If \(p \neq 0\) is a polynomial and \(a \neq 0\) is a complex number, show that \(p(z) - e^{az}\) has infinitely many zeros.
Solution
Consider the meromorphic function \[ F(z)=\frac{e^{az}}{p(z)}. \] At \(\infty\), the function \(F\) has an essential singularity: if it were removable or a pole there, then \(F\) would be rational, which is impossible because \(e^{az}\) is transcendental and \(p\) is a nonzero polynomial.
By the Great Picard theorem, in every punctured neighborhood of \(\infty\) the function \(F\) assumes every complex value, with at most one exception, infinitely often. The value \(0\) is omitted everywhere because \(e^{az}\) never vanishes. Therefore \(0\) is the possible exceptional value, so \(1\) is assumed infinitely often.
At every point where \(F(z)=1\) we have \(p(z)\ne0\) and \[ e^{az}=p(z). \] Hence \(p(z)-e^{az}\) has infinitely many zeros.