Determine the type of isolated single points of a function 
Determine the type of isolated single points of the function
  $$f(z)=\frac{\sin(z)}{z^5+2z^3+z}.$$

I tried:
$$f(z)=\frac{\sin(z)}{z(z^2+1)^2}.$$ So, $f(z)$ has a isolated singularity at $z=0$.
$$\sin(z)=\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)!}z^{2n+1}$$ and
$$\frac{1}{(z^2+1)^2}$$ is already in the form of a Laurent series with $z_0=\pm i$ and $c_{-2}=1$. I don't know how to deal with this term.
Also, $$\frac{\sin(z)}{z}=\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)!}z^{2n}.$$

You're only asked to determine they types of isolated singularities.  
At $z=0$, the singularity is removable since $\lim_{z\to 0}\frac{\sin(z)}{z}=1$.  In fact, one can see from the Taylor series of the sine function that $\frac{\sin(z)}{z}$ is an entire function.
We have isolated singularities at $z=i$ and $z=-i$.  Both are poles of second order. 
