# Finding the Laurent series of $f(z)=1/((z-1)(z-2))$

Let $$f(z)=\frac{1}{(z-1)(z-2)}$$ and let $$R_1=\Bigl\{z\Bigm| 1<|z|<2\Bigr\}\quad\text{ and }\quad R_2=\Bigl\{z\Bigm| |z|>2\Bigr\}.$$

How do you find the Laurent series convergent on $R_1$? Also how do you do it for $R_2$?

I'm having serious trouble with this as I can't see how to expand things into series with n as any integer, not just natural number. Also how to apply Cauchy's integral formula to an annulus. If anyone can explain this to me I will be extremely grateful.

• $f(z) = \frac{1}{z-2}-\frac{1}{z-1}$ So you only need the Laurent series for $\frac{1}{z-2}$ and $\frac{1}{z-1}$ on these two sets. Nov 4, 2011 at 19:13

I assume you want the Laurent series expansion for this function at $0$ (since this is the point those annuli are centered at). The standard way to do this is not to use the integral formula at all--- just the uniqueness of the Laurent series expansion (which I guess is often proved using the integral formula). You do some algebra to find series representations of this function in integer powers of $z$ that converge in the given annuli. By uniqueness these series have to be the Laurent series.

From partial fractions $$\frac{1}{(z - 1)(z - 2)} = -\frac{1}{z-1} + \frac{1}{z-2}$$ so on either annulus it is enough to find the Laurent series expansions of the functions $\frac{1}{z-1}$ and $\frac{1}{z-2}$ and then combine them term by term.

If $|z| > 2$ then note that $$\frac{1}{z - 2} = \frac{1}{z} \frac{1}{1 - 2z^{-1}}$$ and the fact that $|z| > 2$ implies that $|2 z^{-1}| < 1$. If you recall the geometric series formula, $$\frac{1}{1-u} = 1 + u + u^2 + \cdots, \qquad |u| < 1,$$ then putting in $u = 2 z^{-1}$ you see that $$\frac{1}{z - 2} = \frac{1}{z} (1 + (2z^{-1}) + (2z^{-1})^2 + \cdots)$$ holds for all $|z| > 2$. If you just expand this out you get the Laurent series for $\frac{1}{z - 2}$ in this region. Of course it is much cleaner if you use sigma notation to write it out.

On the annulus $1 < |z| < 2$ of course the above method does not work. But we also have $$\frac{1}{z - 2} = -\frac{1}{2} \frac{1}{(1 - \frac{z}{2})}$$ and if $1 < |z| < 2$ then $|\frac{z}{2}| < 1$ so the geometric series formula can be used again in a slightly different way here.

The Laurent series expansion of $\frac{1}{z - 1}$ is easier since $\frac{1}{z-1} = \frac{1}{z} \frac{1}{1 - z^{-1}}$ and $|z^{-1}| < 1$ holds for all $z$ in either annulus. So the Laurent expansion of this function is the same in both regions.

The ideas used here generalizes to finding the Laurent series other rational functions on other annuli of course.

• Thank you, this has really helped my understanding much more than just symbols, not that I don't appreciate that sort of help, this has just helped much more to bring clarity. Nov 4, 2011 at 22:51

The function $f(z)$ can be expanded into two partial fractions $$f(z):=\frac{1}{\left( z-1\right) \left( z-2\right) }=\frac{1}{z-2}-\frac{1}{z-1}.$$

We now expand each partial fraction into a geometric series. On $R_{2}$ these series are $$\begin{eqnarray*} \frac{1}{z-2} &=&\frac{1}{z\left( 1-2/z\right) }=\frac{1}{z} \sum_{n=0}^{\infty }\left( \frac{2}{z}\right) ^{n}\qquad \left\vert z\right\vert >2 \\ &=&\frac{1}{z}\sum_{n=0}^{\infty }2^{n}\frac{1}{z^{n}}=\sum_{n=0}^{\infty }2^{n}\frac{1}{z^{n+1}} \end{eqnarray*}$$ and $$\begin{eqnarray*} \frac{1}{z-1} &=&\frac{1}{z\left( 1-1/z\right) } \\ &=&\frac{1}{z}\sum_{n=0}^{\infty }\left( \frac{1}{z}\right) ^{n}=\sum_{n=0}^{\infty }\frac{1}{z^{n+1}}\qquad \left\vert z\right\vert >1. \end{eqnarray*}$$

And so, the Laurent series is

$$\frac{1}{\left( z-1\right) \left( z-2\right) }=\sum_{n=0}^{\infty }\frac{1}{ z^{n+1}}(2^{n}-1)\qquad \left\vert z\right\vert >2>1.$$

On $R_{1}$, the two geometric series are $$\begin{eqnarray*} \frac{1}{z-2} &=&\frac{-1/2}{1-z/2}=\sum_{n=0}^{\infty }\left( -\frac{1}{2} \right) \left( \frac{z}{2}\right) ^{n}\qquad \left\vert z\right\vert <2 \\ &=&\sum_{n=0}^{\infty }-\frac{1}{2^{n+1}}z^{n} \end{eqnarray*}$$

and

$$\begin{eqnarray*} \frac{1}{z-1} &=&\frac{1/z}{1-1/z}=\sum_{n=0}^{\infty }\frac{1}{z}\left( \frac{1}{z}\right) ^{n}\qquad \left\vert z\right\vert >1 \\ &=&\sum_{n=0}^{\infty }\frac{1}{z^{n+1}}. \end{eqnarray*}$$

We thus get the following Laurent series

$$\frac{1}{\left( z-1\right) \left( z-2\right) }=\sum_{n=0}^{\infty }\left( - \frac{1}{2^{n+1}}z^{n}-\frac{1}{z^{n+1}}\right) \qquad 1<\left\vert z\right\vert <2.$$

• This is great! Thanks for your help, very satisfying to understand this one Nov 4, 2011 at 22:51
• @LHS Glad to help. Nov 4, 2011 at 23:09

Most problems like this can be done with two tools only: partial fractions, and the result $1/(1-w) = 1+w+w^2+\cdots$ for $|w|<1$. So first split your function $f$ into $1/(z-2)-1/(z-1)$. I will show you how to cope with one of these factors, $1/(z-2)$. Writing this as $\frac{-1}{2} \frac{1}{1-z/2}$ is tempting but no good: the "$1/(1-w)$" expansion will converge only for $|z/2|<1$, not on the regions you are supposed to care about. So you take out a factor of $z^{-1}$ instead:

$$1/(z-2) = z^{-1} \frac{1}{1-2/z}$$

The final term can be expanded with the $1/(1-w)$ series, valid for $|2/z|<1$ that is $|z|>2$. So that does give you a Laurent series valid in the right region (once you multiply bu $z^{-1}$. These methods can be used to solve all of your problems.

• Thank you, this is of great help Nov 4, 2011 at 22:52

HINT: For $R_1$ $$\frac{1}{(z-1)(z-2)} = \frac{1}{(z-1)((z-1)-1)}$$ And expand in $w=z-1$. For $R_2$: $$\frac{1}{(z-1)(z-2)} = \frac{1}{z^2} \frac{1}{\left( 1 - \frac{1}{z}\right) \left( 1 - \frac{2}{z}\right)}$$ and expand in $w=\frac{1}{z}$.