laurent serie

**redrose5** · September 24th 2012, 10:41 AM

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

**johnsomeone** · September 24th 2012, 02:30 PM

Yes.

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

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

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

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

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

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

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