Analysis Problem of the Day 10

Today’s problem is Problem 6 on UCSD’s Fall 2022 Complex Analysis Qual:

Problem 6: If $p$ is a nonzero polynomial and $a$ is a nonzero complex number, show that $p(z)-e^{az}$ has infinitely many zeros.

Solution: Notice that $p(z)-e^{az}$ has a zero if and only if $p(z)e^{-az}=1.$ Since $p(z)e^{-az}$ is not a polynomial, it has an essential singularity at $\infty,$ so by the Great Picard theorem, it must attain every complex value (with the exception of at most one) in a neighborhood of infinity an infinite number of times. However, it is clear that $p(z)e^{-az}$ has finitely many zeros, so therefore $p(z)e^{-az}=1$ must have infinitely many solutions, i.e. $p(z)-e^{az}$ must have infinitely many zeros.

Remark: This approach shows that any function of the form $p(z)-e^{g(z)}$ where $p$ is a polynomial and $g$ is a nonconstant holomorphic function will have infinitely many zeros.

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