# Finding the Laurent series (complex numbers)

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.

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 $$z \to \infty$$.

• 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 $z=1$. – Mark Viola Dec 18 '18 at 18:13
• @MarkViola I am working in an edit from earlier since I saw my error. – Rebellos Dec 18 '18 at 18:14
• @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. – Rebellos Dec 18 '18 at 18:21
• I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge. – Mark Viola Dec 18 '18 at 18:24

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}

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}