uniqueness of Laurent expansion

The Laurent series expansion of a function $f(z)$ in an annulus     is unique.

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

$$f(z)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n{(z-z_0)}^n=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{b}_n{(z-z_0)}^n$$

It follows that

$$f(z){(z-z_0)}^{-\nu -1}=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n{(z-z_0)}^{n-\nu -1}=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{b}_n{(z-z_0)}^{n-\nu -1}$$

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.

${\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}{a}_n{\displaystyle {\oint }_{\gamma }}{(z-z_0)}^{n-\nu -1}𝑑z={\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}{b}_n{\displaystyle {\oint }_{\gamma }}{(z-z_0)}^{n-\nu -1}𝑑z.$(1)

But

${\displaystyle {\oint }_{\gamma }}{(z-z_0)}^{n-\nu -1}𝑑z=\{\begin{array}{cc}2i\pi \hspace{1em}\text{if}n=\nu ,\hfill & \\ 0\hspace{1em}\hspace{1em}\text{if}n\ne \nu ,\hfill & \end{array}$

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

$$2i\pi a_{\nu }=\mathrm{\hspace{0.33em}2}i\pi b_{\nu },$$

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