So I'm looking over previous year assignment questions about Laurent expansions and they have this one.
Compute the Laurent expansions of
$$\frac{z^2 - 8z}{(z-2)^2(z+1)}$$
in the region $|z-1|<1$. But they mention that "Remember that if it holds for $z$ that $|z − 1| < 1$ then it certainly holds for this $z$ that $|z − 1| < 2$".
I'm trying to understand this hints relevance to the question. Any guidance would be greatly appreciated.
Thank you.
Note that\begin{align}\frac{z^2-8z}{(z-2)^2(z+1)}&=\frac1{z+1}-\frac4{(z-2)^2}\\&=\frac1{2+(z-1)}-\frac4{(-1+(z-1))^2}.\end{align}And now it is easy to find a series centered at $1$ which converges, when $|z-1|<2$, to $\frac1{z+1}$, and to find a series which converges, when $|z-1|<1$, to $\frac4{(-1+(z-1))^2}$. And, since $|z-1|<1\implies|z-1|<2$, the sum of these two series is the series that you're after.