If $P\in\mathbb{C}[z]$ is a non-zero polynomial, prove $f(z):=e^{z}-P(z)$ has infinitely many zeros.

I've made some progress so far, but I still have a step missing. Here is where I'm at:

Suppose $f$ has finitely many zeros, namely $z=a_1,\,...\,,a_n \neq 0$ and possibly $z=0$. Since $e^z$ and $P(z)$ are entire functions with orders of growth $1$ and $0$ respectively, then $f$ has order of growth $1$.

From Hadamard's factorization:

$$f(z)=e^{\alpha z+\beta}z^m\prod_{k=1}^{n}\left(1-\frac{z}{a_k}\right) e^{\left(\frac{z}{a_k}\right)} \tag{$*$}$$ (where $\alpha$, $\beta$ are complex constants and $m:=$ multiplicity of $z=0$).

Rearranging $(*)$, we get:

$$f(z)=e^{\gamma z+\beta}Q(z)\tag{$**$}$$

(where $\gamma = \alpha +\sum_{k=1}^{n}\frac{1}{a_k}$ is a constant and $Q(z)=z^m\prod_{k=1}^{n}\left(1-\frac{z}{a_k}\right)$ is a polynomial of degree $m+n$)

Now, for $d:=\text{deg}(P)$, notice $f^{(d+1)}(z)=e^{z}$. Taking the $(d+1)$-th derivative in $(**)$, we get $e^{z}=e^{\gamma z+\beta}R(z)$, where $R$ is a polynomial of degree $m+n$. If $m+n\geq 1$, $R$ has at least one root, while $e^{z}$ has none (absurd), so $m+n=0\Rightarrow m=n=0\Rightarrow f$ has no zeros.

Conclusion: either $f$ has infinitely many zeros or none at all.

Now how can I prove it has at least $1$ zero?

  • 5
    $\begingroup$ Umm... You've noticed yourself that $e^z$ has no zeros. So, if $P=0$, then $f(z)=e^z$ has no zeros, and the statement in bold is false. $\endgroup$
    – Dan Shved
    May 12, 2016 at 21:30
  • 1
    $\begingroup$ Just corrected the question. In deed, P cannot be 0. Thank you $\endgroup$
    – rmdmc89
    May 12, 2016 at 21:55

2 Answers 2


$P\neq0$ only has finitely many roots, so the function

$$ g:z\mapsto\frac{\exp(1/z)}{P(1/z)} $$

is well defined on a neighbourhood of $0$ (without $0$), has an essential singularity at $0$, and does not vanish. Since it is an analytic function, Great Picard's theorem holds, and $g$ takes all complex values infinitely often except at most one. In particular, $1$ is reached infinitely many times.

  • $\begingroup$ Good one, thanks! $\endgroup$
    – rmdmc89
    May 12, 2016 at 22:10
  • $\begingroup$ But @zuggg how do you know it's an essential singularity though? $\endgroup$
    – rmdmc89
    May 12, 2016 at 22:23
  • 1
    $\begingroup$ @rentatodias $x\mapsto x^n \exp(1/x)/P(1/x)$ is never bounded on any real neighbourhood of $0$ and for any $n\in\mathbb N$, and as such $0$ is neither a removable singularity nor a pole. $\endgroup$
    – zuggg
    May 12, 2016 at 22:33
  • 1
    $\begingroup$ @mnmn1993 Because 0 is already an exception. There cannot be another one. $\endgroup$
    – zuggg
    Jan 8, 2018 at 14:22
  • 1
    $\begingroup$ This, is a nice proof, but I feel we can do it without Great Picard's theorem as well. I will post it as an answer. $\endgroup$ May 20, 2020 at 13:34

A solution based on the Hadamard factorisation theorem: Assume that $P$ is a polynomial, not identically zero, and that $e^z - P(z)$ has only finitely many zeros. Then, as you already figured out, $$ \tag 1 e^z - P(z) = Q(z) e^{\gamma z} $$ with some polynomial $Q$ and a constant $\gamma$. Taking the $(d+1)$-th derivative of that identity (where $d$ is the degree of $P$) we get $$ \tag 2 e^z = R(z) e^{\gamma z} $$ where $R$ has the same degree as $Q$. It follows that $R$ is constant, and that $\gamma = 1$. Then $Q$ is constant as well, say $Q(z) = q$, and $(1)$ becomes $$ (1-q) e^z = P(z) \, . $$ Note that $q \ne 1$ since $P$ is not identically zero. But then $$ e^z = \frac{P(z)}{1-q} $$ is a contradiction: The exponential function is not a polynomial.

This shows that $e^z - P(z)$ must have infinitely many zeros.


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