Laurent series of $ f(z) = \frac{1}{(z+1)(z-2)} $ in $ 1<|z|<2 $ interval

Question

I tried something and got $f(z)= \frac{A}{(z+1)}+\frac{B}{(z-2)}$ and $A=-\frac{1}{(3)}$; $B=-\frac{1}{(3)}$ which comes out to be $f(z)= \frac{1}{-3(z+1)}+\frac{1}{3(z-2)}$. Unfortunately I don't know what to do from now on. I know the basics of the Laurent series but interval we are given restricts me to some degree.

Answer 1

Hint: One wonderful thing about Laurent series is that they are unique. This means that whichever (correct) way you find the coefficients, the end result will be the unique correct result.

In particular, you can find the Laurent series for each term in your partial fraction decomposition and add them coefficient-wise.

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EDIT: You have found

$$\frac1{z-2} = \frac12 \frac1{1-\frac z2} = \frac12\sum_{k\geqslant 0 }\frac {z^k}{2^k} =\sum_{k\geqslant 0 }\frac {z^k}{2^{k+1}},$$

where the series converges whenever $|z/2| < 1 \iff |z|<2$, by the convergence of geometric series.

You have also found that

$$\frac 1{z+1} = \frac1z \frac 1{1-\left(-\frac1z\right)} = \frac1z\sum_{k\geqslant 0 }\frac{(-1)^k}{z^k} = \sum_{k\geqslant 1 }\frac{(-1)^{k+1}}{z^k} = \sum_{k\geqslant 1 }(-1)^{k+1}z^{-k} ,$$

where much like before the series converges whenver $|-1/z| < 1 \iff |z|>1$.

You can add these up and the result will converge in the intersection of the regions of convergence, which is precisely the region you're interested in. Notice that the second series involves negative powers of $z$, so we can add them up as follows:

$$\sum_{k\in\Bbb Z}c_k z^k$$

where $c_k = 2^{k+1}$ if $k\geqslant 0$ and $c_k = (-1)^{1-k}$ if $k<0$.