Consider the complex valued function
$$f(z)=\frac{\sin(\frac{\pi z}{2})}{\sin(\pi z)}$$
I am trying to investigate the poles at integers points. That is points where $z=n,n\in \mathbb{Z}$
Case 1: When $n$ is odd,
Numerator is non zero and denominator is zero so I can use the following theorem
If a function $f(z)$ is of the form $p(z)/q(z)$, $p(z)$ and $q(z)$
are analytic at $z_0$, $p(z_0)\neq 0$, then at the points $z=z_0$$f(z)$ has pole of order $m\iff$ $q(z)$ has a zero of order $m$
So I conclude that $f$ has simple pole at $z=n$ where $n$ is an odd integer.
Case 2: When $n$ is even
We have the result,
A function $f(z)$ has a pole of order $m$ at $z=a\iff \lim_{z\to
a}f(z)=\infty$
Let us take $a=2k$, an even integer. Then
$$\lim_{z\to a}f(z)$$ It is in $0/0$ form so we can apply L'Hospital rule. It works in complex analysis as L'Hopital's rule is a local statement.
$$\lim_{z\to a}\frac{\frac{\pi}{2} \cos(\frac{\pi z}{2})}{\pi\cos(\pi z)}$$
We can conclude from the above that $\lim_{z\to a}\neq \infty$.
$\therefore$ $f$ does not have pole for even integers.
If I am correct then what kind of singularity $f$ has for even integers. I guess it is removable.
Best Answer
Away from the zeroes of the denominator, $$f(z)=\frac1{2\cos(\pi z/2)}.$$ The cosine just has simple zeros, so $f$ has simple poles at odd integers. Any other singularities of $f$ must be removable.