It is easy to construct surjective locally univalent holomorphic functions $f: {\mathbb D}\to {\mathbb C}$, where ${\mathbb D}$ is the open unit disk.

I am pretty sure that the answer to the following is positive and well-known (to experts) and is somewhere in the literature:

Question. Are there (non-affine) surjective entire functions $f: {\mathbb C}\to {\mathbb C}$ without critical points?

Such functions (up to an additive constant) would have the form $f(z)=\int_0^z e^{h(w)} dw$, where $h(w)$ is another nonconstant entire function. I do not see, however, how to verify surjectivity of such $f$. Of course, it can miss at most one point in ${\mathbb C}$ and, hence, it would be astounding if indeed this was the case for arbitrary $h$. However, I do not see how to rule this out. Applying Cauchy's argument principle in this setting looks extremely messy.


2 Answers 2


Picards theorem will do the trick. Take $$h(z)=z\cdot e^{\int_0^zf(t)dt}$$ where $f(z)=\frac{e^z-1}{z}.$ Then, clearly $$h'(z)=e^{\int_0^zf(t)dt}(1+zf(z))=e^{\int_0^zf(t)dt+z}\ne 0$$ and $h$ is locally univalent. If $h(z)$ is not surjective, then by Picard it omits just one complex value $A$ and we can write $h(z)=e^{g(z)}+A.$ But then equality $$e^{g(z)}+A=z\cdot e^{\int_0^zf(t)dt}$$ is impossible since the LHS takes value $0$ infinitely often while the RHS does not.

In fact, the same proof works for any function of the form $h(z)=g(z)\cdot e^{\int_0^zf(t)dt}$ with $g'(z)+g(z)f(z)\ne 0$ and $g(z)$ is a polynomial.

  • $\begingroup$ Instead of writing $h(z) - A = e^{g(z)}$, you can conclude that on $\mathbb{C} - \{0\}$, $h$ misses both $0$ and $A$ (as $z = 0$ is the only value for which $f(z) = 0$), and use big Picard theorem. $\endgroup$
    – xyzzyz
    Nov 7, 2013 at 23:56

The simplest surjective entire function without critical points is the "probability integral", $\int_0^z e^{-\zeta^2}d\zeta$.

Proof. Suppose it omits $a$. As it is of order 2, it must be of the form $a+e^P$, where $P$ is a polynomial of degree $2$. But this function has critical points:-)

The argument extends to $\int e^Q$ for any polynomial of degree at least $2$.

  • $\begingroup$ But how do you see that this function is surjective? $\endgroup$ Nov 6, 2013 at 21:30
  • $\begingroup$ I see, very nice! $\endgroup$ Nov 6, 2013 at 21:38
  • 2
    $\begingroup$ Here is a sneaky way to answer @Moishe's question on the surjectivity of the entire function $F(z)=\int_0^z e^{-\zeta^2}d\zeta$. Since $F(0)=0$ it suffices to prove that every $0\neq a\in \mathbb C$ is a value of $F$. But if $a$ were omitted, then $-a$ also would be omitted since $F$ is odd. But then $F$ would be a constant by Picard's "little" theorem and this absurdity shows that $F$ actually is surjective. $\endgroup$ Jan 17, 2022 at 19:15

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