Residue theorem Study Guide

📖 Core Concepts Residue – the coefficient \(a{-1}\) of \((z-z0)^{-1}\) in the Laurent series of \(f\) about an isolated singularity \(z0\). Simple pole – a singularity where the Laurent series has only the \((z-z0)^{-1}\) term (no higher negative powers). Higher‑order pole – a pole of order \(m\ge 2\); the Laurent series contains terms up to \((z-z0)^{-m}\). Removable singularity – a point where \(f\) can be re‑defined to become holomorphic; its residue is zero. Winding number \(\operatorname{Ind}{\gamma}(z0)\) – counts how many times a closed curve \(\gamma\) encircles \(z0\) counter‑clockwise; determines the contribution of each residue. Residue theorem (simple curve) – for a positively oriented simple closed curve \(\gamma\), \[ \oint{\gamma} f(z)\,dz = 2\pi i\sum{\text{inside }\gamma}\operatorname{Res}(f,zk). \] Residue at infinity – defined by \[ \operatorname{Res}(f,\infty)= -\operatorname{Res}\!\Bigl(\frac{1}{z^{2}}f\!\bigl(\tfrac{1}{z}\bigr),0\Bigr), \] and for meromorphic \(f\) on \(\mathbb C\): \(\displaystyle \sum{k=1}^{n}\operatorname{Res}(f,zk)+\operatorname{Res}(f,\infty)=0\). --- 📌 Must Remember Residue theorem conditions – open set simply connected, \(f\) holomorphic except at isolated points, \(\gamma\) rectifiable closed curve. Simple‑pole residue formula \[ \operatorname{Res}(f,z0)=\lim{z\to z0}(z-z0)f(z)=\frac{g(z0)}{h'(z0)}\quad\text{if }f=\frac{g}{h},\;h(z0)=0,\;h'(z0)\neq0. \] Higher‑order pole formula \[ \operatorname{Res}(f,z0)=\frac{1}{(m-1)!}\,\lim{z\to z0}\frac{d^{\,m-1}}{dz^{\,m-1}}\bigl[(z-z0)^{m}f(z)\bigr]. \] Residue at a removable singularity = 0. Arc‑vanishing principle – on a large semicircle in the upper (or lower) half‑plane, \(\displaystyle \int{\text{arc}} f(z)\,dz \to 0\) when Jordan’s lemma applies (e.g., \(f(z)=e^{i a z}g(z)\) with \(a>0\)). Sum of all residues (including \(\infty\)) = 0 for meromorphic functions on \(\mathbb C\). --- 🔄 Key Processes Evaluating a real integral with residues Extend the real integrand to a meromorphic \(f(z)\). Choose a closed contour (real line segment + large semicircle). Verify the arc integral tends to 0 (Jordan’s lemma or direct estimate). Identify singularities inside the contour. Compute each residue (use simple‑pole, limit, or derivative formula). Apply the residue theorem: integral = \(2\pi i\) × sum of residues. Take the limit as the semicircle radius \(\to\infty\); the contour integral reduces to the original real integral. Finding a residue at a simple pole Write \(f(z)=\frac{g(z)}{h(z)}\) with \(h(z0)=0,\;h'(z0)\neq0\). Compute \(\displaystyle \operatorname{Res}(f,z0)=\frac{g(z0)}{h'(z0)}\). Finding a residue at a higher‑order pole Determine the order \(m\). Form \((z-z0)^{m}f(z)\). Differentiate \(m-1\) times, then take the limit \(z\to z0\) and divide by \((m-1)!\). Computing residue at infinity Substitute \(w=1/z\), rewrite \(f\) as \(F(w)=\frac{1}{w^{2}}f(1/w)\). Find \(\operatorname{Res}(F,w=0)\) and change sign. --- 🔍 Key Comparisons Simple pole vs. Removable singularity Simple pole: \(\operatorname{Res}\neq0\); Laurent series has a \((z-z0)^{-1}\) term. Removable: \(\operatorname{Res}=0\); function can be made holomorphic. Residue at a finite point vs. at infinity Finite: computed directly from local behavior. Infinity: computed via the transformed function \(\frac{1}{z^{2}}f(1/z)\); serves as a bookkeeping device for the global sum of residues. Positive vs. Negative orientation of \(\gamma\) Positive (counter‑clockwise): integral = \(+2\pi i\) × sum of residues. Negative (clockwise): integral = \(-2\pi i\) × sum of residues (winding numbers become negative). Jordan curve decomposition vs. single simple curve General closed curve → finite collection of simple closed curves (Jordan theorem). Allows reduction to the simple‑curve version of the residue theorem. --- ⚠️ Common Misunderstandings “All singularities contribute.” Only those with non‑zero winding number (i.e., inside the contour) matter; outside singularities give zero contribution. “A removable singularity has a non‑zero residue.” By definition its residue is zero. “The semicircular arc integral is always zero.” Must verify conditions (e.g., Jordan’s lemma) – otherwise the arc may contribute. “Higher‑order poles can use the simple‑pole formula.” The simple limit fails; you need the derivative/limit formula or a Laurent expansion. “Residue at infinity is just the negative of the sum of finite residues.” True only when \(\displaystyle \lim{z\to\infty}f(z)=0\); otherwise compute via the transformed function. “Winding number is always 1 for any interior point.” If the curve loops multiple times, the winding number equals that count. --- 🧠 Mental Models / Intuition Residue = “charge” at a pole – the integral “collects” \(2\pi i\) times that charge each time the contour circles the pole. Winding number = number of times you walk around a pole – think of a rubber band stretched around a nail; each full turn adds one to the count. Residue at infinity = “balance sheet” – the total “charge” of all finite poles must be offset by the charge at infinity, keeping the global sum zero. Arc‑vanishing = “damping” – exponential factors like \(e^{i a z}\) with \(\operatorname{Im}z>0\) shrink the contribution on large arcs, just as friction damps motion. --- 🚩 Exceptions & Edge Cases Singularity lying on the contour – residue theorem does not apply directly; must indent the contour or take a principal value. Non‑simply‑connected domain – the theorem may fail because the integral can depend on homotopy class; need to consider multiple winding numbers. Non‑rectifiable curves – infinite length makes the line integral ill‑defined; the theorem requires rectifiability. Essential singularities – residues are defined, but limit formulas for poles are invalid; usually treat via Laurent series. Arc integral not zero – when the integrand grows or lacks sufficient decay, the semicircle contribution must be computed explicitly. --- 📍 When to Use Which Simple pole → use \(\displaystyle \operatorname{Res}=\frac{g(z0)}{h'(z0)}\) (fastest). Higher‑order pole (order ≤ 3) → use the derivative limit formula; for order ≥ 4, expand the Laurent series if easier. Residue at infinity → use the transformed function when you need the global sum or when the integral is over a closed contour that encloses all finite poles. Real integral over \((-\infty,\infty)\) → close contour in the half‑plane where the exponential factor decays (Jordan’s lemma). Integral over \([0,\infty)\) with branch cuts → use a keyhole contour; account for contributions from both sides of the cut. --- 👀 Patterns to Recognize Rational function + \(\sin\) or \(\cos\) → close in the upper half‑plane for \(\sin\) (because \(\sin z = \frac{e^{iz}-e^{-iz}}{2i}\) and \(e^{iz}\) decays upward). Integrand \(f(x)=\frac{P(x)}{Q(x)}\) with \(\deg Q \ge \deg P+2\) → arc integral will usually vanish; only residues of poles in the chosen half‑plane matter. Even/odd symmetry – if the integrand is even, you can double the integral from 0 to \(\infty\) after contour evaluation. Presence of \(\frac{1}{z^2}f(1/z)\) → hint that the problem involves the residue at infinity. Repeated poles – look for the derivative term in the limit formula; the pattern “order \(m\) → differentiate \(m-1\) times”. --- 🗂️ Exam Traps Missing the factor \(2\pi i\). Many answer choices forget it or use \( \pi i\). Wrong sign for winding number – a clockwise contour flips the sign of the entire result. Assuming zero residue for a pole because the numerator also vanishes. Check the derivative condition \(h'(z0)\neq0\). Treating a removable singularity as a pole – leads to a non‑zero “residue” that is actually zero. Neglecting contributions from poles on the real axis – you must take half‑residue (principal value) or deform the contour. Using residue at infinity without verifying \(\lim{z\to\infty}f(z)=0\). Otherwise the simple “negative sum” rule is invalid. Choosing the wrong half‑plane for the closing arc – e.g., closing upward for \(e^{-i a z}\) (which grows) will give a non‑zero arc integral and an incorrect answer. ---