# Given entire function has infinitely many zeros

Prove that for any $\lambda \neq 0$ and any polynomial $p(z) \not\equiv 0,$ the function $g(z)=e^{\lambda z}-p(z)$ has infinitely many zeros.

My approach: Suppose to the contrary that $g$ has finitely many zeros, at $a_j$'s (say) $$g(z)=\prod_{j=1}^{n} (z-a_j).$$ By Hadamard's factorization $$g(z)=e^{h(z)} \cdot \prod_{j=1}^{n} (z-a_j),$$ for some polynomial $h(z).$ I don't know how to get a contradiction. Any help is much appreicated.

• Seems like great Picard theorem will be useful. – user99914 Sep 1 '16 at 4:05
• @user1952009 thank you. I'm sorry but I'm having a hard time comprehending the asymptotic argument. – user358174 Sep 1 '16 at 4:42
• $g(z) = e^{h(z)} q(z)$ means that its maximum modulus is of the order of $e^{c r^k} r^n$. Note also that as $Re(\lambda z) \to +\infty: \ \log g(z) \to \lambda z$, and as $Re(\lambda z) \to -\infty$ : $\log g(z) \to (n-1)\log z$. it suggests that $\int_{|z| = r} \frac{g'(z)}{g(z)} dz$ grows linearly when $r \to \infty$. – reuns Sep 1 '16 at 4:54

The same approach as in Showing that $e^z=z$ has infinitely many solutions can be used:

$h(z) = p(z) e^{-\lambda z}$ has an essential singularity at $z = \infty$, and has only finitely many zeros. Using Great Picard's Theorem it follows that $h$ takes any non-zero value infinitely often. In particular, $h(z) = 1$ has infinitely many solutions.

This is the desired conclusion because $h(z) = 1 \Longleftrightarrow e^{\lambda z} = p(z)$.

• @ Martin R thank you. This helps. But I was looking for something that avoids Picard theorems. – user358174 Sep 1 '16 at 11:19
• @ManMath: I have provided an alternative solution. Please let me know if you need more information. Otherwise you might consider to accept an answer by clicking on the check mark! – Martin R Sep 16 '16 at 18:35

An alternative proof: If the claim is false then $$\tag{*} e^{\lambda z} - p(z) = q(z) e^{h(z)}$$ with non-zero polynomials $p, q$, and an entire function $h$. To simplify the notation let us assume that $\lambda = 1$, this is no loss of generality. Then $$e^{h(z)} = \frac{e^z - p(z)}{q(z)}$$ and for $|z| = r$ sufficiently large, some $m \in \Bbb N$, and positive real constants $C_1, \ldots, C_4$ $$e^{\operatorname{Re}h(z)} = \lvert e^{h(z)} \rvert \le C_1 (\lvert e^z \rvert + C_2 r^m) \le C_1 ( e^r + C_2 r^m) \le C_3 e^r$$ and therefore $$\operatorname{Re }h(z) \le C_4 + r$$ for $|z| = r > R_0$. So $$A(r) :=\max_{|z|=r}\operatorname{Re}h(z) \le C_4 + r$$ Now use

to estimate the Taylor coefficients of $h$ in terms of $A(r)$, and conclude that $h$ a polynomial of degree at most one: $h(z) = az + b$.

Finally, substitute $h$ back into equation $(*)$ and compare the growth of the different terms.