Suppose that $ f(z) $ is analytic within a region $ {\mathcal {R}} $. If we consider a closed contour $ C $ in $ {\mathcal {R}} $ and a point $ a $ enclosed within $ C $ then the Laurent expansion of $ f(z) $ about the point $ a $ is
given by
$ f(z)=\sum _{n=-\infty }^{\infty }c_{n}(z-a)^{n} $
where for $ n=0,\pm 1,\pm 2,\pm 3,... $
$ c_{n}={\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{(w-a)^{n+1}}}\;dw. $
Thus for $ n=0,1,2,... $
$ c_{n}={\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{(w-a)^{n+1}}}\;dw $
and for $ n=1,2,3,... $
$ c_{-n}={\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{(w-a)^{-n+1}}}\;dw. $
The Residue Theorem follows from the fact that the coefficient of the $ n=-1 $ term is
$ c_{-1}={\frac {1}{2\pi i}}\oint _{C}f(w)\;dw. $
Proof of Laurent's theorem
We consider two nested contours $ C_{1} $ and $ C_{2} $ and points $ z $ contained in the annular region, and the point
$ z=a $ contained within the inner contour. By Cauchy's theorem and the Cauchy Goursat theorem
$ f(z)={\frac {1}{2\pi i}}\oint _{C_{2}}{\frac {f(w)}{w-z}}\;dw-{\frac {1}{2\pi i}}\oint _{C_{1}}{\frac {f(w)}{w-z}}\;dw. $
Integral over $ C_{2} $
We begin the proof by rewriting the integrand in the $ C_{2} $ integral by adding and subtracting $ a $ in the denominator,
$ {\frac {f(w)}{w-z}}={\frac {f(w)}{(w-a-(z-a))}}={\frac {f(w)}{(w-a)}}\left[{\frac {1}{1-\left({\frac {z-a}{w-a}}\right)}}\right]. $
The factor in square brackets $ \left[\right] $ may be expanded into a geometrical series provided that
$ \left|{\frac {z-a}{w-a}}\right|<1. $
Writing up to the $ N $-th term, remainder, we have
$ {\frac {f(w)}{w-z}}=\sum _{n=0}^{N-1}{\frac {f(w)(z-a)^{n}}{(w-a)^{n+1}}}+{\frac {f(w)}{(w-z)}}\left({\frac {z-a}{w-a}}\right)^{N} $
Thus, we may substutute this expansion into the original integral over $ C_{2} $
$ {\frac {1}{2\pi i}}\oint _{C_{2}}{\frac {f(w)}{w-z}}\;dw=\sum _{n=0}^{N-1}\left[{\frac {1}{2\pi i}}\oint _{C_{2}}{\frac {f(w)}{(w-a)^{n+1}}}\;dw\right]\;(z-a)^{n}+R_{N}^{+} $
where
$ R_{N}^{+}={\frac {1}{2\pi i}}\int _{C_{2}}{\frac {f(w)}{(w-z)}}\left({\frac {z-a}{w-a}}\right)^{N}\;dw. $
We may estimate the remainder $ R_{n} $
$ \left|R_{N}^{+}\right|=\left|{\frac {1}{2\pi i}}\oint _{C_{2}}{\frac {f(w)}{(w-z)}}\left({\frac {z-a}{w-a}}\right)^{N}\;dw\right|\leq {\frac {1}{2\pi }}\oint _{C_{2}}\left|{\frac {f(w)}{(w-z)}}\left({\frac {z-a}{w-a}}\right)^{N}\right|\;dw $
We note that
$ |w-z|=|w-a-(z-a)|\geq |w-a|-|z-a| $
$ |z-a|<|w-a| $
$ \gamma ={\frac {z-a}{w-a}}<1 $
which yields the estimate
$ |R_{n}^{+}|<{\frac {1}{2\pi }}M(2\pi |w-a|){\frac {\gamma ^{N}}{|w-a|-|z-a|}}={\frac {M\gamma ^{N}|w-a|}{|w-a|-|z-a|}}\rightarrow 0\qquad {\mbox{as}}\qquad N\rightarrow \infty $ .
Here the constant factor $ M $ follows from the Maximum Modulus Theorem.
Integral over $ C_{1} $
We begin the proof by rewriting the integrand in the $ C_{2} $ integral by adding and subtracting $ a $ in the denominator,
$ {\frac {f(w)}{z-w}}={\frac {f(w)}{(z-a-(w-a))}}={\frac {f(w)}{(w-a)}}\left[{\frac {1}{1-\left({\frac {w-a}{z-a}}\right)}}\right]. $
The factor in square brackets $ \left[\right] $ may be expanded into a geometrical series provided that
$ \left|{\frac {w-a}{z-a}}\right|<1. $
Writing up to the $ N $-th term, remainder, we have
$ -{\frac {f(w)}{w-z}}=\sum _{n=0}^{N-1}{\frac {f(w)(z-a)^{-n}}{(w-a)^{-n+1}}}+{\frac {f(w)}{(z-w)}}\left({\frac {w-a}{z-a}}\right)^{N} $
Thus, we may substitute this expansion into the original integral over $ C_{2} $
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): - \frac{1}{2 \pi i } \oint_{C_1} \frac{f(w)}{w - z } \; dw = \sum_{n=1}^{N-1} \left[\frac{1}{2 \pi i } \int_{C_1} \frac{f(w )}{(w -a)^{-n+1} } \; dw \right] \; (z - a)^{-n} + R_N^{-}
where
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): R_N^{-} = \frac{1}{2 \pi i } \int_{C_1} \frac{ f(w) }{ ( w - z )} \left( \frac{ w- a }{ z - a} \right)^{N} \; dw .
We may estimate the remainder Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): R_n
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \left| R_N^{-} \right| = \left| \frac{1}{2 \pi i } \oint_{C_1} \frac{ f(w) }{ ( w - z )} \left( \frac{ w - a }{ z- a} \right)^{N} \; dw \right| \le \frac{1}{2 \pi} \oint_{C_1} \left| \frac{ f(w) }{ ( z - w)} \left( \frac{ w - a }{ z - a} \right)^{N} \right| \; dw
We note that
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): | z - w | = | z - a - (w - a) | \ge | z - a | - | w - a |
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): |w - a | < | z - a |
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \gamma = \frac{w -a }{ z -a } < 1
which yields the estimate
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): |R_n^{-}| < \frac{1}{2 \pi} M (2 \pi | w - a| ) \frac{\gamma^N}{ | z - a | - |w - a|} =\frac{M\gamma^N |w-a|}{|z-a| - |w -a|} \rightarrow 0 \qquad \mbox{as} \qquad N \rightarrow \infty
.
Here the constant factor Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): M
follows from the Maximum Modulus Theorem.
Combining the $ C_{1} $ and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): C_2
results
Because the respective remainders for the series representations vanish, we may combine these two results to yield
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): f(z) = \frac{1}{2 \pi i } \left\{ \sum_{n=0}^{\infty} \left[ \oint_{C_2} \frac{f(w)}{(w - a)^{n+1} } \; dw \right] (z - a)^{n} + \sum_{n=1}^{\infty} \left[ \oint_{C_1} \frac{f(w)}{(w - a)^{-n+1} } \; dw \right] (z - a)^{-n} \right\}.
Finally, the by Cauchy's theorem, the integrals over the contour Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): C_1
and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): C_2
are equivalent to the integral over any closed contour Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): C
which lies in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\mathcal R}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): f(z) = \frac{1}{2 \pi i } \left\{ \sum_{n=0}^{\infty} \left[ \oint_{C} \frac{f(w)}{(w - a)^{n+1} } \; dw \right] (z - a)^{n} + \sum_{n=1}^{\infty} \left[ \oint_{C} \frac{f(w)}{(w - a)^{-n+1} }\; dw \right] (z - a)^{-n} \right\} = \sum_{n=-\infty}^{\infty} \left[ \frac{1}{2 \pi i } \oint_{C} \frac{f(w)}{(w - a)^{n+1} } \; dw \right] (z - a)^{n}
proving the Laurent's theorem.
It must be mentioned that, like the Taylor's expansion, the Laurent expansion of a function is unique where the function is analytic.