# Laurent Series doubt about the Sum limits

## Main Question or Discussion Point

Hi everyone!

I'm studying Laurent Series, and I thought I understood it but after solving one exercises I get confused. The problem was to get the Laurent series for the following function:

$$\frac{1}{\left(z-1\right)^2\left(z+3\right)}$$

I do it this way:

$$\frac{1}{\left(z-1\right)^2}\cdot \frac{1}{z+3+1-1}=$$
$$\frac{1}{\left(z-1\right)^2}\cdot \frac{1}{4-\left(-\left(z-1\right)\right)}=$$
$$\frac{1}{\left(z-1\right)^3}\cdot \frac{1}{1-\left(-\left(\frac{z-1}{4}\right)\right)}=$$
$$\frac{1}{\left(z-1\right)^3}\cdot \sum_{n=0}^\infty \left(-\frac{4}{z-1}\right)^n=$$
$$\sum_{n=0}^\infty \left(-4\right)^n\cdot \frac{1}{\left(z-1\right)^{n+3}}$$

But when I saw the solution they say that is:

$$\frac{1}{4\left(z-1\right)^2} - \frac{1}{4^2\left(z-1\right)}+\frac{1}{4^3}-\frac{\left(z-1\right)}{4^4}+\frac{\left(z-1\right)^2}{4^5}+\dots$$

which is different from what I get. However I noticed that if I change the limits of the sum to -inf,0 the expression is equal. Why do this happens or what do I did wrong?

Writing

$$\frac{1}{\left(z-1\right)^3}\cdot \frac{1}{1-\left(-\left(\frac{z-1}{4}\right)\right)}=$$

is correct. But when you changed that fraction in a sum, that is incorrect. The correct sum is

$$\frac{1}{\left(z-1\right)^3}\sum_{n=0}^{+\infty}{\left(-\frac{z-1}{4}\right)^n}$$

Also, that 3 in my above post should be a 2.

I didn't notice that. After all it was just an annoying error. Thanks for your help!

Just one more thing: Is the Laurent serie unique?
I'm asking this because I tried different approaches to get the Laurent serie, and each one gave me a different expression.

Yes, the coefficients of the Laurent series are unique. Strange that you got different expressions. I think you've made a mistake somewhere...

For example if I do:

$$\frac{1}{(z-1)^{2}}\cdot\frac{1}{z+3-2+2}=$$
$$\frac{1}{(z-1)^{2}}\cdot\frac{1}{z-1+2}=$$
$$-\frac{1}{(z-1)^{3}}\cdot\frac{1}{1-\frac{2}{(z-1)}}=$$
$$-\frac{1}{(z-1)^{3}}{\displaystyle \sum_{n=0}^{\infty}(\frac{2}{z-1})^{n}=\sum_{n=0}^{\infty}\frac{2^{n}}{(z-1)^{n+3}}}$$

or:

$$\frac{1}{\left(z-1\right)^{2}}\cdot\frac{1}{4-\left(-\left(z-1\right)\right)}=$$
$$-\frac{1}{\left(z-1\right)^{3}}\cdot\frac{1}{1-\frac{4}{(z-1)}}=$$
$$\sum_{n=0}^{\infty}-\frac{4^{n}}{(z-1)^{n+3}}$$

The expressions are different.

Sorry forget the first case. I have already seen the mistake. The second I still don't understand.