uniqueness of Laurent expansion

Proof.  Suppose that $f(z)$ has in the annulus two Laurent expansions:

 $f(z)\;=\;\sum_{n=-\infty}^{\infty}\!a_{n}(z\!-\!z_{0})^{n}\;=\;\sum_{n=-\infty% }^{\infty}\!b_{n}(z\!-\!z_{0})^{n}$

It follows that

 $f(z)(z\!-\!z_{0})^{-\nu-1}\;=\;\sum_{n=-\infty}^{\infty}\!a_{n}(z\!-\!z_{0})^{% n-\nu-1}\;=\;\sum_{n=-\infty}^{\infty}\!b_{n}(z\!-\!z_{0})^{n-\nu-1}$

where $\nu$ is an integer.  Let now $\gamma$ be an arbitrary closed contour in the annulus, going once around $z_{0}$.  Since $\gamma$ is a compact set of points, those two Laurent series converge uniformly (http://planetmath.org/UniformConvergence) on it and therefore they can be integrated termwise (http://planetmath.org/SumFunctionOfSeries) along $\gamma$, i.e.

 $\displaystyle\sum_{n=-\infty}^{\infty}\!a_{n}\oint_{\gamma}(z\!-\!z_{0})^{n-% \nu-1}\,dz\;=\;\sum_{n=-\infty}^{\infty}\!b_{n}\oint_{\gamma}(z\!-\!z_{0})^{n-% \nu-1}\,dz.$ (1)

But

 $\displaystyle\oint_{\gamma}(z\!-\!z_{0})^{n-\nu-1}\,dz\;=\;\begin{cases}2i\pi% \quad\mbox{if}\;\;n\;=\;\nu,\\ 0\qquad\mbox{if}\;\;n\;\neq\;\nu,\\ \end{cases}$

when integrated anticlockwise (see calculation of contour integral).  Thus (1) reads

 $2i\pi a_{\nu}\;=\;2i\pi b_{\nu},$

i.e.  $a_{\nu}=b_{\nu}$,  for any integer $\nu$, whence both expansions are identical.

Title uniqueness of Laurent expansion UniquenessOfLaurentExpansion 2013-03-22 19:14:12 2013-03-22 19:14:12 pahio (2872) pahio (2872) 11 pahio (2872) Theorem msc 30B10 CoefficientsOfLaurentSeries UniquenessOfFourierExpansion UniquenessOfDigitalRepresentation