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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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    @ Ben : Thanks you are correct.2012-09-12

1 Answers 1

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There are none.

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 $1$.

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    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.2012-09-12