<s>[INST] Finding the residue with a Laurent series expansion. I have a question about the following problem:
"Detect the error in the following argument. The function $f(z)=\frac{1}{z(z-1)^2}$ has an isolated singularity at $z=0$. The Laurent series is $f(z)=\frac{1}{(z-1)^3}-\frac{1}{(z-1)^4}+\frac{1}{(z-1)^5}-...$ 
for $|z-1|>1$. Apparently $z=1$ is an essential singularity with residue 0."
Now I know that the error is that one has to compute the Laurent series on the annulus $0<|z-1|<1$, but why is this the case? Do you always have to take the inner annulus to compute the residue?
A clear explanation would be much appreciated!
 [/INST] We  consider the expansion of $f$ at $z_0=1$  since the center $1$ is indicated by the terms $\frac{1}{(z-1)^k}$.

The function 
  \begin{align*}
f(z)=\frac{1}{z(z-1)^2}=\frac{1}{z}-\frac{1}{z-1}+\frac{1}{(z-1)^2}\tag{1}
\end{align*}
   is to expand around the center $z_0=1$. Since there are isolated singularities, namely a single pole at $z=0$ and a double pole at $z=1$ we have to distinguish two regions 
\begin{align*}
D_1:&\quad 0<|z-1|<1\\
D_2:&\quad |z-1|>1
 \end{align*} 
  
  
*
  
*The first region $D_1$ is a punctured disc with center $z_0=1$, radius $1$ and the pole $0$ at the boundary of the disc.
In the interior we have a representation of the fractions with a pole at $z=1$ as principal part of a Laurent series at $z_0=1$, while the fraction with pole at $z=0$ admits a representation as power series.
  
*The other region $D_2$ containing all points outside the closure of $D_1$ admits for all fractions a representation as principal part of a Laurent series at $z=1$.

The expansion of $f$ as Laurent series at $z=1$ in $D_1$:

We obtain
  \begin{align*}
\color{blue}{f(z)}&\color{blue}{=\frac{1}{z(z-1)^2}}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{z}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{1+(z-1)}\\
&\color{blue}{=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\sum_{n=0}^\infty (-1)^n(z-1)^n}\tag{2}\\
\end{align*}

The expansion of $f$ as Laurent series at $z=1$ in $D_2$:

We  obtain 
  \begin{align*}
\color{blue}{f(z)}&\color{blue}{=\frac{1}{z(z-1)^2}}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{z}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{1+(z-1)}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{z-1}\frac{1}{1+\frac{1}{z-1}}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{z-1}\sum_{n=0}^\infty (-1)^n\frac{1}{(z-1)^n}\\
&\color{blue}{=\sum_{n=3}^\infty (-1)^{n+1}\frac{1}{(z-1)^n}}\tag{3}\\
\end{align*}

Conclusion:


*

*We see two valid series expansions (2) and (3) of $f$ at $z=1$. One is in the punctured disc $D_1$ and the other in the region  $D_2$.

*In order to determine the type of singularity at $z=1$ we have to consider the region near the singularity. This means we can check the Laurent series expansion in $D_1$ but not that in $D_2$. From the series expansion (2) we clearly see that $z=1$ is a pole of order $2$. This can also be immediately deduced from the representation (1).
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