# Entire functions $f$ such that $f(f(z))=f(z)$

Find all entire functions such that $$\forall z\in \mathbb{C}$$ : $$f(f(z))=f(z)$$

I am given the following answer but I do not fully understand it, it goes:

Let $$I$$ be the image of $$f$$

$$\forall a\in I \exists z_{a}\in \mathbb{C}: f(z_{a})=a$$

$$f(a)=f\bigl(f(z_{a})\bigr)=f(z_{a})=a$$ therefore $$\forall a\in I f(a)=a$$

If $$f$$ is constant so we are done, let assume it is not.

The image of a non constant entire function is dense in $$\mathbb{C}$$ so $$I$$ is dense in $$\mathbb{C}$$, so every point in $$\mathbb{C}$$ is an accumulation point so $$I$$ is a set with accumulation points and therefore $$\forall z\in \mathbb{C} ; f(z)=z$$

Why can we conclude that $$f(z)=z$$

You proved that the equality $f(a)=a$ is true for each $a\in I$ and that $I$ is dense. So, if $z\in\mathbb C$, you take a sequence $(z_n)_{n\in\mathbb N}$ such that $(\forall n\in\mathbb{N}):z_n\in I$ and that $\lim_{n\in\mathbb N}z_n=z$, and then\begin{align*}f(z)&=f\left(\lim_{n\in\mathbb N}z_n\right)\\&=\lim_{n\in\mathbb N}f(z_n)\\&=\lim_{n\in\mathbb N}z_n\\&=z.\end{align*}
• Maybe you should specify that $\{z_n\}_n\subset I$; btw nice answer, +1 – Joe Jun 27 '17 at 21:06
If two entire functions agree on a set with an accumulation point, then they agree on all of $\mathbb C$. Since $I$ has been shown to have accumulation points; $f(z) = z$ on all of $I$; and $f(z)$ and $z$ are both entire functions, we must have $f(z) = z$ for all $z \in \mathbb C$.