A composite square root of a function $g$ is a function $f$ such that $f(f(z)) = g(z)$. Not surprisingly, for arbitrary $g$ a function like this is hard to find. Specifically I am looking at functions $g$ such that $g$ has finite nonzero order, so that

$$0 < \limsup_{r\to\infty} \dfrac{\log\log M(r)}{\log r} = \rho < \infty$$

where $M(r)$ is the max of $g(z)$ on the disk of radius $r$.

Now naturally $\cos(\cos(z))$ has a square root function $\cos$, or $e^{e^z}$ has one, but these functions are of infinite order. Similarly with polynomials, any polynomial $p(p(z))$ has a square root function, but these are of zero order. The special case of functions of finite order, is that a possible square root function is of zero order. Which is an argument simple enough (I'll provide it if asked).

The evidence I am seeing piling is that there exists no entire composite square roots of functions $g$ of finite order. Now if $I$ is the immediate basin about a geometrically attracting fixed point of $g$ (and $I$ is simply connected) then there is an $f$; but $f$ only sends $I \to I$ and is not entire. I'm wondering if this is always the case; that a composite square root cannot be entire.

Taking $g(z)=e^z$ for example, if there were such an $f$, then $f(f(z))$ can't equal zero, so that $f(z)$ misses $f^{-1}(0)$ but then $f(f(z))$ must miss $0$ and $f^{-1}(0)$ and must therefore be constant by Picard's theorem. If $f^{-1}(0)$ is empty than $f(f(z)) = e^z$ misses $0$ and $f(0)$. This argument generalizes for all $e^{P(z)}$ where $P$ is a polynomial.

Therefore the only possible cases where there is an $f$ for arbitrary $g$ are when $g$ is surjective on $\mathbb{C}$. This doesn't work for $\cos$ though, because $\cos$ has all its zeroes in $I$, the immediate basin about $z_0 \in \mathbb{R}^+\,\,\, z_0 \approx 0.739085$. It can be shown that if $f$ were entire all it's zeroes are described by the sequence $\{f(\pi/2 + n\pi)\}_{n=-\infty}^\infty$, but $f$ is necessarily periodic (again an argument I will supply if asked) so that $f$ would only have two zeroes $\pi/2$ and $3\pi/2$ and thus by Hadamard's factorization theorem $f(z) = C(z-f(\pi/2))(z-f(3\pi/2))$ which is obviously wrong. Thus no entire square root of $\cos$. The same argument applies for $\sin$ just as well but is a bit more finicky.

As much as I look, the only possible candidates are when we have an entire function $f$ of zero order and we compose it with itself $f(f(z)) = g$ and $g$ is of finite nonzero order. I've yet to come across an example of such a function though. Can anyone even name such a function?

So all in all my question is rather simple. For an entire function $g$ of finite nonzero order, can there exist an entire function $f$ such that $f(f(z)) = g(z)$?

  • $\begingroup$ @ChristianRemling Assume $f^{-1}(0)$ is empty, then $f$ never attains $0$ and $f(f(z)) = e^z$ never attains $0$ and $f(0)$; again falling victim to the same argument. I was being quick and didn't want to go into too much detail--I'll fix it as it may seem sloppy. And for the $\limsup$ I'll fix that now, that is what I meant. $\endgroup$
    – user78249
    Nov 3, 2016 at 0:20

1 Answer 1


Such functions were constructed by I. N. Baker:

The Iteration of Entire Transcendental Functions and the Solution of the Functional Equation f{f(z)} =F(z). Math. Ann. 129 (1955), 174-180.


It is Theorem 2 (iii) of this paper.

  • 1
    $\begingroup$ Aww this is beautiful! Exactly what I was looking for. It's rather surprising to be honest, I was almost convinced order zero functions can't self compose to make a finite nonzero order function. $\endgroup$
    – user78249
    Nov 3, 2016 at 19:37
  • $\begingroup$ @james nion: your intuition was wrong: the growth scale of entire functions is "continuous" and has no gaps. A function like $e^x-1$ has a square root on the positive ray, and this "square root" can be approximated by some entire function whose max modulus is on the positive ray. This is the natural idea of baker's construction. $\endgroup$ Nov 3, 2016 at 22:09

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