Residues #
If \( z_0 \) is an isolated singularity, and \( f(z) \) can be written as a series
\[f(z) = \sum_{n=-\infty} ^\infty a_n (z - z_0)^n\]which is valid in a punctured neighborhood of \( z_0 \), then
\[\oint_C f(z) \dd z = 2 \pi i a_{-1}\]where \( C \) is a simple closed curve in the neighborhood enclosing \( z_0 \). This is because
\[\oint_C (z - z_0)^n \dd z = \begin{cases} 0 & \quad n \neq -1 \\ 2 \pi i & \quad n = -1 \end{cases}\]We call \( a_{-1} \) the residue at \( z_0 \).
Now we can write down the residue theorem:
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Cauchy’s Residue Theorem
Let \( C \) be a simple closed contour within and on which a function \( f(z) \) is analytic except for a finite number of isolated singularities \( z_1, \ldots, z_m \). Then
\[\oint_C f(z) \dd z = 2 \pi i \sum_{n=1} ^m \text{Res}[z_n]\]where \( \text{Res}[z_n] \) is the residue of \( f(z) \) at \( z_n \).
To make use of this very powerful result to perform contour integrals, the name of the game is completing a contour that properly encloses all singularities while remaining analytic. This is the approach we will take to find residues.
If a function \( f(z) \) can be expanded in a Laurent series about \( z_0 \)
\[f(z) = \sum_{n=-\infty} ^\infty a_n (z - z_0)^n\]then the residue at \( z_0 \) is just the coefficient \( a_{-1} \). But suppose we don’t have a Laurent series expansion for \( f(z) \) at \( z_0 \)? Constucting the series isn’t the only way to find the residue. If we consider \( (z - z_0)f(z) \):
\[(z - z_0) f(z) = \ldots + a_{-1} + a_0(z - z_0) + a_1(z - z_0)^2 + \ldots\]If \( z_0 \) is a simple pole (of order 1), then we can take the limit
\[\lim_{z \rightarrow z_0} (z - z_0) f(z) = \lim_{z \rightarrow z_0} a_{-1} + a_0(z - z_0) + a_1(z - z_0)^2 + \ldots = a_{-1}\]What if \( z_0 \) is a double pole?
\[f(z) = a_{-2} (z - z_0)^{-2} + a_{-1} (z - z_0) + a_0 + a_1(z - z_0) + \ldots\]In that case we can multiply by \( (z - z_0)^2 \) and differentiate to pick out the residue
\[\lim_{z \rightarrow z_0} \left( \pdv{}{z} \left( (z - z_0)^2 f(z) \right) \right) = a_{-1}\]So, the general recipe is: If \( f(z) \) has an \( m^{th} \) order pole at \( z_0 \), then
\[\text{Res}[z_0] = \lim_{z \rightarrow z_0} \left[ \frac{1}{(m - 1)!} \pdv{^m}{z^m} \left((z - z_0)^m f(z) \right) \right]\]In the case of an essential singularity with infinite order, we have no choice but to find the Laurent series expansion and pick out the \( a_{-1} \) term directly. No differentiation tricks will help us there.
\[\]Residue of rational functions
This gives us a useful formula to find the residue at a simple pole of \( z_0 \) of a rational function
\[f(z) = P(z) / Q(z)\]where \( P(z) \) is analytic and \( Q(z) \) has a simple pole at \( z_0 \):
\[\text{Res}(z_0) = \frac{P(z_0)}{Q'(z_0)}\]