1. ## laurent serie

hi, i have to find out what is the principal part of the laurent serie of f(z)=z^4/(z^2-1) about infinity.
is z^2 true?
regards

2. ## Re: laurent serie

Yes.

Let $\displaystyle f(z) = \frac{z^4}{z^2 - 1}$, let $\displaystyle w = 1/z, z = 1/w$, and get

Let $\displaystyle g(w) = f(1/w) = \frac{(1/w)^4}{(1/w)^2 - 1} = \frac{1}{w^2 - w^4} = \frac{1}{w^2}\frac{1}{1 - w^2}$.

$\displaystyle = \frac{1}{w^2}\left(1 + (-w^2) + (-w^2)^2 + (-w^2)^3 + ... \right) = \frac{1}{w^2}\left(1 - w^2 + w^4 - w^6 +- ... \right)$

$\displaystyle = \frac{1}{w^2} - 1 + w^2 - w^4 +- ... \right)$ which is holomorphic on $\displaystyle 0<\lVert w \rVert < 1$, hence after $\displaystyle w \leftrightarrow 1/z$, for $\displaystyle 1 < \lVert z \rVert < \infty.$

That $\displaystyle \frac{1}{w^2}$ is the principlal part of $\displaystyle g(w) = f(1/w)$ at $\displaystyle w = 0$, and so is the principle part, after $\displaystyle w \leftrightarrow 1/z$, of $\displaystyle f$ at infinity.

Thus $\displaystyle \frac{1}{(1/z)^2} = z^2$ is the principle part of $\displaystyle f$ at infinity.