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Derivation

The Residue Theorem

Statement

Let \(f\) be holomorphic on an open set containing a simple closed positively-oriented contour \(C\) and its interior, except at finitely many isolated singularities \(z_1,\dots,z_N\) lying inside \(C\). Then \(\displaystyle \oint_C f(z)\,dz = 2\pi i \sum_{k=1}^{N}\operatorname{Res}(f,z_k)\), where each residue \(\operatorname{Res}(f,z_k)=a_{-1}^{(k)}\) is the coefficient of \((z-z_k)^{-1}\) in the Laurent expansion of \(f\) about \(z_k\).

Why it matters

The residue theorem collapses an entire contour integral onto a finite algebraic problem: it reduces the analytic operation of integration to reading off one Laurent coefficient per singularity. It is the computational engine behind Cauchy's integral formula, the argument principle, and the evaluation of vast classes of real integrals and infinite series that resist elementary methods.

In physics it is ubiquitous. It underlies the Kramers-Kronig relations linking the real and imaginary parts of response functions, the evaluation of propagators and Green's functions in field theory, dispersion relations, the analytic structure of the S-matrix, and countless integrals in statistical mechanics and optics where poles encode resonances and bound states.

Assumptions
\(f\) is holomorphic on and inside \(C\) except at isolated singularities.If singularities accumulate or \(f\) fails to be complex-differentiable on an open region, no Laurent expansion exists there and the residue sum is undefined. The singularities \(z_k\) are isolated and finite in number inside \(C\).A branch point (e.g. \(\sqrt{z}\), \(\log z\)) is not isolated — every punctured neighbourhood still contains the cut — so the theorem does not apply without first choosing a branch and excising the cut. \(C\) is a rectifiable simple closed curve with positive (counter-clockwise) orientation.Reversing orientation flips the sign; a self-intersecting curve requires the winding number \(n(C,z_k)\) as a weight, giving \(\oint_C f\,dz = 2\pi i\sum_k n(C,z_k)\operatorname{Res}(f,z_k)\). Each Laurent series converges uniformly on the small circles used, so term-by-term integration is legitimate.Without uniform convergence on a compact set inside the annulus, interchanging \(\sum\) and \(\oint\) is not justified and the derivation breaks at Step 5.
Derivation
1
\[ \oint_C f(z)\,dz = \sum_{k=1}^{N}\oint_{\gamma_k} f(z)\,dz \]
Enclose each singularity \(z_k\) in a small positively-oriented circle \(\gamma_k=\{|z-z_k|=\rho_k\}\), disjoint and interior to \(C\). The region between \(C\) and the \(\gamma_k\) is holomorphic, so by the Cauchy-Goursat theorem on this multiply-connected domain the outer integral equals the sum of the inner ones. B
2
\[ f(z) = \sum_{n=-\infty}^{\infty} a_n^{(k)}\,(z-z_k)^{n}, \qquad 0<|z-z_k|<R_k \]
On the punctured disc about the isolated singularity \(z_k\), the Laurent expansion exists and is unique. Choose \(\rho_k<R_k\) so \(\gamma_k\) lies inside the annulus of convergence. A
3
\[ \oint_{\gamma_k} f(z)\,dz = \oint_{\gamma_k}\sum_{n=-\infty}^{\infty} a_n^{(k)}(z-z_k)^{n}\,dz = \sum_{n=-\infty}^{\infty} a_n^{(k)}\oint_{\gamma_k}(z-z_k)^{n}\,dz \]
The Laurent series converges uniformly on the compact circle \(\gamma_k\) (a closed subset of the open annulus), so integration and summation may be interchanged. C
4
\[ z = z_k + \rho_k e^{i\theta},\quad dz = i\rho_k e^{i\theta}\,d\theta \;\Rightarrow\; \oint_{\gamma_k}(z-z_k)^{n}\,dz = i\,\rho_k^{\,n+1}\!\int_0^{2\pi} e^{i(n+1)\theta}\,d\theta \]
Parametrise the circle and substitute. This isolates each power \((z-z_k)^n\) into a single elementary angular integral. A
5
\[ \int_0^{2\pi} e^{i(n+1)\theta}\,d\theta = \begin{cases} 2\pi, & n=-1\\[2pt] 0, & n\neq -1 \end{cases} \qquad\Longrightarrow\qquad \oint_{\gamma_k}(z-z_k)^{n}\,dz = 2\pi i\,\delta_{n,-1} \]
For \(n\neq-1\) the integrand \(e^{i(n+1)\theta}\) is a full-period complex exponential and integrates to zero; only \(n=-1\) gives \(\int_0^{2\pi}d\theta=2\pi\), and \(\rho_k^{\,0}=1\) removes the radius dependence. This is the orthogonality that projects out the single coefficient \(a_{-1}\). B
6
\[ \oint_{\gamma_k} f(z)\,dz = \sum_{n} a_n^{(k)}\cdot 2\pi i\,\delta_{n,-1} = 2\pi i\,a_{-1}^{(k)} = 2\pi i\,\operatorname{Res}(f,z_k) \]
Every term vanishes except \(n=-1\), leaving the residue by definition. The result is independent of \(\rho_k\), confirming the singular part alone carries the contour integral. A
7
\[ \oint_C f(z)\,dz = \sum_{k=1}^{N} 2\pi i\,\operatorname{Res}(f,z_k) = 2\pi i\sum_{k=1}^{N}\operatorname{Res}(f,z_k) \]
Insert Step 6 into the decomposition of Step 1 and factor the common \(2\pi i\). A
Result
\[ \boxed{\;\oint_C f(z)\,dz = 2\pi i \sum_{k=1}^{N}\operatorname{Res}(f,z_k)\;} \]

Reading. The whole contour integral depends only on the sum of residues — the \(a_{-1}\) Laurent coefficients — of the singularities the curve encloses. Regular parts and all other Laurent powers contribute nothing; the loop "counts" singularities weighted by their \((z-z_k)^{-1}\) strength, each worth exactly \(2\pi i\) times its residue.

Units check. The statement is dimensionless in the analytic sense: \([f(z)\,dz]\) and \([\operatorname{Res}(f,z_k)]\) carry the same dimensions, since a residue is a value of \(f\times(\text{length in }z)\), matching \(f\,dz\). If \(z\) carries a physical unit (say \(\mathrm{s}^{-1}\) as a frequency), both sides scale identically because \(2\pi i\) is a pure number, so dimensional consistency is automatic.

Limiting cases
  • No enclosed singularity (\(N=0\)): the sum is empty and \(\oint_C f\,dz=0\), recovering the Cauchy-Goursat theorem.
  • Single simple pole: \(\operatorname{Res}(f,z_0)=\lim_{z\to z_0}(z-z_0)f(z)\), and \(\oint_C f\,dz = 2\pi i\,(z-z_0)f\big|_{z_0}\).
  • \(f=g/(z-z_0)\) with \(g\) holomorphic: residue is \(g(z_0)\), giving \(\oint_C \frac{g(z)}{z-z_0}\,dz = 2\pi i\,g(z_0)\) — Cauchy's integral formula.
  • Pole of order \(m\): \(\operatorname{Res}(f,z_0)=\dfrac{1}{(m-1)!}\lim_{z\to z_0}\dfrac{d^{\,m-1}}{dz^{\,m-1}}\big[(z-z_0)^m f(z)\big]\).
  • Residue at infinity: for the whole extended plane, \(\sum_{\text{all }k}\operatorname{Res}(f,z_k) + \operatorname{Res}(f,\infty)=0\).
Breaks when
  • Branch points or cuts inside \(C\). Functions like \(z^{1/2}\), \(\log z\), or \(z^{\alpha}\) with non-integer \(\alpha\) have non-isolated singularities; no Laurent series exists, so there is no residue and the theorem fails. One must instead choose a branch and deform \(C\) around the cut.
  • Essential singularity approached without care, or infinitely many poles. The theorem still holds for an essential singularity (the residue is well-defined as \(a_{-1}\)), but if \(C\) encloses infinitely many poles or singularities accumulate on \(C\), the finite sum diverges or is ill-defined and the hypotheses collapse.
  • A singularity lies exactly on the contour \(C\). Then \(f\) is not holomorphic on \(C\), the integral is not even convergent as an ordinary contour integral, and one must resort to a principal value plus a semicircular indentation (each half-residue).
Failure modes
  • Counting singularities outside \(C\). Only enclosed singularities contribute; students routinely include poles that lie outside the contour and get a spurious nonzero answer.
  • Using the wrong-order residue formula. Applying the simple-pole formula \(\lim (z-z_0)f\) to a double pole gives \(\infty\) or a wrong finite value; the order-\(m\) derivative formula is required.
  • Forgetting the winding number for non-simple contours. A curve that loops twice weights the residue by \(n=2\); dropping this halves the answer.
  • Ignoring orientation. A clockwise contour introduces a sign; \(\oint = -2\pi i\sum\operatorname{Res}\).
  • Confusing residue with the function value. \(\operatorname{Res}(f,z_0)\) is the \(a_{-1}\) coefficient, not \(f(z_0)\) (which is infinite at a pole).
  • Mis-handling a removable singularity. A removable point has \(a_{-1}=0\); students sometimes assign it a nonzero residue from an incorrect limit.
Discussion

The theorem's power comes from the orthogonality relation in Step 5: integrating \((z-z_0)^n\) around a closed loop annihilates every power except \(n=-1\). Geometrically, only the \((z-z_0)^{-1}\) term has a nonzero winding contribution — its phase advances by exactly \(2\pi\) as \(z\) circles \(z_0\), while every other power returns to its starting value with net-zero angular sweep. The residue is thus the unique Laurent coefficient with topological content.

Physically, poles encode resonances and particle content. In a scattering amplitude a simple pole at \(z=E_0\) signals a bound state or resonance of energy \(E_0\), and its residue fixes the coupling strength. Closing a contour in the complex energy or frequency plane and summing residues is exactly how one converts a spectral integral into a discrete sum over modes — the mathematical skeleton of the Matsubara sum in finite-temperature field theory and of mode expansions in cavity QED.

The theorem also unifies apparently distinct results. Setting \(f=g/(z-z_0)\) yields Cauchy's integral formula; differentiating under the integral yields the formula for derivatives; applying it to \(f'/f\) yields the argument principle, counting zeros minus poles. Each is a special case of reading off one residue.

At the deepest level the residue is a de Rham cohomology pairing: \(\frac{1}{2\pi i}\oint\) is the period of the meromorphic 1-form \(f\,dz\) against the homology class of \(C\). Because \(f\,dz\) is closed away from its poles, the integral depends only on the homology class of the contour in the punctured plane — precisely why deformation in Step 1 is legitimate. The residue theorem is then the statement that the period equals the sum of local residues, a finite-dimensional pairing between \(H^1\) of the punctured surface and \(H_1\) of the cycles. This viewpoint generalises directly to residues of forms on Riemann surfaces and to the Grothendieck residue in several complex variables.

Common misconceptions. The value of \(\oint_C f\,dz\) does not depend on the size or precise shape of \(C\), only on which singularities it encloses — a source of surprise for students used to real integrals whose value depends on the path. Equally, a residue can be nonzero at a point where \(f\) is finite in a limiting sense (an order-\(m\) pole whose \(a_{-1}\) survives), and can be zero at a genuine singularity (e.g. a double pole of an even function).

Worked examples

Example 1 — two simple poles enclosed. Evaluate \(\displaystyle \oint_{|z|=2}\frac{2z+1}{z(z-1)}\,dz\).

1
\[ f(z)=\frac{2z+1}{z(z-1)},\qquad \text{poles at } z=0,\ z=1,\ \text{both inside } |z|=2. \]
Identify the singularities and confirm both lie inside the contour \(|z|=2\). A
2
\[ \operatorname{Res}(f,0)=\lim_{z\to0} z\,f(z)=\frac{2z+1}{z-1}\bigg|_{z=0}=\frac{1}{-1}=-1. \]
Simple-pole formula: multiply by \((z-0)\) and evaluate. A
3
\[ \operatorname{Res}(f,1)=\lim_{z\to1}(z-1)f(z)=\frac{2z+1}{z}\bigg|_{z=1}=\frac{3}{1}=3. \]
Simple-pole formula at the second pole. A
4
\[ \oint_{|z|=2} f(z)\,dz = 2\pi i\big(-1+3\big)=2\pi i\cdot 2. \]
Apply the residue theorem: sum residues, multiply by \(2\pi i\). A
\[ \oint_{|z|=2}\frac{2z+1}{z(z-1)}\,dz = 4\pi i \]

Reading. The two enclosed simple poles contribute residues \(-1\) and \(3\); their sum \(2\) times \(2\pi i\) gives \(4\pi i\). Units. Pure number times \(dz\)-dimension, consistent on both sides.

Example 2 — a third-order pole. Evaluate \(\displaystyle \oint_{|z|=1}\frac{e^{z}}{z^{3}}\,dz\).

1
\[ f(z)=\frac{e^{z}}{z^{3}},\qquad \text{pole of order } m=3 \text{ at } z=0\ (\text{inside }|z|=1). \]
Only singularity is \(z=0\); a triple pole because \(e^z\) is nonzero there. A
2
\[ \operatorname{Res}(f,0)=\frac{1}{(m-1)!}\lim_{z\to0}\frac{d^{\,m-1}}{dz^{\,m-1}}\big[z^{m}f(z)\big]=\frac{1}{2!}\lim_{z\to0}\frac{d^{2}}{dz^{2}}e^{z}. \]
Order-\(m\) residue formula with \(m=3\); \(z^3 f(z)=e^z\). B
3
\[ \frac{d^{2}}{dz^{2}}e^{z}=e^{z}\ \Rightarrow\ \operatorname{Res}(f,0)=\frac{1}{2}\,e^{0}=\frac12. \]
Differentiate twice, evaluate at \(0\). Equivalently, the \(a_{-1}\) coefficient is the \(z^2\) term of \(e^z=1+z+\tfrac{z^2}{2}+\cdots\) divided into \(z^{-3}\). A
4
\[ \oint_{|z|=1}\frac{e^{z}}{z^{3}}\,dz = 2\pi i\cdot\frac12. \]
Residue theorem with a single enclosed singularity. A
\[ \oint_{|z|=1}\frac{e^{z}}{z^{3}}\,dz = \pi i \]

Reading. The Laurent coefficient \(a_{-1}=\tfrac12\) (the \(z^2/2\) term of \(e^z\) shifted by \(z^{-3}\)) alone determines the integral. Units. Dimensionless coefficient times \(2\pi i\), consistent.

Problems
  1. Evaluate \(\displaystyle \oint_{|z|=1}\frac{dz}{z}\).
    Solution

    Single simple pole at \(z=0\), inside \(|z|=1\). \(\operatorname{Res}(1/z,0)=a_{-1}=1\). Hence \(\oint = 2\pi i\cdot 1 = \boxed{2\pi i}\). This is the canonical generating integral of Step 5.

  2. Evaluate \(\displaystyle \oint_{|z|=3}\frac{z}{z^{2}-4}\,dz\).
    Solution

    Poles where \(z^2=4\), i.e. \(z=\pm2\), both inside \(|z|=3\). Residues: \(\operatorname{Res}(z=2)=\dfrac{z}{2z}\big|_{2}=\dfrac{z}{z+2}\cdot\dfrac{1}{1}\)... use \(\dfrac{z}{(z-2)(z+2)}\): \(\operatorname{Res}(2)=\dfrac{2}{2+2}=\tfrac12\), \(\operatorname{Res}(-2)=\dfrac{-2}{-2-2}=\tfrac12\). Sum \(=1\). \(\oint = 2\pi i\cdot 1 = \boxed{2\pi i}\).

  3. Evaluate \(\displaystyle \oint_{|z|=1}\frac{\cos z}{z^{2}}\,dz\).
    Solution

    Order-2 pole at \(z=0\). \(\operatorname{Res}=\dfrac{1}{1!}\dfrac{d}{dz}\big[\cos z\big]_{0}=-\sin 0=0\). Equivalently \(\cos z = 1-\tfrac{z^2}{2}+\cdots\) has no \(z^1\) term, so \(a_{-1}=0\). Hence \(\oint = \boxed{0}\). A genuine singularity with zero residue.

  4. Evaluate \(\displaystyle \oint_{|z|=1}\frac{dz}{z^{2}-5z+6}\).
    Solution

    Factor: \(z^2-5z+6=(z-2)(z-3)\), poles at \(z=2,3\). Both satisfy \(|z|>1\), so neither lies inside \(|z|=1\): no enclosed singularity. By the residue theorem (empty sum) \(\oint = \boxed{0}\). This is Cauchy-Goursat in disguise.

  5. Use residues to evaluate the real integral \(\displaystyle I=\int_{0}^{2\pi}\frac{d\theta}{2+\cos\theta}\).
    Solution

    Set \(z=e^{i\theta}\), so \(\cos\theta=\tfrac12(z+z^{-1})\) and \(d\theta=\dfrac{dz}{iz}\), with \(z\) tracing \(|z|=1\). Then \[ I=\oint_{|z|=1}\frac{1}{2+\frac12(z+z^{-1})}\frac{dz}{iz}=\oint_{|z|=1}\frac{2}{i\,(z^{2}+4z+1)}\,dz. \] Roots of \(z^2+4z+1=0\): \(z=-2\pm\sqrt{3}\). Only \(z_+=-2+\sqrt3\approx-0.27\) lies inside \(|z|=1\). Residue there: \[ \operatorname{Res}=\frac{2}{i}\cdot\frac{1}{2z+4}\bigg|_{z_+}=\frac{2}{i}\cdot\frac{1}{2\sqrt3}=\frac{1}{i\sqrt3}. \] Thus \(I=2\pi i\cdot\dfrac{1}{i\sqrt3}=\boxed{\dfrac{2\pi}{\sqrt3}}\approx 3.6276\). The imaginary unit cancels, correctly returning a real, positive value.