Residues
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1 The integral over the boundary of a domain with singularities
Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).
Plan: TikZ: a contour/illustration for residues.
Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/220/010_residue_contour_fig.tex
Let \(D \subset \mathbb {C}\) be a simply connected domain whose boundary \(\partial D =: \Gamma\) is a piecewise-smooth curve. Let \(f\) be holomorphic in
\[ D \setminus \left\{ z_1, z_2, \ldots , z_n\right\} \]
We need to find
\[ \oint _{\Gamma }f(z) dz \]
If there were no points \(z_{1}, z_{2}, \ldots\), the integral would be zero by Cauchyβs theorem. Let us inflate an open ball around each point \(z_{i}\) so that it lies entirely in \(D\) and does not intersect the other balls; denote it \(U(z_i)\). Denote its boundary \(\gamma_{i}\) (oriented counterclockwise, i.e. so that the residue point stays on the left as it is traversed). The function \(f\) is holomorphic in the domain \(D \setminus \bigsqcup_{i=1}^n U(z_i) =: \tilde{D}\). It is connected, but not simply connected, i.e. its boundary does not consist of a single curve:
\[ \partial \tilde{D} = \Gamma + (-\gamma _1) + \ldots + (-\gamma _n) \]
The minus sign denotes the clockwise orientation (so that the domain stays on the left as it is traversed). Cauchyβs integral theorem for this domain:
\[ 0 = \oint _{\partial \tilde{D}} f(z) dz = \oint _{\Gamma }f(z) dz - \sum _{i=1}^n \oint _{\gamma _i}f(z) dz \]
Hence,
\[ \oint _{\Gamma }f(z) dz = \sum _{i=1}^n \oint _{\gamma _i}f(z) dz \]
2 The residue at a finite point
Let \(z_0\) be an isolated singular point for \(f\), i.e. \(f\) is holomorphic in \(\dot{U}_{\varepsilon }(z_0)\). The residue of the function \(f\) at the point \(z_0\) is
\[ \mathop{\mathrm{res}}_{z = z_0}f(z) = \frac{1}{2\pi i} \oint _{\gamma _{\rho }}f(z) dz \]
where \(\gamma_{\rho } = \left\{ z \; : \; \left|z-z_0\right| = \rho \right\}\), \(0 < \rho < \varepsilon\), oriented counterclockwise. The integral:
\[ \oint _{\Gamma }f(z) dz = 2\pi i \sum _{i=1}^{n}\mathop{\mathrm{res}}_{z_i}f \]
2.1 Quick computation of the residue
If \(f\) is holomorphic in \(\dot{U}_{\varepsilon }(z_0)\), then it expands there into a Laurent series:
\[ f(z) = \sum _{n=-\infty }^{+\infty }c_n(z - z_0)^{n} \]
This series converges on every compact subset within the punctured neighborhood, so when integrating over the contour \(\gamma_{\rho }\), \(\rho < \varepsilon\), we may interchange integration and summation:
\[ \mathop{\mathrm{res}}_{z = z_0}f(z) = \frac{1}{2\pi i} \oint _{\gamma _{\rho }}f(z) dz = \frac{1}{2\pi i} \oint _{\gamma _{\rho }} \sum _{n=-\infty }^{+\infty }c_n(z - z_0)^{n} dz = \frac{1}{2\pi i} \sum _{n=-\infty }^{+\infty }c_n\oint _{\gamma _{\rho }}(z - z_0)^{n} dz \]
Further, recall that \(\oint_{\gamma_{\rho }}(z - z_0)^{n} dz = 2\pi i\) if \(n = -1\). In every other case this integral equals zero. In total,
\[ \mathop{\mathrm{res}}_{z = z_0}f(z) = \frac{1}{2\pi i} \cdot c_{-1} \cdot 2\pi i = c_{-1}, \]
- i.e. the residue simply equals the coefficient of the Laurent series at the power \(-1\).
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Computing residues
3 The residue at infinity
Let \(f\) be holomorphic in some punctured \(\varepsilon\)-neighborhood of infinity \(\dot{U}_{\varepsilon }(\infty ) = \left\{ z \; : \; \left|z\right| > \frac{1}{\varepsilon }\right\}\). Denote by \(\gamma_{1/\rho }\) the circle of radius \(\frac{1}{\rho }\), \(\frac{1}{\rho } > \frac{1}{\varepsilon }\). The residue at infinity is the integral over this curve (with coefficient \(\frac{1}{2\pi i}\)), where the curve is traversed clockwise:
\[ \mathop{\mathrm{res}}_{\infty }f(z) = \frac{1}{2\pi i} \oint ^{\curvearrowright }_{\gamma _{1/\rho }}f(z) dz \]
This is exactly the orientation needed because, when traversing the boundary, the residue point must stay on the left. If we traversed the circle \(\gamma_{1/\varepsilon }\) counterclockwise, zero would stay on the left instead.
Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).
Plan: TikZ: the orientation of the contour for the residue at infinity β the circle is traversed CLOCKWISE, with the shaded exterior (the side of infinity) staying on the left. Interactive: an orientation toggle with an animated point running along the contour and highlighting which region stays on the left; a slider for the radius of the circle.
Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/220/020_residue_at_infinity_fig.tex
3.1 Quick computation of the residue
3.1.1 Reduction to the residue at zero
We use the substitution \(w = \frac{1}{z} = \frac{1}{\left|z\right|}e^{-i\operatorname {arg}z}\). Note that under this substitution the circle of radius \(\frac{1}{\rho }\) about zero maps to the circle of radius \(\rho\) about zero. Further, note that under this substitution the orientation of traversal reverses (the angle used to be \(\operatorname {arg}z\), and becomes \(-\operatorname {arg}z\)). Also note that if \(f(z)\) was holomorphic in \(\dot{U}_{\varepsilon }(\infty )\), then \(g(w) = f\left(\frac{1}{w}\right)\) will be holomorphic in \(\dot{U}_{\varepsilon }(0) = \left\{ z \; : \; 0 < \left|z\right| < \varepsilon \right\}\), with \(\rho < \varepsilon\). Then
\[ \oint ^{\curvearrowright }_{\gamma _{1/\rho }}f(z) dz = \oint _{\gamma _{\rho }}f\left(\frac{1}{w}\right) d\frac{1}{w} = -\oint _{\gamma _{\rho }}\frac{1}{w^2}f\left(\frac{1}{w}\right) dw \]
Consequently,
\[ \mathop{\mathrm{res}}_{\infty }f(z) = -\mathop{\mathrm{res}}_0\left(\frac{1}{w^2}f\left(\frac{1}{w}\right)\right) \]
3.1.2 Via the Laurent series
If \(f\) is holomorphic in \(\dot{U}_{\varepsilon }(\infty )\), then it expands there into a Laurent series:
\[ f(z) = \sum _{n=-\infty }^{+\infty }c_n\left(z^{-1}\right)^{n} \]
We may interchange integration and summation (for justification see above, for the finite point):
\[ \begin{align} \mathop{\mathrm{res}}_{\infty }f(z) & = \frac{1}{2\pi i} \oint ^{\curvearrowright }_{\gamma _{1/\rho }}f(z) dz = \frac{1}{2\pi i} \oint ^{\curvearrowright }_{\gamma _{1/\rho }} \sum _{n=-\infty }^{+\infty }c_nz^{-n} dz = \\ & = \frac{1}{2\pi i} \sum _{n=-\infty }^{+\infty }c_n \cdot \oint ^{\curvearrowright }_{\gamma _{1/\rho }}z^{-n} dz = \frac{1}{2\pi i} \sum _{n=-\infty }^{+\infty }c_n\cdot \left(-\oint _{\gamma _{1/\rho }}z^{-n} dz\right) = \\ & =-c_{1} \end{align} \]
i.e. the residue simply equals the coefficient of the Laurent series at \(\frac{1}{z} = \left(z^{-1}\right)^1\), taken with a minus sign. The minus sign appears because the circle is traversed clockwise in the definition.
3.1.3 General table
: Computing residues at infinity
Be careful when using the substitution \(w = \frac{1}{z}\): \[ \boxed {\mathop{\mathrm{res}}_{z = \infty } f(z) \neq \mathop{\mathrm{res}}_{w = 0}f\left(\frac{1}{w}\right)} \] In fact: \[ \boxed {\mathop{\mathrm{res}}_{z = \infty } f(z) = -\mathop{\mathrm{res}}_{w = 0}\frac{1}{w^2}f\left(\frac{1}{w}\right)} \]
4 Sum of residues
Let \(f\) be holomorphic in \(\mathbb {C} \setminus \left\{ a_1, \ldots , a_n\right\}\), i.e. in the entire complex plane except for finitely many points. Then, by Cauchyβs theorem, the sum of the residues at all points (including \(\infty\)) equals zero:
\[ \sum _{k=1}^n \mathop{\mathrm{res}}_{a_k} f + \mathop{\mathrm{res}}_{\infty }f = 0 \]