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:

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

Rearranging $(*)$, we get:

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

(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 '16 at 21:30
  • 1
    $\begingroup$ Just corrected the question. In deed, P cannot be 0. Thank you $\endgroup$ – rmdmc89 May 12 '16 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.

  • $\begingroup$ Good one, thanks! $\endgroup$ – rmdmc89 May 12 '16 at 22:10
  • $\begingroup$ But @zuggg how do you know it's an essential singularity though? $\endgroup$ – rmdmc89 May 12 '16 at 22:23
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    $\begingroup$ @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 '16 at 22:33
  • $\begingroup$ How do you know $1$ is not the exception? $\endgroup$ – mnmn1993 Jan 2 '18 at 11:27
  • $\begingroup$ @mnmn1993 Because 0 is already an exception. There cannot be another one. $\endgroup$ – zuggg Jan 8 '18 at 14:22

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