# Laurent expansion of rational function

We first have the partial fraction decomposition

 $\displaystyle f(z)\;=\;\frac{1}{z\!-\!i}+\frac{1}{z\!+\!i}$ (1)

whence the principal part of the Laurent expansion contains $\displaystyle\frac{1}{z\!-\!i}$.  Taking into account the poles  $z=\pm i$  of $f$ we see that there are two possible annuli for the Laurent expansion:

a)  The annulus  $\displaystyle 0\,<\,|z\!-\!i|\,<\,2$.  We can write

 $\frac{1}{z\!+\!i}\;=\;\frac{1}{2i\!+\!(z\!-\!i)}\;=\;\frac{1}{2i}\cdot\frac{1}% {1\!-\!\left(-\frac{z-i}{2i}\right)}\;=\;\frac{1}{2i}-\frac{z\!-\!i}{(2i)^{2}}% +\frac{(z\!-\!i)^{2}}{(2i)^{3}}-+\ldots$

Thus

 $\frac{2z}{1\!+\!z^{2}}\;=\;\frac{1}{z\!-\!i}-\sum_{n=0}^{\infty}\left(\frac{i}% {2}\right)^{n+1}(z\!-\!i)^{n}\quad\qquad(0\,<\,|z\!-\!i|\,<\,2).$

b)  The annulus  $\displaystyle 2\,<\,|z\!-\!i|\,<\,\infty$.  Now we write

 $\frac{1}{z\!+\!i}\;=\;\frac{1}{(z\!-\!i)\!+\!2i}\;=\;\frac{1}{z\!-\!i}\cdot% \frac{1}{1\!-\!\left(-\frac{2i}{z-i}\right)}\;=\;\frac{1}{z\!-\!i}-\frac{2i}{(% z\!-\!i)^{2}}+\frac{(2i)^{2}}{(z\!-\!i)^{3}}-+\ldots$

Accordingly

 $\frac{2z}{1\!+\!z^{2}}\;=\;\frac{2}{z\!-\!i}+\sum_{n=2}^{\infty}\frac{(-2i)^{n% -1}}{(z\!-\!i)^{n}}\quad\qquad(2\,<\,|z\!-\!i|\,<\,\infty).$

This latter Laurent expansion consists of negative powers only, but  $z=i$  isn’t an essential singularity of $f$, though.

Title Laurent expansion of rational function LaurentExpansionOfRationalFunction 2013-03-11 19:16:06 2013-03-11 19:16:06 pahio (2872) (0) 5 pahio (0) Example msc 30B10