Question about understanding something in laurent series

[Genericcoder]

So the question I got the represention for both partial fractions after I broke the functions into two partial fraction one I got as 1/3(z + 1) + 2/3(z - 2) and I got laurent series represention for both but I was wondering for |z| < 1 how can they both converge for |z| < 1 are we acctually minusing both series from each other when we did that ? but when I got second series representation for second partial fraction I got - so that's how I understood since we are minusing |z| < 2 from |z| < 1 we get back to |z| < 1 I don't know if my intuition here is correct or wrong. but what if for second partial fraction we get a positive instead of a negative maybe here the negative that we got was by luck or something.

 

[lippyka]

Your intuition is correct. The Laurent series for a rational function can be written as a sum of the Laurent series for its partial fractions. To obtain the Laurent series for the partial fraction, you divide the numerator and denominator by their greatest common factor, then obtain the coefficients of the partial fraction by solving a linear system of equations. In your example, the partial fractions are $\frac{1}{3(z+1)} + \frac{2}{3(z-2)}$. When you obtain the coefficients of these partial fractions, you will get $1/3$ and $2/3$, respectively. Then, you can calculate the Laurent series for each partial fraction separately. For the first one, you have$$\frac{1}{3(z+1)} = \sum_{n=-\infty}^{\infty}\frac{(-1)^n}{3^{n+1}}z^n.$$For the second one, you have$$\frac{2}{3(z-2)} = \sum_{n=-\infty}^{\infty}\frac{2(-1)^n}{3^{n+1}}(z-2)^n.$$Finally, adding the two Laurent series together, you get$$\frac{1}{3(z+1)} + \frac{2}{3(z-2)} = \sum_{n=-\infty}^{\infty}\left(\frac{(-1)^n}{3^{n+1}}z^n + \frac{2(-1)^n}{3^{n+1}}(z-2)^n\right).$$For $|z|<1$, this series converges since both series for the partial fractions individually converge for that region.