Complex Analysis: Zeros of an analytic function

Question

What approach should I take to solve the attached problem. I was looking along the lines of 'Great Picard Theorem', which states that 'If an analytic function f has an essential singularity at a point w, then on any punctured neighborhood of w, f(z) takes on all possible complex values, with at most a single exception, infinitely often.' Any help is much appreciated.

Answer 1

This can be done without Picard's theorem.

Hint for the first part: assume that $g$ has finitely many roots, and use Hadamard's factorization theorem to reach a contradiction.

For the second part, let $w\in\mathbb C$, and set $q(z)=p(z)+w$. Then $q$ is a polynomial, so $e^{\lambda z}-q(z)$ has infinitely many roots, therefore $e^{\lambda z}-p(z)=w$ for infinitely many $z$.

Answer 2

Proof using Picard's theorem:

First, we note that the state problem is equivalent to the same problem with $\lambda$ fixed at $1$, since by substituting $w = \lambda z$ we have for some re-scaled polynomial $T(w)$ that $$ e^{\lambda z} -P(z) = e^{w} - T(w) $$ (This observation is not essential to the proof but it simplifies the statements.)

Let the roots (with multiple roots listed as multiple entries in this sequence) of the $n$-th degree polynomial $P(z)$ be $$ (\zeta_1, \zeta_2, \ldots, \zeta_m, q_{1},q_{2},\ldots, q_{n-m}) $$ where $0\leq m\leq n$, all the $\zeta_i$ are the zero roots, and all the $q_i \neq 0$. Thus $$ P(z) = z^m Q(z) = z^m\prod_{i=1}^{n-m}(z-q_i) $$ Let $$ k(z) = z^m g\left( \frac1z \right) = z^m e^{1/z} - z^m \left( z^{-m} \prod_{i=1}^{n-m}\frac1{(z-q_i)} \right) = z^m e^{1/z} - \prod_{i=1}^{n-m}\frac1{(z-q_i)} $$ Let $r = \min\{ |q_i|\}$. Since no $q_i$ is zero, $r > 0$. In the open neighborhood $|z|<r$, $k(z)$ has no singularities arising from $\prod_{i=1}^{n-m}\frac1{(z-q_i)}$.

So in this neighborhood, $k(z)$ only has the singularity at $z=0$, which is an essential singularity. The conditions for Picard's theorem are satisfied, so there is an infinite sequence of non-zero roots $\alpha_j$ satisfying $$ k(\alpha_j) = \alpha_j^m e^{1/\alpha_j} - \prod_{i=1}^{n-m}\frac1{(\alpha_j-q_i)}=0 $$ Divide each such equation (for each $j$) by the non-zero quantity $\alpha_j^m$: $$ e^{1/\alpha_j} - \alpha_j^{-m}\prod_{i=1}^{n-m}\frac1{(\alpha_j-q_i)} = 0 $$ Now let $r_j = 1/\alpha_j$. For each $r_j$ we have $$ e^{r_j} - r_j^m\prod_{i=1}^{n-m}(r_j-q_i) = 0 \\ e^{r_j} - r_j^m\,Q(r_j) = 0 \\ e^{r_j} - P(r_j) = 0 $$ thus demonstrating an infinite sequence of roots of $ e^z - P(z) = 0 $.