A Laurent series
is a representation of a complex function \(f(z)\) as a power series that include both positive and negative powers of \(z-z_0\). It is a generalization of the Taylor series used to represent functions in regions where singularities exist, specifically within an annular (ring-shaped) domain.
If a function \(f(z)\) is analytic in an annular domain \(D\) defined by \(R_1 < |z - z_0| < R_2\), then for any point \(z\) in \(D\), \(f(z)\) can be expanded as:
\[ f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n \]
Where the coefficients \(a_n\) are defined by the contour integral:
\[ a_n = \frac{1}{2\pi i} \oint_C \frac{f(z')}{(z' - z_0)^{n+1}} \, dz' \]
Here, \(C\) is any simple closed path lying entirely within the annulus and encircling \(z_0\).
Detailed Proof
Step 1: Application of Cauchy's Integral Formula
Let \(C_1\) and \(C_2\) be two concentric circles centered at \(z_0\) with radii \(r_1\) and \(r_2\) respectively, such that \(R_1 < r_2 < |z - z_0| < r_1 < R_2\). By applying Cauchy's Integral Formula to the multiply connected region bounded by these circles, we can write \(f(z)\) as the sum of integrals over the outer and inner boundaries:
\[ f(z) = \frac{1}{2\pi i} \left[ \oint_{C_1} \frac{f(z')}{z' - z} \, dz' - \oint_{C_2} \frac{f(z')}{z' - z} \, dz' \right] \]
Step 2: The Analytic Part (Positive Powers)
Consider the first integral over \(C_1\). For points \(z'\) on \(C_1\), we have \(|z' - z_0| > |z - z_0|\). We can rewrite the kernel as:
\[ \frac{1}{z' - z} = \frac{1}{(z' - z_0) - (z - z_0)} = \frac{1}{z' - z_0} \left( \frac{1}{1 - \frac{z - z_0}{z' - z_0}} \right) \]
Using the geometric series expansion \(\frac{1}{1-u} = \sum u^n\), which converges because \(| \frac{z - z_0}{z' - z_0} | < 1\):
\[ \frac{1}{z' - z} = \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(z' - z_0)^{n+1}} \]
Integrating term by term yields the positive power terms: \[ \sum_{n=0}^{\infty} a_n (z - z_0)^n \]
Step 3: The Principal Part (Negative Powers)
Consider the second integral over \(C_2\). For points \(z'\) on \(C_2\), we have \(|z' - z_0| < |z - z_0|\). We rewrite the kernel differently:
\[ -\frac{1}{z' - z} = \frac{1}{z - z'} = \frac{1}{(z - z_0) - (z' - z_0)} = \frac{1}{z - z_0} \left( \frac{1}{1 - \frac{z' - z_0}{z - z_0}} \right) \]
Using the geometric series expansion again:
\[ \frac{1}{z - z'} = \sum_{k=0}^{\infty} \frac{(z' - z_0)^k}{(z - z_0)^{k+1}} \]
Integrating term by term and substituting \(n = -(k+1)\) yields the negative power terms: \[ \sum_{n=-\infty}^{-1} a_n (z - z_0)^n \]
Conclusion
Adding both parts together gives the full Laurent Series expansion:
\[ f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n \]