# Laurent's theorem

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 $C_{2}$ $-{\frac {1}{2\pi i}}\oint _{C_{1}}{\frac {f(w)}{w-z}}\;dw=\sum _{n=1}^{N-1}\left[{\frac {1}{2\pi i}}\int _{C_{1}}{\frac {f(w)}{(w-a)^{-n+1}}}\;dw\right]\;(z-a)^{-n}+R_{N}^{-}$ where

$R_{N}^{-}={\frac {1}{2\pi i}}\int _{C_{1}}{\frac {f(w)}{(w-z)}}\left({\frac {w-a}{z-a}}\right)^{N}\;dw.$ We may estimate the remainder $R_{n}$ $\left|R_{N}^{-}\right|=\left|{\frac {1}{2\pi i}}\oint _{C_{1}}{\frac {f(w)}{(w-z)}}\left({\frac {w-a}{z-a}}\right)^{N}\;dw\right|\leq {\frac {1}{2\pi }}\oint _{C_{1}}\left|{\frac {f(w)}{(z-w)}}\left({\frac {w-a}{z-a}}\right)^{N}\right|\;dw$ We note that

$|z-w|=|z-a-(w-a)|\geq |z-a|-|w-a|$ $|w-a|<|z-a|$ $\gamma ={\frac {w-a}{z-a}}<1$ which yields the estimate

$|R_{n}^{-}|<{\frac {1}{2\pi }}M(2\pi |w-a|){\frac {\gamma ^{N}}{|z-a|-|w-a|}}={\frac {M\gamma ^{N}|w-a|}{|z-a|-|w-a|}}\rightarrow 0\qquad {\mbox{as}}\qquad N\rightarrow \infty$ .

Here the constant factor $M$ follows from the Maximum Modulus Theorem.

## Combining the $C_{1}$ and $C_{2}$ results

Because the respective remainders for the series representations vanish, we may combine these two results to yield

$f(z)={\frac {1}{2\pi i}}\left\{\sum _{n=0}^{\infty }\left[\oint _{C_{2}}{\frac {f(w)}{(w-a)^{n+1}}}\;dw\right](z-a)^{n}+\sum _{n=1}^{\infty }\left[\oint _{C_{1}}{\frac {f(w)}{(w-a)^{-n+1}}}\;dw\right](z-a)^{-n}\right\}.$ Finally, the by Cauchy's theorem, the integrals over the contour $C_{1}$ and $C_{2}$ are equivalent to the integral over any closed contour $C$ which lies in ${\mathcal {R}}$ $f(z)={\frac {1}{2\pi i}}\left\{\sum _{n=0}^{\infty }\left[\oint _{C}{\frac {f(w)}{(w-a)^{n+1}}}\;dw\right](z-a)^{n}+\sum _{n=1}^{\infty }\left[\oint _{C}{\frac {f(w)}{(w-a)^{-n+1}}}\;dw\right](z-a)^{-n}\right\}=\sum _{n=-\infty }^{\infty }\left[{\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{(w-a)^{n+1}}}\;dw\right](z-a)^{n}$ 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.