What is the Laurent series of $f(z)=1/z^2$ I have just looked at about 10 different Laurent series in the past days, but I cannot solve  $f(z)=\frac{1}{z^2}$.
I attempt, $|z|>0$, with u=$z^2+1$:
\begin{equation}
\frac{1}{z^2}=\frac{1}{z^2-1+1}=\frac{1}{u-1}=-\frac{1}{1-u}
\end{equation}
From here, assuming there exists substitution method at all for Laurent series, I get:
\begin{equation}
-\frac{1}{1-u}=\sum_{n=-\infty}^\infty(-1)^nu^n=\sum_{n=-\infty}^\infty(-1)^n(z^2+1)^n
\end{equation}
but this is not correct. The correct result is :
\begin{equation}
-\frac{1}{1-u}=\sum_{n=0}^\infty(-1)^n(1+n)(-1+z)^n
\end{equation}
It seems there is a rule of differentiation here, so I try this:
Put $g(z)=\frac{1}{z}$, and $f(z)=\frac{1}{z^2}$, we see that $g'(z)=-f(z)$
So
\begin{equation}
g(z)=\frac{1}{z}=\frac{1}{1-z-1}=-\frac{1}{z+1}\frac{1}{-\frac{1}{z+1}+1}=\frac{1}{z+1}\frac{1}{1-(-\frac{1}{z+1})}
\end{equation}
This gives
\begin{equation}
\frac{1}{z+1}\sum_{n=0}^{\infty}(-1)^n\bigg(\frac{1}{z+1}\bigg)^n = \sum_{n=0}^{\infty}(-1)^{n-1}\frac{1}{(z+1)^n}
\end{equation}
Since $g'(z)=-f(z)$, then $-g'(z)=f(z)$, so I differentiate the series:
\begin{equation}
f(z)=\frac{1}{z^2}=\sum_{n=0}^{\infty}\bigg((-1)^{n-1}\frac{1}{(z+1)^n}\bigg)' =\sum_{n=0}^{\infty}(-1)^{n-1}(-n)(z+1)^{-n-1}
\end{equation}
But this is not entirely correct, though we are getting closer. What is the wrong in this last procedure here?
 A: As noted in the comments (except I lost a crucial not while typing), a function doesn't have " a Laurent series," it has "a Laurent series in an annulus" that depends on the annulus.
In more detail, since the comments have clarified the issue, the function $f(z) = 1/z^{2}$ is holomorphic on the punctured plane, and can be expanded as a Laurent series in every annulus not containing $0$. If $z_{0}$ denotes the center of the annulus, the Laurent series has the form
$$
\sum_{k=-\infty}^{\infty} a_{k} (z - z_{0})^{k}
= \sum_{k=1}^{\infty} \frac{a_{-k}}{(z - z_{0})^{k}}
+ \sum_{k=0}^{\infty} a_{k} (z - z_{0})^{k}
$$
for some complex coefficients $(a_{k})_{k=-\infty}^{\infty}$. If this doubly-infinite series converges for some $z \neq z_{0}$, i.e., each of the series (negative degree and non-negative degree) converges for some $z \neq z_{0}$, then each series has positive radius. Consequently, there exist a real number $r$ and an extended real number $R$ with $0 \leq r < R \leq \infty$ such that the preceding series converges for $r < |z - z_{0}| < R$.


*

*Using the convention that terms with coefficient $0$ are omitted, $1/z^{2}$ is already a Laurent series of $f(z)$ about $z_{0} = 0$.


*The substitution $z^{2} = 1 - (1 - z^{2})$ does not yield a Laurent series about any $z_{0}$, because $1 - z^{2}$ has two distinct complex roots, so is not itself written in powers of $z - z_{0}$ regardless of $z_{0}$.


*Instead, to expand about $z_{0} = 1$, we need to use $z = 1 - (1 - z)$, so
$$
\frac{1}{z^{2}} = \frac{1}{[1 - (1 - z)]^{2}}.
$$
To expand the square, we can use the termwise-differentiation trick
$$
\frac{1}{(1 - u)^{2}}
= \frac{d}{du} \biggl[\frac{1}{1 - u}\biggr]
= \frac{d}{du} \biggl[\sum_{k=0}^{\infty} u^{k}\biggr]
= \sum_{k=1}^{\infty} ku^{k-1}
= \sum_{k=0}^{\infty} (k + 1) u^{k}
$$
with $u = 1 - z$, obtaining
$$
\frac{1}{z^{2}} = \sum_{k=0}^{\infty} (k + 1) (1 - z)^{k}
= \sum_{k=0}^{\infty} (-1)^{k} (k + 1) (z - 1)^{k}.
$$
This series converges where $|z - 1| < 1$, i.e., in the open disk of radius $1$ about $z_{0} = 1$.
