The residue theorem
An integral over an entire curve is reduced to adding up a few numbers, one per singularity. And you already knew those numbers: they are the coefficients of the partial fraction expansion.
01It all comes down to one coefficient
When the Laurent series is integrated term by term over a closed contour, all the terms give zero except one: the \( 1/(z - z_0) \) term, which gives \( 2\pi j \). That coefficient, the \( b_1 \) of the expansion, is the residue, and it is the only thing about the function that survives the integral.
It is a result of remarkable economy: an integral over an entire curve is reduced to adding up a few numbers, one per singularity inside. Everything else —the shape of the contour, the behavior of the function far from the poles— plays no part.
When a rational function is decomposed into partial fractions, the coefficients obtained are the residues. The expansion \( \frac{1}{(s+1)(s+2)} = \frac{1}{s+1} - \frac{1}{s+2} \) says that the residue at \( -1 \) is 1 and the one at \( -2 \) is \( -1 \). The residue method is the general version of a calculation you already did in Calculus.
02The three ways to compute it
There is no need to expand the full series. Depending on the case:
| Case | Formula |
|---|---|
| Simple pole | \( \operatorname{Res} = \lim_{z \to z_0} (z - z_0)\,f(z) \) |
| Simple pole with \( f = g/h \) | \( \operatorname{Res} = \dfrac{g(z_0)}{h'(z_0)} \) |
| Pole of order \( m \) | \( \operatorname{Res} = \dfrac{1}{(m-1)!}\lim_{z \to z_0} \dfrac{d^{\,m-1}}{dz^{\,m-1}}\Big[(z - z_0)^{m} f(z)\Big] \) |
| Essential singularity | No shortcut: you have to expand the Laurent series and read \( b_1 \) |
The second row is the one most used in practice and worth keeping at hand: with \( f(z) = 1/(z^3 - 1) \), the denominator has derivative \( 3z^2 \), so the residue at each cube root of unity is \( 1/(3z_k^2) \). And since \( z_k^3 = 1 \), that simplifies to \( z_k/3 \). The three residues add up to zero, and that is why a contour that encloses all three gives zero.
\( 1/z^2 \) has a double pole at the origin and residue zero, because its expansion has no \( 1/z \) term. The pole is there, but it contributes nothing to the integral. Confusing the two things leads to wrong conclusions about which singularities matter.
03Each pole contributes its share
On the left, the contour and the poles; on the right, the contributions \( 2\pi j \operatorname{Res} \) chained one after another. The end of the chain is the value of the integral, and the yellow point is what the numerical calculation over the contour gives. Click or tap the canvas to move the center.
04What it is really for: inverse transforms
The inverse Laplace transform is defined as a line integral along a vertical line to the right of all the poles —the Bromwich line—:
Nobody computes it that way. What is done is to close the contour with a semicircle on the left, which by the ML bound contributes zero as the radius grows, and apply the residue theorem to the closed contour that results. The outcome is the formula that is always used:
With \( F(s) = \dfrac{1}{(s+1)(s+2)} \), the residue at \( s = -1 \) is \( e^{-t} \) and the one at \( s = -2 \) is \( -e^{-2t} \), so \( f(t) = e^{-t} - e^{-2t} \). It is the same answer that the table gives, obtained without a table.
And here the circle closes, the one that began with poles and zeros: each pole of the transfer function contributes an exponential term to the response, and the residue is precisely how much amplitude it does so with. A pole with a small residue is there but is barely noticeable.
05In the lab
Compute \( \operatorname{Res}\big(1/(z^2+1),\, j\big) \) in the three ways: by the limit, by \( g/h' \) and by expanding in partial fractions. Verify that all three give \( -j/2 \).
For \( F(s) = \dfrac{s+3}{s(s+1)(s+2)} \), compute the three residues of \( F(s)e^{st} \) and build \( f(t) \). Check the initial and final values with the corresponding theorems.
Inverse-transform \( F(s) = \dfrac{1}{(s+1)^2(s+2)} \) using the formula for a pole of order 2. Verify that a \( t\,e^{-t} \) term appears and compare it with the measured response of a critically damped RLC.
06Common mistakes
- Adding residues of poles that ended up outside. Only those enclosed by the contour count.
- Using the simple-pole formula on a multiple pole. The limit gives infinity; you have to differentiate \( m-1 \) times.
- Forgetting the \( (m-1)! \) in the formula for a pole of order \( m \).
- Confusing a zero residue with the absence of a pole. \( 1/z^2 \) has a pole and residue zero.
- Inverse-transforming without checking that the semicircle contributes zero. If the degree of the numerator is not less than that of the denominator, the polynomial part has to be separated first.
07Self-assessment
What is the residue, in one line?
The coefficient of \( 1/(z-z_0) \) in the Laurent series: the only term that survives integration over a closed contour.
What is \( \operatorname{Res}(1/(z^2+1),\, j) \)?
\( 1/(2j) = -j/2 \), using \( g/h' \) with \( h' = 2z \).
What does \( \oint dz/(z^3-1) \) give over \( |z| = 2 \)?
Zero: it encloses the three poles, and their residues \( z_k/3 \) add up to zero because the three cube roots of unity add up to zero.
How is the residue computed at a pole of order 2?
Multiply by \( (z-z_0)^2 \), differentiate once and take the limit. The \( (m-1)! \) is 1 in this case.
What is the relationship between residues and partial fractions?
They are the same thing: the coefficients of the partial fraction decomposition are the residues at each pole.
How do you inverse-transform with residues?
\( f(t) = \sum \operatorname{Res}[F(s)e^{st}] \) at the poles of \( F \). Each pole contributes its exponential, with the amplitude given by its residue.
08Further reading
- Ruel V. Churchill and James W. Brown. Complex Variables and Applications. 7th ed. (Spanish edition, “Variable compleja y aplicaciones”), McGraw-Hill, 2004. The residue theorem with its computation formulas and the chapter on integral transforms.
- Murray R. Spiegel. Complex Variables. Schaum’s Outline Series (Spanish edition, “Variable compleja”), McGraw-Hill, 2011. Residues applied to dozens of integrals, organized by type of problem.
- Alan V. Oppenheim and Alan S. Willsky. Signals and Systems. 2nd ed. (Spanish edition, “Señales y sistemas”), Prentice Hall, 1998. The inverse Laplace transform, with the relationship between residues and partial fractions.