# $e^z-P(z)$ has infinitely many zeros

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 $$1$$.

$$f(z)=e^{\alpha z+\beta}z^m\prod_{k=1}^{n}\left(1-\frac{z}{a_k}\right) e^{\left(\frac{z}{a_k}\right)} \tag{*}$$ (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?

• 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. May 12, 2016 at 21:30
• Just corrected the question. In deed, P cannot be 0. Thank you May 12, 2016 at 21:55

$P\neq0$ 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.

• Good one, thanks! May 12, 2016 at 22:10
• But @zuggg how do you know it's an essential singularity though? May 12, 2016 at 22:23
• @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. May 12, 2016 at 22:33
• @mnmn1993 Because 0 is already an exception. There cannot be another one. Jan 8, 2018 at 14:22
• 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. 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.