Im trying to figure out what type of singularity $z=1$ is for the function $f(z)=\frac{z^{4}(z-1)^{2}}{\sin^{2}(\pi z)}$.
In my solution manual it says that it is a pole of order 1 but Im trying to understand how they come to this conclusion.
If I set that \begin{align*} f(z)=\frac{g(z)}{z-1}, \end{align*} which means that $g(z)=(z-1)f(z)$ and $g$ is holomorphic in the neighborhood of $z=1$. We have that \begin{align*} \lim_{z\to 1}g(z)=\lim_{z\to 1}(z-1)f(z)=\lim_{z\to 1}\frac{(z-1)^{3}}{\sin^{2}(\pi z)}= 0. \end{align*} But since $g(1)=0$ then $f$ do NOT have a pole at $z=1$? It seems like $f$ has a removable singularity at $z=1$ not a pole of order 1.
Can someone help me? Is it correct that $z=1$ is a pole of order 1 to $f$? If so, then how?
One of the Riemann's theorems concerning removable singularities states it needs to be $$\lim_{z\rightarrow1}(z-1)f(z)=0$$ And, just to be sure the calculations are right using this: $$\lim_{z\rightarrow1}(z-1)f(z)=\lim_{z\rightarrow1}\frac{z^4(z-1)^3}{\sin^2{(\pi z)}}=\lim_{z\rightarrow1}\frac{z^4(\pi z-\pi)^3}{\pi^3 \sin^3{(\pi z-\pi)}}\cdot \frac{\sin^3{(\pi z-\pi)}}{\sin^2{(\pi z)}}=\\ \frac{1}{\pi^3}\lim_{z\rightarrow1}\frac{\sin^3{(\pi z-\pi)}}{\sin^2{(\pi z)}}=\frac{1}{\pi^3}\lim_{z\rightarrow1}\frac{\left(\sin{\pi z \cos{\pi} - \cos{\pi z} \sin{\pi}}\right)^3}{\sin^2{(\pi z)}}=\frac{-1}{\pi^3}\lim_{z\rightarrow1}\sin{\pi z}=0$$