How do we know that $\tan z$ or $\cos z$ don't have singularities off of the real axis?

Question

Since $\sin z$ is bounded, $\tan z = \frac{\sin z}{ \cos z}$ has singularities when $\cos z = 0$. We know $\cos z = 0$ for $z = \frac{\pi}{2} + n\pi$ for $n \in \mathbb{Z}$, but could it not also be zero for values of $z$ with a non-zero imaginary component?

Answer 1

As it turns out, no.

The trick is to expand $\cos z$ into its exponential form and substitute $z = x + iy$:

$\cos z = \frac{e^{iz} + e^{-iz}}{2} = 0 \Rightarrow e^{i(x + iy)} = -e^{-i(x + iy)} \Rightarrow e^{-y}e^{ix} = -e^ye^{-ix}$. For these quantities to be equal, certainly their magnitudes must be equal as well. Thus we have $|e^{-y}e^{ix}| = |-e^ye^{-ix}| \Rightarrow e^{-y} = e^y \Rightarrow \frac{1}{e^y} - e^y = \frac{1 - e^{2y}}{e^y} = 0 \Rightarrow e^{2y} = 1$ (since $e^y \neq 0$). This occurs precisely when $y = 0$. Thus the imaginary part of $z$ must be equal to zero for $\cos z = 0$.

A similar analysis may be carried out for $\sin z$ as well. As noted in the question, this also restricts zeros of other trig functions like $\tan z$ to lie on the real axis.