2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ How would you propose computing integrals around $z=1$ using the series that you have, given the region of convergence that you have? $\endgroup$ – DisintegratingByParts Oct 27 '18 at 16:02
0
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your explanation! It is all clear now. $\endgroup$ – Jesper Nov 5 '18 at 17:46
  • $\begingroup$ @Jesper: You're welcome. $\endgroup$ – Markus Scheuer Nov 5 '18 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.