Classify the singularities (removable, poles, essential) of the function
$$f(z) = \dfrac{\sin\left(\dfrac{z}{z+1}\right)}{z(z-1)^2(z^2+1)}$$
First we consider the isolated singularity at $z=0$.
\begin{align} f(z) = \dfrac{\sin\left(\dfrac{z}{z+1}\right)}{z(z-1)^2(z^2+1)} &=z^{-1}\sin{\left(1-\dfrac{1}{z+1}\right)}g(z)\\ &=z^{-1}\left(\sin1\cos\dfrac{1}{z+1}-\sin\dfrac{1}{z+1}\cos1\right)g(z)\\ &= g(z)z^{-1}\left(\sin1\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(z+1)^{2n}(2n)!}-\cos1\sum_{n=0}^\infty \frac{(-1)^n}{(z+1)^{2n+1}(2n+1)!}\right) \end{align}
Note that
$$g(z)=\frac{1}{(z-1)^2(z^2+1)}\to 1 \quad \text{ as }\quad z\to0.$$
So $z=0$ is an essential singularity. However this seems incredibly lengthy if I'm going to have to do this for every singularity.. Is there a better way to do this?
$$ \frac1z\,\sin\frac{z}{1+z}=\frac1z\,\frac{z}{1+z}\frac{\sin\frac{z}{1+z}}{\frac{z}{1+z}}=\frac1{1+z}\,\frac{\sin\frac{z}{1+z}}{\frac{z}{1+z}}. $$ Sinze $\frac{z}{1+z}\to0$ as $z\to0$, we see that $$ \lim_{z\to0}\frac1{1+z}\,\frac{\sin\frac{z}{1+z}}{\frac{z}{1+z}}=\lim_{z\to0}\frac1{1+z}\times\lim_{z\to0}\frac{\sin\frac{z}{1+z}}{\frac{z}{1+z}}=1. $$ As for $$ \frac{1}{(z-1)^2(z^2+1)}, $$ it is holomorphic at $z=0$. Thus, $z=0$ is an avoidable singularity.