I am trying to classify the singularity of $f(z)=\frac{\sin(z)}{(z-\pi)^2}$ at $z=\pi$. I need to know whether this function converges to a point when $z\to \pi$. But I am unable to figure out whether it has a removable singularity, a pole or essential singularity at $z=\pi$.I was thinking of making a power series of $\sin z$ at $z=\pi$. Can someone help me with this problem?
My attempt
$\sin z=-(z-\pi)+\frac{1}{3!}(z-\pi)^3-\frac{1}{5!}(z-\pi)^5+....$
so that $\frac{\sin z}{(z-\pi)^2}=\frac{-1}{z-\pi}+G(z)$ where $G(z)$ is analytic. So, $f(z)$ has a pole of order $1$ i.e. a simple pole at $z=\pi$.
A number $z_0$ is a pole of order $k$ of the function $f(z),$ which is analytic in $0<|z-z_0|<r$ if an only if the following limit exists and is not equal $0.$ $$\lim_{z\to z_0}(z-z_0)^k f(z)\neq 0$$ The function $f(z)=\displaystyle {\sin z\over (z-\pi)^2}$ satisfies $$\lim_{z\to \pi} (z-\pi)f(z)=\lim_{z\to \pi} {\sin z\over z-\pi}=-\lim_{z\to \pi} {\sin (z-\pi)\over z-\pi}=-\lim_{u\to 0}{\sin u\over u}=-1$$