all possible Taylor's and Laurent series of $\frac{2z-3}{z^2-3z+2} $ about $z=0$

Question

Find all possible Taylor's and Laurent series of $\frac{2z-3}{z^2-3z+2} $ about $z=0$

I solved it as

$$\frac{2z-3}{z^2-3z+2} = \frac{1}{z-2} + \frac{1}{z-1}$$

$$= \frac{-1}{2}{(1+\frac{z}{2}+\frac{z^2}{4}+.....)} - (1+z+z^2+z^3)$$

Is this answer correct ? Is the above series both Taylor as well as Laurent ?

Answer 1

What you wrote is correct. I would add just one more step, merging your series into a single one:$$-\left(1+\frac12\right)-\left(1+\frac14\right)z-\left(1+\frac18\right)z^2+\cdots$$This is the Taylor series of your function at $z=0$ and it is its Laurent series too.

Answer 2

Your result is only for $|z|<1$ and you are missing two other cases.

To expand $$\frac{2z-3}{z^2-3z+2}=-\left(\frac{1}{1-z} +\frac{1}{2-z}\right)$$ there are two singularities at $z=1$ and $z=2$. So we can have three different cases.

It is all about the region of convergence (RoC) of the geometric series, since $$\displaystyle\sum_{n=a}^\infty r^n=\frac{r^a}{1-r},\color{red} {\text{when } |r|<1}$$

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• If $|z|<1$, then

$$\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$$ $$\frac{1}{2-z}=\frac{1/2}{1-z/2}=\frac12\sum_{n=0}^{\infty}\left(\frac z2\right)^n$$

• If $1<|z|<2$, then

$$\frac{1}{1-z}=-\frac{1/z}{1-1/z}=-\sum_{n=1}^{\infty}\left(\frac 1z\right)^n$$ $$\frac{1}{2-z}=\frac{1/2}{1-z/2}=\frac12\sum_{n=0}^{\infty}\left(\frac z2\right)^n$$

• If $|z|>2$, then

$$\frac{1}{1-z}=-\frac{1/z}{1-1/z}=-\sum_{n=1}^{\infty}\left(\frac 1z\right)^n$$ $$\frac{1}{2-z}=-\frac{1}{2}\cdot\frac{2/z}{1-2/z}=-\frac12\sum_{n=1}^{\infty}\left(\frac 2z\right)^n$$