How many zeroes does this transcendental equation have?

Question

The equation is

$$ (2i z - 1)^2 - e^{i a z } = 0 . $$

Here $a$ is a positive real number. Is there any algorithm for this kind of equation?

Answer 1

The equation $$ (2iz-1)^2-e^{iaz}=0 $$ has infinitely many roots.

In general, for any $\lambda \ne 0$ and any polynomial $p(z)\not\equiv 0$ the equation $e^{\lambda z}-p(z)=0$ has infinitely many roots.
Proof. $$ e^{\lambda z}-p(z)=0 \iff e^{\lambda z}=p(z) \iff p(z)e^{-\lambda z}=1. $$ Now $p(z)e^{-\lambda z}$ has finitely many zeros since $e^{-\lambda z}\ne 0 $ for all $z$ and $p(z)$ has finitely many zeros. In other words, the value $0$ is the single exception in Great Picard's Theorem. Therefore $p(z)e^{-\lambda z}$ attains the value $1$ infinitely often. This implies that $p(z)e^{-\lambda z}=1$ has infinitely many roots and $e^{\lambda z}-p(z)=0$ has infinitely many roots.

Great Picard's Theorem: 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.