z)=-z$. This is easy to see with some complex analysis, but not purely through power series. Here is an argument: If $f$ is a polynomial of degree $d$, then $f(f(z))$ is a polynomial of degree $d^2$. So you get $d=1$ and the conclusion follows easily. If $f$ is transcendental, then $f(f(z))$ is also transcendental (this can be argued using the Casorati-Weierstrass theorem), so it can't be a solution.2012-10-16
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    Ok! So at least my intuition was correct! Thank you very much! You'll find the upvote for the comment into the answer! ;)2012-10-16