What is the Laurent series of $f(z)=1/z^2$

Question

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 function does have " a Laurent series," it has "a Laurent series in an annulus" that depends on the annulus. <> If you're expanding in the annulus $z \neq 0$ (i.e., $|z| > 0$), your work is done: The original representation of $f$ is already a Laurent series in the punctured plane. — Andrew D. Hwang, Jan 06 '22 at 16:06
On what annulus are you finding your Laurent series? Because this matters. It seems as though you are trying to find your Laurent series with an annulus either centered around $z=-1$ or $z=0,$ but it is unclear, hence why i ask. — Angel, Jan 06 '22 at 16:31
@AndrewD.Hwang thanks for that. So I just expanded it as a geometric series instead, since the region is $|z|>0$. Note the correction. Still Wolfram gives another result. — Luthier415Hz, Jan 07 '22 at 09:29
The Laurent series of $1/z^2$ in the region $|z|>0$ is $1/z^2$, as Andrew pointed out in his comment — jjagmath, Jan 07 '22 at 09:30
I thought n has to be infinite for that, but here it is only 2. So why does Mathematica develop it into a series similar to mine? — Luthier415Hz, Jan 07 '22 at 09:31
You can write it as $0/z +1/z^2 + 0/z^3 + 0/z^4 + \cdots$ if that makes you feel better. — jjagmath, Jan 07 '22 at 09:33
:) that did it. But why is Mathematica developing it into a Power series? — Luthier415Hz, Jan 07 '22 at 09:34
Search for "Laurent series calculator" at Wolframalpha, then insert the function $1/z^2$ with z as variable, expansion point $0$. — Luthier415Hz, Jan 07 '22 at 09:35

score 1 · Accepted Answer · answered Jan 07 '22 at 13:55

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 $z_{0} = 1$.

is an annulus also $|z|>2$, or is it only squeezed between an infinite domain and a circle? — Luthier415Hz, Jan 07 '22 at 20:19
The set $2 < |z|$ is an annulus ($2 < |z| < \infty$) in the sense of Laurent series. :) — Andrew D. Hwang, Jan 08 '22 at 00:20
I see, so the annulus has to be defined with an upper and lower limit. The region $|z|\geq 2$ is just a punctured infinite region? — Luthier415Hz, Jan 08 '22 at 08:37
In the same way we usually omit terms equal to $0$ in a series, an infinite upper bound on an inequality is often omitted without changing meaning (because $|z| < \infty$ is true for all complex $z$). The term punctured disk usually refers to a disk without its center, i.e., points $z$ with $0 < |z - z_{0}| < R \leq \infty$. In the complex plane, the set $2 \leq |z|$ has no puncture; in the Riemann sphere, $\infty$ is a puncture. We could call this set a closed annulus, but to some that term signifies a compact set bounded by two concentric circles, like $2 \leq |z| \leq 5$. — Andrew D. Hwang, Jan 08 '22 at 13:43

What is the Laurent series of $f(z)=1/z^2$

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