Finding the zeros of a complex function on a disc

Question

I have encountered the following problem:

Find the number of zeros of $f(x)$ on the disk $|z|$ < $1/2$ where $f(x)$ = $z^2$+$cosh(iz)$

How would one compute the solution?

Answer 1

For $z \in \mathbb{C}$ with $|z| < \frac12$, write $z$ as $x + iy$ where $x,y \in (-\frac12,\frac12)$. We have

$$\begin{align} & \cosh(iz) = \cosh(-y+ix) = \cosh(y)\cos(x) - \sinh(y)\sin(x) i\\ \implies & \Re \cosh(iz) = \cosh(y)\cos(x) \ge \cos\left(\frac12\right)\\ \implies & \Re( z^2 + \cosh(iz) ) \ge \cos\left(\frac12\right) - \frac14 \approx 0.6275825618903728 > 0 \end{align} $$ As a result $z^2 + \cosh(iz) \ne 0$ for $|z| < \frac12$.