Find Laurent series of $\frac{x^2}{(1+x^2)^2}$

Question

I would like to find the Laurent series of the function $$ f(x)=\frac{x^2}{(1+x^2)^2}. $$

I already know that there are two poles of order $2$, namely $x=\pm i$. And I also know that, by partial fraction decomposition, $$ \frac{x^2}{(1+x^2)^2}=\frac{1}{x^2+1}-\frac{1}{(1+x^2)^2}. $$

Answer 1

Here are two variations to calculate the residue of $f$ at $x=i$. We already know the pole of $f$ at $x=i$ is of order $2$. A common way is to use the

Limit formula for higher order poles:

We obtain \begin{align*} \color{blue}{\mathrm{res}_{x=i}f(x)}&=\mathrm{res}_{x=i}\frac{x^2}{\left(1+x^2\right)}\\ &=\lim_{x\to i}\frac{d}{dx}\left((x-i)^2f(x)\right)\\ &=\lim_{x\to i}\frac{d}{dx}\left(\frac{x^2}{(x+i)^2}\right)\\ &=\lim_{x\to i}\frac{2ix}{(x+i)^3}\\ &=\frac{2i^2}{(2i)^3}\\ &\,\,\color{blue}{=-\frac{1}{4}i}\tag{1} \end{align*}

It is somewhat more cumbersome, but also not so difficult, to calculate the remainder of $f$ at $x=i$ by consequent expansion at $x=i$.

Manual expansion in terms of $(x-i)$:

We start with \begin{align*} f(x)&=\frac{x^2}{\left(1+x^2\right)^2}=\frac{x^2}{(x+i)^2(x-i)^2}\\ &=\frac{(x-i+i)^2}{(x-i+2i)^2(x-i)^2}\\ &=\frac{(x-i)^2+2i(x-i)-1}{(2i)^2\left(1+\frac{x-i}{2i}\right)^2(x-i)^2}\\ &=-\frac{1}{4}\frac{1}{\left(1+\frac{x-i}{2i}\right)^2} +\frac{1}{2i(x-i)}\,\frac{1}{\left(1+\frac{x-i}{2i}\right)^2}\\ &\qquad+\frac{1}{4(x-i)^2}\,\frac{1}{\left(1+\frac{x-i}{2i}\right)^2}\tag{2} \end{align*} The expansion of the square of the geometric series in (2) gives \begin{align*} \frac{1}{\left(1+\frac{x-i}{2i}\right)^2} &=\left(1-\frac{x-i}{2i}+\cdots\right)\left(1-\frac{x-i}{2i}+\cdots\right)\\ &=1-\frac{x-i}{i}+\mathrm{higher\ order\ terms}\tag{3} \end{align*}

We derive from (2) and (3) \begin{align*} \color{blue}{\mathrm{res}_{x=i}f(x)}=0+\frac{1}{2i}+\frac{1}{4}\left(-\frac{1}{i}\right)\,\,\color{blue}{=-\frac{1}{4}i} \end{align*} in accordance with (1).