when finding the Laurent series for $\frac{1}{(z-1)(z-2)}$ in the region $|z| <1$, I thought that one should first evaluate where the poles are, inside or outside the given region. z=2 is outside the region, but still, Saff and Snyder include that term in the Laurent series obtaining:

\begin{equation} \frac{1}{(z-1)(z-2)}=\frac{1}{z-2}-\frac{1}{z-1} \end{equation}

Such that they get for $|z|<1$ the following series for the first term:

\begin{equation} \frac{1}{z-2}=-\frac{1}{2}\frac{1}{1-\frac{z}{2}}=\frac{1}{2}\sum_{n=0}^{\infty}\big(\frac{z}{2}\big)^n=-\sum_{n=0}^{\infty}\big(\frac{z^n}{2^{n+1}}\big) \end{equation}

for the second term they get:

$\frac{1}{z-1}= -\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$

which they combine by subtraction to get :

\begin{equation} \sum_{n=0}^{\infty}\bigg(-\frac{1}{2^{n+1}}+1\bigg)z^n \end{equation}

But for another case $\frac{z^2-2z+3}{z-2}$, Saff and Snyder write "Notice that the region $|z-1| <1$ excludes $z=2$."

So isn't this as the residue theorem, where poles outside of the region are simply not included in the calculation - hence, one should not include them in the Laurent series?

This was confusing indeed.

Any comments?


  • 2
    $\begingroup$ $z=1$ is also outside the region. There are no poles. $\endgroup$ Jan 3, 2022 at 11:31
  • $\begingroup$ Ok, so limit points are not included ? I thought limit points were in general part of a region. $\endgroup$ Jan 3, 2022 at 11:33

1 Answer 1


The function is analytic in $|z| <1$ and its power series expansion is its Laurent series also. So the correct answer is $\sum (1-2^{-n-1})z^{n}$ [You missed $z^{n}$ in the series]. Note that poles and risidues play no role in this.

  • $\begingroup$ Thanks Kavi! The error is corrected, and thanks for confirmation on that poles have no role in the Laurent series. $\endgroup$ Jan 3, 2022 at 11:32

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