Residues of a complex function measure its behavior at singularities (poles), representing the coefficient of the term \( (z-z_0)^-1 \) in its Laurent series, crucial for the Residue Theorem, which relates contour integrals to the sum of residues (multiplied by \(2\pi i \) ) within the contour, simplifying complex integration and solving real integrals. To find them, use limit formulas for simple poles \( \lim_{z \to z_0} (z - z_0)f(z) \) or higher-order poles (involving derivatives).
Derivation of Residue Formulae
Calculating the residue of a complex function is a fundamental step in applying the Residue Theorem for contour integration. While the definition relies on the Laurent Series, we often use specific formulae depending on the order of the pole.
1. The General Definition
The residue of a function \(f(z)\) at an isolated singularity \(z_0\) is defined as the coefficient \(a_{-1}\) in its Laurent Series expansion:
\[ f(z) = \dots + \frac{a_{-1}}{z - z_0} + a_0 + a_1(z - z_0) + \dots \]
2. Derivation for a Simple Pole
A simple pole means the pole is of order \(m=1\). The Laurent series terminates at the term \((z-z_0)^{-1}\).
Step 1: Write the series: \[ f(z) = \frac{a_{-1}}{z - z_0} + a_0 + a_1(z - z_0) + \dots \]
Step 2: Multiply by \((z - z_0)\) to isolate the coefficient: \[ (z - z_0)f(z) = a_{-1} + a_0(z - z_0) + a_1(z - z_0)^2 + \dots \]
Step 3: Take the limit as \(z \to z_0\). All terms containing \((z-z_0)\) vanish, leaving only \(a_{-1}\).
\[ \text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z) \]
3. Derivation for a Pole of Order \(m\)
If the pole has order \(m > 1\), the series contains negative powers up to \(-m\).
Step 1: Write the series: \[ f(z) = \frac{a_{-m}}{(z - z_0)^m} + \dots + \frac{a_{-1}}{z - z_0} + a_0 + \dots \]
Step 2: Multiply by \((z - z_0)^m\) to clear the denominator: \[ (z - z_0)^m f(z) = a_{-m} + \dots + a_{-1}(z - z_0)^{m-1} + a_0(z - z_0)^m + \dots \]
Step 3: Differentiate \(m-1\) times. This eliminates the constants \(a_{-m} \dots a_{-2}\) and brings down factorials for the \(a_{-1}\) term. \[ \frac{d^{m-1}}{dz^{m-1}}[(z - z_0)^m f(z)] = (m-1)! a_{-1} + (\text{terms with } z-z_0) \]
Step 4: Taking the limit as \(z \to z_0\) isolates the residue.
\[ \text{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z - z_0)^m f(z) \right] \]