# Use Picard's Theorem to prove infinite zeros for $\exp(z)+Q(z)$

Suppose that $Q(z)$ is a nonconstant polynomial. Then show that the function $$f(z)=\exp(z)+Q(z)$$ has infinitely zeros.

My idea is to show that $\infty$ is an essential singularity thus by Picard's theorem $f(z)$ assumes every complex number infinitely times except on possible value. I was stuck with the possibility that $0$ may be the exception, if $Q(z)=z$,thus we use the periodic $2\pi i$ and Little Picard's theorem to get the result. But for general polynomial I can't find the periodic.

Suppose that there are only a finite number of zeros. Since $f(z)$ is an entire function of genus 1, the Hadamard factorization theorem allows us to write $$f(z)=e^z+Q(z)=e^{\alpha z+\beta} P(z) \tag{1}$$ where $P$ is a polynomial.
Let $n=\deg Q+1$, we have $$Q^{(n)}(z)=\frac{d^n}{dx^n} \left[e^{\alpha z+\beta} P(z)-e^z \right]$$ which simplifies to (using the product rule) $$e^z = \sum_{k=0}^n \binom{n}{k}\alpha^k P^{(n-k)}(z) e^{\alpha z+\beta}$$ or $$e^{z-\alpha z-\beta}=\sum_{k=0}^n \binom{n}{k}\alpha^k P^{(n-k)}(z)$$ Since the LHS has no zeros, we must have $\deg P=0$ (otherwise the polynomial in the RHS will have a root). We moreover find that the LHS is constant which means $$e^{z}=C e^{\alpha z+\beta} .$$ Plugging this into $(1)$ we get $$Q(z)=(P(z)-C)e^{\alpha z+\beta}$$ but a function of the form $A e^{\alpha z+\beta}$ (don't forget that $P$ is constant) is a polynomial iff it's a constant. This is a contradiction.
• Is $1+\frac{Q'}{Q}$ the logarithmic derivative of $e^z+Q$? Commented Jan 9, 2014 at 15:46
• Sorry I'm not quite familiar with genus and rank,order stuff,but at the beginning how can you assert that $f$ is of genus 1?I recall that finite genus means that finite rank and Weierstass Factorization gives $g$ is polynomial,but how do you know for sure in this case that $g$ is a polynomial? Commented Jan 10, 2014 at 5:27
• @DanielS. Firstly, you determine the order of the function, which is the smallest number $\lambda$ such that $\max_{|z|=r} |f(z)| \leq e^{r^{\lambda+\epsilon}}$ for any $\epsilon>0$ for sufficiently large $r$. In our example, $\lambda=1$. The next step is Hadamard's inequality, which says that the genus $h$ satisfies $h \leq \lambda \leq h+1$, thus $h \leq 1$, and the genus as an integer can be either $0$ or $1$. Is this explanation sufficient, or do you want me to expand further? Commented Jan 10, 2014 at 8:57