# How to find all Laurent series for $\frac{1}{z^2 - z}$ centered at $1$?

I have two questions. The definition I learned defined a laurents for a function analytic in some annalus? I guess this question is asking annali centered at $1$? How do I determine all the annali?

Also, is there a quicker way to find coefficients $a_n$ of the laurent series without applying the definition directly (i.e $a_n = 1/2\pi i \int f(\zeta)/(\zeta - 1?)^{n+1} d\zeta$)

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The function has two simple poles at $z=0$ and $z=1$. If we want the Laurent series of $f$ at $1$ then we have to consider two annuli, $0<\left|z-1\right|<1$ and $1<\left|z-1\right|$. For the first annulus, i.e. $0<\left|z-1\right|<1$ we have that $$\frac{1}{z(z-1)}=\frac{1}{z-1}\frac{1}{1+(z-1)}=\frac{1}{z-1}\sum_{n=0}^{\infty}(-1)^n(z-1)^n=\sum_{n=-1}^{\infty}(-1)^{n+1}(z-1)^n$$ For the second annulus $1<\left|z-1\right|$, $$\frac{1}{z(z-1)}=\frac{1}{(z-1)^2}\frac{1}{1+\frac{1}{z-1}}=\frac{1}{(z-1)^2}\sum_{n=0}^{\infty}(-1)^n\frac{1}{(z-1)^n}=\sum_{-\infty}^{n=-2}(-1)^{n}(z-1)^n$$ In both cases we used that $$\frac{1}{1+\zeta}=\frac{1}{1-(-\zeta)}=\sum_{n=0}^{\infty}(-\zeta)^n=\sum_{n=0}^{\infty}(-1)^{n}\zeta^n$$ for $\left|\zeta\right|<1$.

Final notes: When you are asked to compute all Laurent series at a given point $z_0$ you first determine the non-removable singularities of the function. You choose your annuli $\left\{z\in \mathbb{C}:r<\left|z-z_0\right|<R\right\}$ ($0\le r<R\le +\infty$) so that your function has no non-removable singularites in them. Then you use the geometric series $$\frac{1}{1-\zeta}=\sum_{n=0}^{\infty}\zeta^n$$ for $\left|\zeta\right|<1$ or use the Taylor series to expand the other functions (exponential, trigonometric etc.) that may be involved. Your goal is to have only terms of the form $c_n(z-z_0)^n$ in your expression. If you find one Laurent series at an annuli then it is unique. Finally, we rarely use the integral formula of $a_n$ for computations. I think this should be of help Laurent Series

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Here is a related problem. First notice that $z=1$ is a simple pole of the function. The residue of the function at $z=1$, which is the coefficient of $(z-1)^{-1}$ in the Laurent series, is $1$ and you will see this from the series once it is derived. Here is how you compute the series

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

$$=\frac{1}{z-1}(1-(z-1)+(z-1)^2-(z-1)^3-\dots)$$

$$\frac{1}{z-1}-1+(z-1)-(z-1)^2+\dots$$

$$= \frac{1}{z-1} - \sum_{k=0}^{\infty}(-1)^k(z-1)^k, \quad 0<|z-1|<1$$

You can see from the above series that the coefficient of $(z-1)^{-1}$ is one as we said in the beginning. For the other series, $|z-1|>1$, we have

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

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

$$=\frac{1}{(z-1)^2}+\sum_{k=0}^{\infty}(z-1)^{-k}, \quad |z-1|>1.$$

Note that, the series

$$\frac{1}{1+x}=1-x+x^2-x^3+\dots\,,$$

was used in the above derivation.

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Note the "all" part of the question. You're missing one. – WimC Dec 1 '12 at 7:32
@WimC: Thanks for the reminder. – Mhenni Benghorbal Dec 4 '12 at 3:36