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I have $$ f(z)={\frac{1}{z(1-z)}} $$ Need to find the Laurent series around $z=0, z=1, z=\infty$. I did $$ {\frac{1}{z(1-z)}} = {\frac{A}{z}}+{\frac{B}{1-z}} $$ and found $A=1, B=1$. Therefore we get $$ {\frac{1}{z}}+{\frac{1}{1-z}} = {\frac{1}{z}} + \sum z^n $$ But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.

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Hints :

For the $z=1$ case :

You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :

$$f(z) = \frac{1}{z(1-z)} = \frac{1}{z} + \frac{1}{1-z} = \frac{1}{1+(z-1)} + \frac{1}{1-z} $$ $$=$$ $$\frac{1}{(z-1)\left(\frac{1}{z-1} + 1\right)} - \frac{1}{z-1} = \frac{1}{z-1}\left(\frac{1}{\frac{1}{z-1} + 1}\right)$$

Now, recall the geometric series $\frac{1}{1+w} = \sum_{n=1}^\infty (-1)^nw^n$. Let $w = \frac{1}{z-1}$. Thus :

$$f(z) = \frac{1}{z-1}\sum_{n=0}^\infty (-1)^n \left(\frac{1}{z-1}\right)^n =\sum_{n=1}^\infty (-1)^{n-1}\left(\frac1{z-1}\right)^{n+1}$$

For the $\infty$ case :

Recall the geometric series $\frac{1}{1-w} = \sum_{n=1}^\infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :

$$f(z) = \frac{1}{z(1-z)}= -\frac{1}{z^2(1-\frac{1}{z})}=-\sum_{n=0}^{\infty}z^{-n-2}$$

Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z \to \infty$.

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  • $\begingroup$ That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$. $\endgroup$ – Mark Viola Dec 18 '18 at 18:13
  • $\begingroup$ @MarkViola I am working in an edit from earlier since I saw my error. $\endgroup$ – Rebellos Dec 18 '18 at 18:14
  • $\begingroup$ @MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially. $\endgroup$ – Rebellos Dec 18 '18 at 18:21
  • $\begingroup$ I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge. $\endgroup$ – Mark Viola Dec 18 '18 at 18:24
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In the annulus $1<|z|<\infty$, we have

$$\begin{align} \frac{1}{z(1-z)}&=\frac{1}{z}+\frac1{1-z}\\\\ &=\frac{1}{1+(z-1)}+\frac1{1-z}\\\\ &=\frac1{z-1}\frac{1}{1+\frac1{z-1}}-\frac1{z-1}\\\\ &=\frac1{z-1}\sum_{n=0}^\infty (-1)^n \left(\frac{1}{z-1}\right)^n-\frac1{z-1}\\\\ &=\sum_{n=1}^\infty (-1)^{n-1}\left(\frac1{z-1}\right)^{n+1} \end{align}$$

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We have $$ \eqalign{ & {1 \over {z\left( {1 - z} \right)}} = \cr & = \left\{ \matrix{ - \left( {{1 \over z} + {1 \over {\left( {1 - z} \right)}}} \right)\quad \Rightarrow \quad - {1 \over z} - \sum\limits_{0\, \le \,n} {z^{\,n} } \quad \left| {\,z \to 0} \right. \hfill \cr {1 \over {\left( {z - 1} \right)}} - {1 \over {\left( {1 + \left( {z - 1} \right)} \right)}}\quad \Rightarrow \quad {1 \over {\left( {z - 1} \right)}} - \sum\limits_{0\, \le \,n} {\left( { - 1} \right)^{\,n} \left( {z - 1} \right)^{\,n} } \quad \left| {\,z \to 1} \right. \hfill \cr - \left( {{1 \over z}} \right)\left( {1 - {1 \over {\left( {1 - {1 \over z}} \right)}}} \right)\quad \Rightarrow \quad \sum\limits_{0\, \le \,n} {\left( {{1 \over z}} \right)^{\,n + 2} } \quad \left| {\,z \to \infty } \right. \hfill \cr} \right. \cr} $$

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