# 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!

• How would you propose computing integrals around $z=1$ using the series that you have, given the region of convergence that you have? – DisintegratingByParts Oct 27 '18 at 16:02

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).

• Thank you for your explanation! It is all clear now. – Jesper Nov 5 '18 at 17:46
• @Jesper: You're welcome. – Markus Scheuer Nov 5 '18 at 17:48