Laurent's theorem

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Suppose that is analytic within a region . If we consider a closed contour in and a point enclosed within then the Laurent expansion of about the point is given by

where for

Thus for

and for

The Residue Theorem follows from the fact that the coefficient of the term is

Proof of Laurent's theorem

We consider two nested contours and and points contained in the annular region, and the point contained within the inner contour. By Cauchy's theorem and the Cauchy Goursat theorem


Integral over

We begin the proof by rewriting the integrand in the integral by adding and subtracting in the denominator,

The factor in square brackets may be expanded into a geometrical series provided that

Writing up to the -th term, remainder, we have

Thus, we may substutute this expansion into the original integral over

where

We may estimate the remainder


We note that





which yields the estimate


.


Here the constant factor follows from the Maximum Modulus Theorem.

Integral over

We begin the proof by rewriting the integrand in the integral by adding and subtracting in the denominator,

The factor in square brackets may be expanded into a geometrical series provided that

Writing up to the -th term, remainder, we have

Thus, we may substitute this expansion into the original integral over

where

We may estimate the remainder


We note that





which yields the estimate


.


Here the constant factor follows from the Maximum Modulus Theorem.

Combining the and results

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


Finally, the by Cauchy's theorem, the integrals over the contour and are equivalent to the integral over any closed contour which lies in


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.