Residue and removable singularity I have the function $\displaystyle f(z)=\frac{z^2+z+1}{z^2(z-1)}$. 
I have to calculate residue in isolated singularities (including infinity). 
I calculated residue in $z = 0$ and $z = 1$, but I don't know how to calculate it in infinity. I don't understand if infinity is removable singularity or not.

The Residue at Infinity of $f(z)$ is given by 
$$\text{Res}\left(f(z),z=\infty\right)=\text{Res}\left(-\frac1{z^2}f\left(\frac1z\right),z=0\right)$$ 
So, for $f(z)=\frac{z^2+z+1}{z^2(z-1)}$, we have
$$\begin{align}
\text{Res}\left(f(z),z=\infty\right)&=\text{Res}\left(-\frac1{z^2}\frac{1/z^2+1/z+1}{(1/z^2)(1/z-1)},z=0\right)\\\\
&=-\text{Res}\left(\frac{z^2+z+1}{z(1-z)},z=0\right)\\\\
&=-\lim_{z\to 0}\left(z\,\frac{z^2+z+1}{z(1-z)}\right)\\\\
&=-1
\end{align}$$
