My problem is:
Which type of singularities has these functions: \begin{align*} 1:f(z)=\frac{z(z-1)^{2}}{\sin^{2}(\pi z)} \end{align*} \begin{align*} 2:f(z)=\frac{z^{4}(z-1)^{2}}{\sin^{2}(\pi z)} \end{align*} \begin{align*} 3:f(z)=\frac{z^{3}(z-1)^{6}}{\sin^{5}(\pi z)}. \end{align*}
My solution:
All these three functions has zeros at $z=0,1,n$.
1: For $z=0$: $\lim_{z\to 0} |\frac{z(z-1)^{2}}{\sin^{2}(\pi z)}|=\infty$ $\Rightarrow$ $z=0$ is a pole of order 1.
For $z=1$: $\lim_{z\to 1} \frac{z(z-1)^{2}}{\sin^{2}(\pi z)}=\frac{1}{\pi^{2}}$ $\Rightarrow$ $z=1$ is a removable singularity.
For $z=n$: $\lim_{z\to n} \frac{z(z-1)^{2}}{\sin^{2}(\pi z)}=\frac{2(n-1)+n}{\pi^{2}\cos(2\pi n)}$ $\Rightarrow$ $z=n$ is a removable singularity?
2: For $z=0$: $\lim_{z\to 0} \frac{z^{4}(z-1)^{2}}{\sin^{2}(\pi z)}=0$ $\Rightarrow$ $z=0$ is a removable singularity.
For $z=1$: $\lim_{z\to 1} \frac{z^{4}(z-1)^{2}}{\sin^{2}(\pi z)}=\frac{1}{\pi^{2}}$ $\Rightarrow$ $z=1$ is a removable singularity.
For $z=n$: $\lim_{z\to n} \frac{z^{4}(z-1)^{2}}{\sin^{2}(\pi z)}=\frac{6n^{2}(n-1)^{2}+8n^{3}(n-1)+n^{4}}{\pi^{2}\cos(2\pi n)}$ $\Rightarrow$ $z=n$ is a removable singularity?
3:For $z=0$: $\lim_{z\to 0} |\frac{z^{3}(z-1)^{6}}{\sin^{5}(\pi z)}|=\infty$ $\Rightarrow$ $z=0$ is a pole of order 3.
For $z=1$: $\lim_{z\to 1} \frac{z^{3}(z-1)^{6}}{\sin^{5}(\pi z)}=0$ $\Rightarrow$ $z=1$ is a removable singularity.
For $z=n$: $\lim_{z\to n} \frac{z(z-1)^{2}}{\sin^{2}(\pi z)}=something\quad that\quad exists$ $\Rightarrow$ $z=n$ is a removable singularity?
Is this correct? I am not sure if it is correct for $z=n$. How come non of them has an essential singularity? Can someone please help me.
$1.$ Let $$g(z)=\frac{(z-n)^2}{\sin ^2(\pi (z-n))}$$
$$f(z)=g(z)\frac{z(z-1)^2}{(z-n)^2}$$
Since $g(z)$ is analytic at $z=n$, the singularities of $f(z)$ at $z=n$ can be obtained by studying the singularities of the rational function
$$\frac{z(z-1)^2}{(z-n)^2}$$