# Laurent Series doubt about the Sum limits

• cathode-ray
In summary, the conversation revolves around solving for the Laurent series of the function \frac{1}{\left(z-1\right)^2\left(z+3\right)}. The individual is confused about their solution and the given solution, but eventually realizes their mistake in calculating the sum. They also ask if the Laurent series is unique and are told that it is, despite getting different expressions in their attempts.

#### cathode-ray

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:

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

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.