If $f$ is an entire function which of the following is/are correct?

Question

Consider the function $$f(z)=\frac {\sin (\frac {\pi z} {2})} {\sin (\pi z)}.$$ Then $f$ has poles at

$1.$ $\ $ all integers.

$2.$ $\ $ all even integers.

$3.$ $\ $ all odd integers.

$4.$ $\ $ all integers of the form $4k+1$, $\ k \in \Bbb Z$.

What I have thought is as follows $:$

If $z_0$ is a pole of $f$ then it is not a zero of $\sin \left (\frac {\pi z} {2} \right )$. Because if $z_0$ was a zero of $\sin \left (\frac {\pi z} {2} \right )$ then it can't be a zero of $\cos \left (\frac {\pi z} {2} \right )$ and hence $z_0$ becomes a removable singularity of $f$ which contradicts the fact that $z_0$ is a pole of $f$. Hence $z_0$ is indeed not a zero of $\sin \left (\frac {\pi z} {2} \right )$. Hence $$\lim_{z \rightarrow z_0} f(z) = \lim_{z \rightarrow z_0} \frac {1} {\cos \left (\frac {\pi z} {2} \right )}.$$ On the other hand since $z_0$ is a pole of $f$ so $$\lim_{z \rightarrow z_0} f(z) = \infty.$$ i.e. $$\lim_{z \rightarrow z_0} \frac {1} {\cos \left (\frac {\pi z} {2} \right )}=\infty.$$ This shows that $z_0$ is a pole of $\frac {1} {\cos \left (\frac {\pi z} {2} \right )}$ i.e. $z_0$ is a zero of $\cos \left (\frac {\pi z} {2} \right )$. Therefore $\frac {\pi z} {2} = (2n+1)\frac {\pi} {2},\ n \in \Bbb Z$ $\implies$ $z = 2n+1,\ n \in \Bbb Z$. This shows that the poles of $f$ are all the odd integers.

Answer 1

Hint: $\sin(\pi z) = 2 \sin(\pi z/2) \cos(\pi z/2)$.

Answer 2

We have:

$ \mathbb Z= \{z \in \mathbb C: \sin(\pi z)=0\}$

and

$2 \mathbb Z=\{z \in \mathbb C: \sin(\pi z/2)=0\}$.

Hence $f$ has a pole in $z \iff z \in \mathbb Z \setminus 2 \mathbb Z$.