Analysis Problem of the Day 59

Today’s problem appeared as Problem 12 on the UCLA Fall 2022 Analysis Qual:

Problem 12. Let $f$ be an entire function not of the form $f(z)=z+a$ for some $a \in \mathbb{C}.$

a) Show that $f \circ f$ has a fixed point.

b) Show that $f$ need not have a fixed point.

Solution: a) We consider the function $g(z) = \frac{f(f(z))-z}{f(z)-z}.$ Note that $g$ is well-defined if $f$ does not have a fixed point, and otherwise $f \circ f$ has a fixed point. Also notice that if $g$ attains the values 0 or 1, then $f \circ f$ has a fixed point, since then $f(f(z))=z$ or $f(f(z))=f(z),$ i.e. $f$ has a fixed point. By Little Picard, it follows that since $g$ is entire that $g$ must be a constant that is not 0 or 1, i.e. $f(f(z))-z = c(f(z)-z).$ Taking derivatives on both sides yields $f'(f(z))f'(z)=cf'(z) - c+1,$ so $f'$ never vanishes since $c \not =1$ and $f'(z) \not = \frac{c-1}{c}$ since the left hand side never vanishes. Thus, $f'$ is an entire function omitting two distinct values and therefore constant, i.e. $f(z) = az+b.$ But clearly, if $f$ is linear, then $f \circ f$ has a fixed point iff $a \not = 1.$ Thus, if $f(z) \not = z+b,$ then $f \circ f$ has a fixed point.

b) Clearly, $f(z)-z$ is never zero when $f(z)-z=e^z,$ i.e. $f(z)=z+e^z$ is entire and does not have any fixed points.

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