# No fixed point problem for iterations

Let $f(z)$ be an entire function that is not a polynomial of degree 1 or degree 0 , where $z$ is a complex number.

Let $f(z,1) = f(z)$ and let $f(z,n) = f(f(z,n-1))$.

Let $g(f,1)$ be the amount of distinct complex fixpoints of $f(z,1)$.

More general let $g(f,n)$ be the amount of distinct complex fixpoints of $f(z,n)$.

Let $G(f,n)$ = $g(f,1)$ + $g(f,2)$ $+ ... +$ $g(f,n)$

What is an example $f(z)$ such that $G(f,30)$ = 0 ?

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By "nonlinear" do you mean in the sense of linear maps, or linear polynomials? – Ben Millwood Sep 11 '12 at 22:10
Briefly: a linear map is a function $f$ such that $f(\lambda x + \mu y) = \lambda f(x) + \mu f(y)$, whereas a linear polynomial is just a polynomial of degree 1. I assume you mean the latter (the only linear maps that are also entire functions are multiplication by a constant), but you ought to be clear. – Ben Millwood Sep 11 '12 at 22:11
@ Ben : Thanks you are correct. – mick Sep 12 '12 at 13:44

Suppose $f$ is an entire function and both $f$ and $f(\cdot,2)$ lack fixed points. Consider $$q(z) = \frac{f(z) - f(z,2)}{f(z) - z}$$ Since the denominator is never $0$, this is an entire function. It never takes the values $0$ (because $f(z) - f(z,2) = 0$ would mean $f(z)$ is a fixed point of $f$) or $1$ (because $f(z) - f(z,2) = f(z) - z$ would mean $z$ is a fixed point of $f(\cdot,2)$). By Little Picard it is constant: $q(z) \equiv c$ where $c \notin \{0,1\}$. But then $z = c^{-1} (f(f(z)) + (c-1) f(z))$ implies that $f$ is one-to-one, and the only one-to-one entire functions are polynomials of degree $1$.
Thanks for the reply. I see that $q(z)=c$. Could you explain why f is one-to-one ? – mick Sep 12 '12 at 13:29
If $f(z_1) = f(z_2)$, then $$z_1 = c^{-1} (f(f(z_1)) + (c-1) f(z_1)) = c^{-1} (f(f(z_2)) + (c-1) f(z_2)) = z_2$$ – Robert Israel Sep 12 '12 at 16:06
You have to worry about domains. For example, a branch of $\sqrt{z}$ with branch cut on $[0,\infty)$ maps ${\mathbb C} \backslash [0,\infty)$ into itself and has no fixed or periodic points. – Robert Israel Sep 12 '12 at 22:09