Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
@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: en.wikipedia.org/wiki/Power_series at the paragraph "Analytic functions". Did I get your question correctly @studeth? –  Giovanni De Gaetano Oct 11 '12 at 9:55
add comment

2 Answers

up vote 4 down vote accepted

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 $f(f(z))=z$.

share|improve this answer
Can the downvoter please explain what is wrong with my answer? –  Lukas Geyer Oct 15 '12 at 21:33
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! –  Giovanni De Gaetano Oct 16 '12 at 14:51
Alright, thanks Giovanni! I hope the example I added clarified the point I was trying to make. –  Lukas Geyer Oct 16 '12 at 14:55
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 $f(z)$. –  Giovanni De Gaetano Oct 16 '12 at 14:57
add comment

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 $f(z)= \pm z$.

share|improve this answer
The OP wants $f(f(z))=z$ exactly, not just $f(f(z))=z+O(z^2)$ –  Ewan Delanoy Oct 11 '12 at 9:59
Ok, evidently I read it wrong! Thank you for pointing it out! I just changed the answer to be relative to the correct question! –  Giovanni De Gaetano Oct 11 '12 at 10:11
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. $\bar z$. –  Alex Becker Oct 11 '12 at 10:14
@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. –  Giovanni De Gaetano Oct 11 '12 at 10:28
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. –  studeth Oct 11 '12 at 10:59
show 7 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.