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!
Best Answer
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 expansion of $f$ as Laurent series at $z=1$ in $D_1$:
The expansion of $f$ as Laurent series at $z=1$ in $D_2$:
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).