I am trying to understand how to evaluate $\frac{1}{z^2}$ around $z_0 = 1$
so $$\frac{1}{z^2}=\left(\frac{1}{z}\right)^2=\sum_{n=0}^{\infty}(-1)^n(z-1)^n \cdot \sum_{m=0}^{\infty}(-1)^m(z-1)^m$$ $$=\sum_{n=0}^{\infty}\sum_{m=0}^{n}(-1)^n(z-1)^n$$
and the explanation in the book is the word "convolution." I don't understand how is it related to that step; I tried to examine to definition to convolution again, but still nothing came up to my mind.
To simplify the notation let $x=(-1)(z-1)$; we want to show that
$$\left(\sum_{n\ge 0}x^n\right)\left(\sum_{m\ge 0}x^m\right)=\sum_{n\ge 0}\sum_{m=0}^nx^n\,.\tag{1}$$
If you multiply out the lefthand side as if it were a product of two polynomials, you get a term $x^n\cdot x^m=x^{n+m}$ for each pair $\langle n,m\rangle$ of non-negative integers. If $k$ is a non-negative integer, how many of these terms are equal to $x^k$? The terms equal to $x^k$ are precisely the terms $x^{n+m}$ such that $n+m=k$, and there is one for each value of $n$ from $0$ through $k$: these terms are $x^{0+k},x^{1+(k-1)},x^{2+(k-2)},\ldots,x^{(k-1)+1}$, and $x^{k+0}$. Thus,
$$\begin{align*} \left(\sum_{n\ge 0}x^n\right)\left(\sum_{m\ge 0}x^m\right)&=\sum_{k\ge 0}\sum_{n=0}^kx^{n+(k-n)}\\ &=\sum_{k\ge 0}\sum_{n=0}^kx^k\,. \end{align*}$$
And we can rename the index variables $k$ and $n$ to $n$ and $m$, respectively, without changing the value, so we have
$$\begin{align*} \left(\sum_{n\ge 0}x^n\right)\left(\sum_{m\ge 0}x^m\right)&=\sum_{k\ge 0}\sum_{n=0}^kx^k\\ &=\sum_{n\ge 0}\sum_{m=0}^nx^n\,. \end{align*}$$
Note that $\sum_{m=0}^nx^n$ is just the sum of $n+1$ copies of $x^n$, so in fact the whole thing can be further simplified to
$$\sum_{n\ge 0}(n+1)x^n\,.$$