Laurent's theorem

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Suppose that ${\displaystyle f(z)}$ is analytic within a region ${\displaystyle {\mathcal {R}}}$. If we consider a closed contour ${\displaystyle C}$ in ${\displaystyle {\mathcal {R}}}$ and a point ${\displaystyle a}$ enclosed within ${\displaystyle C}$ then the Laurent expansion of ${\displaystyle f(z)}$ about the point ${\displaystyle a}$ is given by

${\displaystyle f(z)=\sum _{n=-\infty }^{\infty }c_{n}(z-a)^{n}}$

where for ${\displaystyle n=0,\pm 1,\pm 2,\pm 3,...}$

${\displaystyle c_{n}={\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{(w-a)^{n+1}}}\;dw.}$

Thus for ${\displaystyle n=0,1,2,...}$

${\displaystyle c_{n}={\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{(w-a)^{n+1}}}\;dw}$

and for ${\displaystyle n=1,2,3,...}$

${\displaystyle c_{-n}={\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{(w-a)^{-n+1}}}\;dw.}$

The Residue Theorem follows from the fact that the coefficient of the ${\displaystyle n=-1}$ term is

${\displaystyle c_{-1}={\frac {1}{2\pi i}}\oint _{C}f(w)\;dw.}$

Proof of Laurent's theorem

We consider two nested contours ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ and points ${\displaystyle z}$ contained in the annular region, and the point ${\displaystyle z=a}$ contained within the inner contour. By Cauchy's theorem and the Cauchy Goursat theorem

${\displaystyle f(z)={\frac {1}{2\pi i}}\oint _{C_{2}}{\frac {f(w)}{w-z}}\;dw-{\frac {1}{2\pi i}}\oint _{C_{1}}{\frac {f(w)}{w-z}}\;dw.}$

Integral over ${\displaystyle C_{2}}$

We begin the proof by rewriting the integrand in the ${\displaystyle C_{2}}$ integral by adding and subtracting ${\displaystyle a}$ in the denominator,

${\displaystyle {\frac {f(w)}{w-z}}={\frac {f(w)}{(w-a-(z-a))}}={\frac {f(w)}{(w-a)}}\left[{\frac {1}{1-\left({\frac {z-a}{w-a}}\right)}}\right].}$

The factor in square brackets ${\displaystyle \left[\right]}$ may be expanded into a geometrical series provided that

${\displaystyle \left|{\frac {z-a}{w-a}}\right|<1.}$

Writing up to the ${\displaystyle N}$-th term, remainder, we have

${\displaystyle {\frac {f(w)}{w-z}}=\sum _{n=0}^{N-1}{\frac {f(w)(z-a)^{n}}{(w-a)^{n+1}}}+{\frac {f(w)}{(w-z)}}\left({\frac {z-a}{w-a}}\right)^{N}}$

Thus, we may substutute this expansion into the original integral over ${\displaystyle C_{2}}$

${\displaystyle {\frac {1}{2\pi i}}\oint _{C_{2}}{\frac {f(w)}{w-z}}\;dw=\sum _{n=0}^{N-1}\left[{\frac {1}{2\pi i}}\oint _{C_{2}}{\frac {f(w)}{(w-a)^{n+1}}}\;dw\right]\;(z-a)^{n}+R_{N}^{+}}$

where

${\displaystyle R_{N}^{+}={\frac {1}{2\pi i}}\int _{C_{2}}{\frac {f(w)}{(w-z)}}\left({\frac {z-a}{w-a}}\right)^{N}\;dw.}$

We may estimate the remainder ${\displaystyle R_{n}}$

${\displaystyle \left|R_{N}^{+}\right|=\left|{\frac {1}{2\pi i}}\oint _{C_{2}}{\frac {f(w)}{(w-z)}}\left({\frac {z-a}{w-a}}\right)^{N}\;dw\right|\leq {\frac {1}{2\pi }}\oint _{C_{2}}\left|{\frac {f(w)}{(w-z)}}\left({\frac {z-a}{w-a}}\right)^{N}\right|\;dw}$

We note that

${\displaystyle |w-z|=|w-a-(z-a)|\geq |w-a|-|z-a|}$

${\displaystyle |z-a|<|w-a|}$

${\displaystyle \gamma ={\frac {z-a}{w-a}}<1}$

which yields the estimate

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

.

Here the constant factor ${\displaystyle M}$ follows from the Maximum Modulus Theorem.

Integral over ${\displaystyle C_{1}}$

We begin the proof by rewriting the integrand in the ${\displaystyle C_{2}}$ integral by adding and subtracting ${\displaystyle a}$ in the denominator,

${\displaystyle {\frac {f(w)}{z-w}}={\frac {f(w)}{(z-a-(w-a))}}={\frac {f(w)}{(w-a)}}\left[{\frac {1}{1-\left({\frac {w-a}{z-a}}\right)}}\right].}$

The factor in square brackets ${\displaystyle \left[\right]}$ may be expanded into a geometrical series provided that

${\displaystyle \left|{\frac {w-a}{z-a}}\right|<1.}$

Writing up to the ${\displaystyle N}$-th term, remainder, we have

${\displaystyle -{\frac {f(w)}{w-z}}=\sum _{n=0}^{N-1}{\frac {f(w)(z-a)^{-n}}{(w-a)^{-n+1}}}+{\frac {f(w)}{(z-w)}}\left({\frac {w-a}{z-a}}\right)^{N}}$

Thus, we may substitute this expansion into the original integral over ${\displaystyle C_{2}}$

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

where

${\displaystyle R_{N}^{-}={\frac {1}{2\pi i}}\int _{C_{1}}{\frac {f(w)}{(w-z)}}\left({\frac {w-a}{z-a}}\right)^{N}\;dw.}$

We may estimate the remainder ${\displaystyle R_{n}}$

${\displaystyle \left|R_{N}^{-}\right|=\left|{\frac {1}{2\pi i}}\oint _{C_{1}}{\frac {f(w)}{(w-z)}}\left({\frac {w-a}{z-a}}\right)^{N}\;dw\right|\leq {\frac {1}{2\pi }}\oint _{C_{1}}\left|{\frac {f(w)}{(z-w)}}\left({\frac {w-a}{z-a}}\right)^{N}\right|\;dw}$

We note that

${\displaystyle |z-w|=|z-a-(w-a)|\geq |z-a|-|w-a|}$

${\displaystyle |w-a|<|z-a|}$

${\displaystyle \gamma ={\frac {w-a}{z-a}}<1}$

which yields the estimate

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

.

Here the constant factor ${\displaystyle M}$ follows from the Maximum Modulus Theorem.

Combining the ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ results

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

${\displaystyle f(z)={\frac {1}{2\pi i}}\left\{\sum _{n=0}^{\infty }\left[\oint _{C_{2}}{\frac {f(w)}{(w-a)^{n+1}}}\;dw\right](z-a)^{n}+\sum _{n=1}^{\infty }\left[\oint _{C_{1}}{\frac {f(w)}{(w-a)^{-n+1}}}\;dw\right](z-a)^{-n}\right\}.}$

Finally, the by Cauchy's theorem, the integrals over the contour ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ are equivalent to the integral over any closed contour ${\displaystyle C}$ which lies in ${\displaystyle {\mathcal {R}}}$

${\displaystyle f(z)={\frac {1}{2\pi i}}\left\{\sum _{n=0}^{\infty }\left[\oint _{C}{\frac {f(w)}{(w-a)^{n+1}}}\;dw\right](z-a)^{n}+\sum _{n=1}^{\infty }\left[\oint _{C}{\frac {f(w)}{(w-a)^{-n+1}}}\;dw\right](z-a)^{-n}\right\}=\sum _{n=-\infty }^{\infty }\left[{\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{(w-a)^{n+1}}}\;dw\right](z-a)^{n}}$

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.