$f(f(z))=f(z)$ for all $z\in \mathbb{C}$ implies that $f(z)=z$ for all $z\in \mathbb{C}$

Asked Aug 13 '20 at 18:25

Active Aug 13 '20 at 18:42

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As in the title, I would like to show that if $f$ is an entire, non-constant function, then $f(f(z))=f(z)$ for all $z\in \mathbb{C}$ implies that $f(z)=z$.

I tried to think about what kind of singularity $f(z)$ has at $\infty$, since it can not be removable and if it is a pole then $f$ is a polynomial and we would be done very quickly. However, if it is an essential singularity I don't really know what to do there.

I also tried to use the chain rule on the initial equation $f(f(z))=f(z)$ and got that for all $z\in \mathbb{C}$, we either have $f'(z)=0$ or $f'(f(z))=1$. If you could show that we always have the second condition and also that $f$ is onto, then we would also be done.

Would appreciate any help.

asked Aug 13 '20 at 18:25

MSA2016

1

Since $f$ is entire, then $f'(w)=1$ for all $w$. hence $f(z)=z$ (or $f(z)=c$). – Chrystomath Aug 13 '20 at 18:29
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$f'(z)=0$ is an isolated set unless $f$ constant so $f'(w)=1$ ae and by density everywhere – Conrad Aug 13 '20 at 18:30

$f(f(z))=f(z)$ for all $z\in \mathbb{C}$ implies that $f(z)=z$ for all $z\in \mathbb{C}$

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