5
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ How would you propose computing integrals around $z=1$ using the series that you have, given the region of convergence that you have? $\endgroup$ Oct 27, 2018 at 16:02

1 Answer 1

1
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ Thank you for your explanation! It is all clear now. $\endgroup$
    – Jesper
    Nov 5, 2018 at 17:46
  • $\begingroup$ @Jesper: You're welcome. $\endgroup$ Nov 5, 2018 at 17:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .