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.
