I'm kind of new to complex analysis, and I've come across this problem I want to solve:
Problem: Find all singularities and zeroes of the following functions. State type of singularity as well as order of zeroes and poles.
1) $f(z) = \tan^2 (\pi z)$
2) $f(z) = \frac{\sin^2 (z)}{z^2 (z^2 – \pi^2)} $
Attempt:
1) We have $$ f(z) = \tan^2 (\pi z) = \frac{\sinh^2 (\pi z)}{ \cosh^2 (\pi z)} = \frac{ \frac{1}{4} (e^{\pi z} – e^{- \pi z})^2 }{\frac{1}{4} (e^{\pi z} + e^{- \pi z})^2 } $$ We have a singularity when the denominator becomes zero, i.e. when $e^{\pi z} = – e^{- \pi z}$. Now, since $(-1) = e^{i (\pi + 2 \pi k)}$, with $k$ some integer, the previous equation is equivalent to $$ e^{\pi z} = e^{i(\pi + 2\pi k)} e^{-\pi z} $$ If we equate the arguments of the exponentials, this becomes $$ \pi z = i (\pi + 2 \pi k) – \pi z$$ or $$ z = i/ 2 + ik. $$ However, I'm not sure what type of singularity this is.
We learned about:
1) Isolated: removable, poles, essential singularities
2) Non-isolated: branch cuts.
For the second example, 2), I would say that there are only two obvious singularities, i.e. when $z = 0$ and $z^2 = \pi^2$.
I would appreciate it if you could point out any mistakes I made, and/or provide other answers.
Best Answer
For 1), $\tan \pi z = \dfrac{\sin \pi z}{\cos \pi z}$ (not $\sinh$ and $\cosh$). The singularites are given by solutions to $\cos \ pi z = 0$ which turn out to be $z = \pm \frac12 + 2k$ for $k \in \mathbb{Z}$ (or $z = \frac12 + k$ if you prefer).
You can check that all zeros of $\cos z$ are simple (if $f(z) = \cos \pi z$, then $f'(z) = -\pi\sin \pi z$, so $f'(z) \neq 0$ whenever $f(z) = 0$). Hence all zeros of $\cos^2 \pi z$ are double, and again the numerator $\sin^2 \pi z$ doesn't vanish where $\cos^2 \pi z = 0$, so there can be no cancellation. Hence, all the singularities of $\tan^2 \pi z$ are double poles.
For 2) the singularities look good: $z = 0$, $z = \pm \pi$. Let $g(z) = z^2(z^2-\pi^2) = z^2(z-\pi)(z+\pi)$. Hence $g$ has a double zero at $z=0$ and simple zeros at $z = \pm \pi$.
Furthermore, $\sin z$ is zero for $z=0$, and $z=\pm \pi$. Hence $\sin^2 z$ has (at least) double zeros there, and consequently the zeros in the denominator can be cancelled against the zeros in the numerator. I.e. all the singularities are removable. (In fact, your function even has simple zeros at $z=\pm \pi$.)