The Central Idea
Suppose \(f\) has an isolated singularity at \(z_0\) with Laurent expansion
\[ f(z) = \cdots + \frac{a_{-2}}{(z-z_0)^2} + \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0) +\cdots. \]The residue of \(f\) at \(z_0\) is
\[ \boxed{ \operatorname{Res}(f;z_0)=a_{-1}. } \]Why the Coefficient \(a_{-1}\) Matters
Around a positively oriented small circle centered at \(z_0\),
\[ \oint_C (z-z_0)^n\,dz \]equals zero for every integer \(n\neq-1\).
But
\[ \boxed{ \oint_C\frac1{z-z_0}\,dz=2\pi i. } \]Therefore only the \((z-z_0)^{-1}\) term contributes to the contour integral.
Residue and a Small Contour
If \(C\) is a sufficiently small positively oriented circle containing no singularity other than \(z_0\), then
\[ \boxed{ \oint_C f(z)\,dz = 2\pi i\operatorname{Res}(f;z_0). } \]This formula gives the local meaning of the residue.
Example: Read the Residue from a Laurent Series
Consider
\[ f(z) = \frac{3}{(z-1)^3} - \frac{2}{(z-1)^2} + \frac{5}{z-1} + 4 + 7(z-1)+\cdots. \]The coefficient of
\[ \frac1{z-1} \]is \(5\).
Therefore,
\[ \boxed{ \operatorname{Res}(f;1)=5. } \]Residue at a Removable Singularity
A removable singularity has no negative-power terms in its Laurent series.
Therefore,
\[ \boxed{ \operatorname{Res}(f;z_0)=0 } \]at every removable singularity.
Residue at a Simple Pole
Suppose \(z_0\) is a simple pole of \(f\).
Then
\[ f(z) = \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0) +\cdots. \]Multiplying by \(z-z_0\),
\[ (z-z_0)f(z) = a_{-1} + a_0(z-z_0) +\cdots. \]Therefore,
\[ \boxed{ \operatorname{Res}(f;z_0) = \lim_{z\to z_0} (z-z_0)f(z). } \]Example: Simple-Pole Limit Formula
Find the residue of
\[ f(z)=\frac{e^z}{z-2} \]at \(z=2\).
Since \(z=2\) is a simple pole,
\[ \operatorname{Res}(f;2) = \lim_{z\to2} (z-2)\frac{e^z}{z-2}. \]Therefore,
\[ \boxed{ \operatorname{Res}(f;2)=e^2. } \]A Quotient Formula for Simple Poles
Suppose
\[ f(z)=\frac{g(z)}{h(z)}, \]where \(g\) and \(h\) are analytic near \(z_0\), and suppose
\[ h(z_0)=0, \qquad h'(z_0)\neq0, \qquad g(z_0)\neq0. \]Then \(z_0\) is a simple pole and
\[ \boxed{ \operatorname{Res}(f;z_0) = \frac{g(z_0)}{h'(z_0)}. } \]Example: Quotient Formula
Find the residue of
\[ f(z)=\frac{1}{z^2+1} \]at \(z=i\).
Let
\[ g(z)=1, \qquad h(z)=z^2+1. \]Then
\[ h'(z)=2z. \]Hence,
\[ \operatorname{Res}(f;i) = \frac1{2i}. \]Therefore,
\[ \boxed{ \operatorname{Res}(f;i) = \frac1{2i} = -\frac{i}{2}. } \]The Other Pole of \(\frac1{z^2+1}\)
At \(z=-i\),
\[ \operatorname{Res}\left(\frac1{z^2+1};-i\right) = \frac1{-2i}. \]Therefore,
\[ \boxed{ \operatorname{Res}\left(\frac1{z^2+1};-i\right) = \frac{i}{2}. } \]Residues at Higher-Order Poles
If \(z_0\) is a pole of order \(m\), then
\[ \boxed{ \operatorname{Res}(f;z_0) = \frac{1}{(m-1)!} \lim_{z\to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right]. } \]This formula extracts the coefficient of \((z-z_0)^{-1}\) without requiring the entire Laurent series.
Example: A Second-Order Pole
Find the residue of
\[ f(z)=\frac{e^z}{z^2} \]at \(z=0\).
Here \(m=2\). Thus
\[ \operatorname{Res}(f;0) = \lim_{z\to0} \frac{d}{dz} \left[ z^2\frac{e^z}{z^2} \right]. \]Therefore,
\[ \operatorname{Res}(f;0) = \lim_{z\to0}e^z = \boxed{1}. \]Example: A Third-Order Pole
Find the residue of
\[ f(z)=\frac{e^z}{z^3} \]at \(z=0\).
Here \(m=3\), so
\[ \operatorname{Res}(f;0) = \frac1{2!} \left. \frac{d^2}{dz^2} e^z \right|_{z=0}. \]Since the second derivative is again \(e^z\),
\[ \boxed{ \operatorname{Res}(f;0)=\frac12. } \]The Series Method
Sometimes the fastest method is simply to expand the function and read off the \((z-z_0)^{-1}\) coefficient.
For example,
\[ \frac{e^z}{z^3} = \frac1{z^3} + \frac1{z^2} + \frac1{2z} + \frac1{3!} +\cdots. \]Therefore,
\[ \boxed{ \operatorname{Res}\left(\frac{e^z}{z^3};0\right)=\frac12. } \]Partial Fractions as a Residue Tool
Rational functions often become transparent after partial fractions.
Consider
\[ \frac1{z(z-1)}. \]Decompose:
\[ \frac1{z(z-1)} = -\frac1z+\frac1{z-1}. \]Therefore,
\[ \boxed{ \operatorname{Res}(f;0)=-1 } \]and
\[ \boxed{ \operatorname{Res}(f;1)=1. } \]Residues at Essential Singularities
Residues are not limited to poles.
An essential singularity has infinitely many negative-power terms, but the coefficient of \((z-z_0)^{-1}\) is still well-defined.
Thus essential singularities can also have residues.
Example: Residue of \(e^{1/z}\)
Expand:
\[ e^{1/z} = 1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots. \]The coefficient of \(1/z\) is \(1\).
Therefore,
\[ \boxed{ \operatorname{Res}(e^{1/z};0)=1. } \]The Residue Theorem
Let \(C\) be a positively oriented simple closed contour, and suppose \(f\) is analytic on and inside \(C\) except for finitely many isolated singularities
\[ z_1,z_2,\ldots,z_n \]inside the contour.
Then
\[ \boxed{ \oint_C f(z)\,dz = 2\pi i \sum_{k=1}^{n} \operatorname{Res}(f;z_k). } \]Why the Residue Theorem Works
Conceptually, remove small circles around each singularity.
The remaining region contains no singularities, so Cauchy's theorem applies.
The outer contour integral is then equal to the sum of the small-circle integrals around the singularities.
Each small-circle integral contributes
\[ 2\pi i\operatorname{Res}(f;z_k). \]Adding the local contributions gives the theorem.
Example: Two Enclosed Poles
Evaluate
\[ \oint_{|z|=2} \frac1{z^2+1}\,dz \]counterclockwise.
The poles \(i\) and \(-i\) both lie inside \(|z|=2\).
Their residues are
\[ -\frac{i}{2} \]and
\[ \frac{i}{2}. \]Their sum is zero.
Therefore,
\[ \boxed{ \oint_{|z|=2}\frac1{z^2+1}\,dz=0. } \]Example: Only One Pole Is Enclosed
Suppose \(C\) encloses \(z=i\) but not \(z=-i\).
Then
\[ \oint_C\frac1{z^2+1}\,dz = 2\pi i \left( -\frac{i}{2} \right). \]Therefore,
\[ \boxed{\pi}. \]Example: A Rational Function
Evaluate
\[ \oint_{|z|=3} \frac{z+1}{z(z-2)}\,dz. \]Both poles \(0\) and \(2\) lie inside.
At \(z=0\),
\[ \operatorname{Res}(f;0) = \lim_{z\to0} z\frac{z+1}{z(z-2)} = -\frac12. \]At \(z=2\),
\[ \operatorname{Res}(f;2) = \lim_{z\to2} (z-2)\frac{z+1}{z(z-2)} = \frac32. \]The residue sum is
\[ -\frac12+\frac32=1. \]Therefore,
\[ \boxed{ \oint_{|z|=3} \frac{z+1}{z(z-2)}\,dz = 2\pi i. } \]Residue Theorem Versus Cauchy's Integral Formula
Cauchy's integral formula is usually fastest when the integrand already has the form
\[ \frac{f(z)}{(z-z_0)^{n+1}}. \]The residue theorem is more flexible when:
- several singularities are enclosed;
- the integrand has a more complicated rational form;
- essential singularities are involved;
- local residue formulas are easier than direct contour integration.
Choosing a Residue Method
Common methods include:
- read the coefficient from a Laurent series;
- use \(\lim_{z\to z_0}(z-z_0)f(z)\) for a simple pole;
- use \(g(z_0)/h'(z_0)\) for a simple zero of the denominator;
- use the derivative formula for higher-order poles;
- use partial fractions for rational functions.
Residue Sums Can Simplify Dramatically
Sometimes several residues cancel.
This can make a complicated-looking contour integral equal to zero even though the contour encloses singularities.
Always compute the sum before multiplying by \(2\pi i\).
A Preview of the Residue at Infinity
On the extended complex plane, one can define a residue at infinity so that the sum of all residues, including the one at infinity, is zero.
This provides a powerful alternative for functions with many finite singularities.
Residues and Real Integrals
One of the most famous applications of the residue theorem is evaluating real improper integrals.
The strategy is to embed a real integrand into a complex function and choose a contour whose real-axis segment reproduces the desired integral.
The contour contribution is then evaluated through residues.
Example: A Classic Real Integral
Consider
\[ \int_{-\infty}^{\infty}\frac{dx}{x^2+1}. \]The corresponding complex function
\[ f(z)=\frac1{z^2+1} \]has poles at
\[ z=\pm i. \]Using a semicircular contour in the upper half-plane encloses only \(z=i\).
Since
\[ \operatorname{Res}(f;i)=-\frac{i}{2}, \]the contour integral is
\[ 2\pi i\left(-\frac{i}{2}\right)=\pi. \]After showing the arc contribution tends to zero,
\[ \boxed{ \int_{-\infty}^{\infty}\frac{dx}{x^2+1}=\pi. } \]Contour Choice Matters in Real-Integral Applications
To use residues for a real integral, one must justify:
- the chosen complex extension;
- which singularities lie inside the contour;
- the behavior on auxiliary arcs;
- the limiting process as the contour expands;
- any indentation around real-axis singularities.
Residues and Trigonometric Integrals
Integrals involving
\[ \sin\theta \]and
\[ \cos\theta \]can often be transformed by
\[ z=e^{i\theta}, \qquad d\theta=\frac{dz}{iz}. \]This converts an integral over \(0\le\theta\le2\pi\) into a contour integral around the unit circle.
Orientation Still Matters
The residue theorem is normally stated for positive, counterclockwise orientation.
For clockwise orientation,
\[ \boxed{ \oint_{-C}f(z)\,dz = -2\pi i \sum \operatorname{Res}(f;z_k). } \]What If a Singular Point Lies on the Contour?
The ordinary residue theorem does not apply directly when a singularity lies on the contour.
The contour must be modified, or a generalized interpretation such as a principal value may be required.
Never silently treat an on-contour pole as inside or outside.
Common Errors
- Summing residues outside the contour: include only enclosed singularities.
- Forgetting orientation: clockwise contours reverse the sign.
- Using the simple-pole formula at a higher-order pole: first identify pole order.
- Forgetting the factorial in the higher-order formula: use \((m-1)!\).
- Assuming residues exist only at poles: essential singularities also have residues.
- Ignoring poles on the contour: the standard theorem then requires modification.
A Reliable Procedure for a Simple Pole
- Confirm that \(z_0\) is a simple pole.
- Compute \[ \lim_{z\to z_0}(z-z_0)f(z). \]
- If \(f=g/h\) and \(h\) has a simple zero, consider \[ \frac{g(z_0)}{h'(z_0)}. \]
- Check the result against a Laurent expansion when convenient.
A Reliable Procedure for a Pole of Order \(m\)
- Confirm the pole order \(m\).
- Form \[ (z-z_0)^m f(z). \]
- Differentiate \(m-1\) times.
- Evaluate at \(z_0\).
- Divide by \((m-1)!\).
A Reliable Procedure for the Residue Theorem
- Identify the contour and orientation.
- Locate every isolated singularity of the integrand.
- Determine which singularities lie inside the contour.
- Compute the residue at each enclosed singularity.
- Add the enclosed residues.
- Multiply the sum by \(2\pi i\).
Worked Example: Three Simple Poles
Evaluate
\[ \oint_{|z|=4} \frac{1}{z(z-1)(z-2)}\,dz. \]All three poles \(0,1,2\) lie inside.
At \(z=0\),
\[ \operatorname{Res}(f;0) = \frac{1}{(-1)(-2)} = \frac12. \]At \(z=1\),
\[ \operatorname{Res}(f;1) = \frac{1}{(1)(-1)} = -1. \]At \(z=2\),
\[ \operatorname{Res}(f;2) = \frac{1}{(2)(1)} = \frac12. \]The residue sum is
\[ \frac12-1+\frac12=0. \]Therefore,
\[ \boxed{0}. \]Worked Example: Higher-Order Pole
Find
\[ \operatorname{Res} \left( \frac{\cos z}{z^3};0 \right). \]Here \(m=3\):
\[ \operatorname{Res} = \frac1{2!} \left. \frac{d^2}{dz^2}\cos z \right|_{z=0}. \]Since
\[ \frac{d^2}{dz^2}\cos z=-\cos z, \]we obtain
\[ \boxed{-\frac12}. \]Worked Example: Essential Singularity
Find the residue of
\[ \sin\left(\frac1z\right) \]at \(z=0\).
Expand:
\[ \sin\left(\frac1z\right) = \frac1z - \frac1{3!z^3} + \frac1{5!z^5} -\cdots. \]The coefficient of \(1/z\) is \(1\).
Therefore,
\[ \boxed{1}. \]Practice Problems
Practice 1
Find the residue of
\[ \frac{1}{z-3} \]at \(z=3\).
Practice 2
Find the residue of
\[ \frac{e^z}{z} \]at \(z=0\).
Practice 3
Find the residue of
\[ \frac1{z^2+4} \]at \(z=2i\).
Practice 4
Find the residue of
\[ \frac{e^z}{z^2} \]at \(z=0\).
Practice 5
Find the residue of
\[ \frac{e^z}{z^3} \]at \(z=0\).
Practice 6
Find the residue of
\[ e^{1/z} \]at \(z=0\).
Practice 7
Evaluate
\[ \oint_{|z|=2}\frac1{z(z-1)}\,dz. \]Practice 8
Evaluate
\[ \oint_{|z|=3}\frac{z+1}{z(z-2)}\,dz. \]Practice 9
If a function has Laurent term
\[ -\frac{7}{z-z_0}, \]what is its residue at \(z_0\)?
Practice 10
State the residue theorem for finitely many isolated singularities inside a positively oriented simple closed contour.
Practice Answers
Practice 1
\[ \boxed{1}. \]Practice 2
\[ \boxed{1}. \]Practice 3
With \(h(z)=z^2+4\),
\[ h'(z)=2z. \]Thus
\[ \operatorname{Res} = \frac1{4i} = \boxed{-\frac{i}{4}}. \]Practice 4
\[ \boxed{1}. \]Practice 5
\[ \boxed{\frac12}. \]Practice 6
\[ \boxed{1}. \]Practice 7
Both poles \(0\) and \(1\) lie inside. Their residues are \(-1\) and \(1\), so the sum is zero.
\[ \boxed{0}. \]Practice 8
The enclosed residues are
\[ -\frac12 \]and
\[ \frac32. \]The sum is \(1\), so
\[ \boxed{2\pi i}. \]Practice 9
\[ \boxed{-7}. \]Practice 10
\[ \boxed{ \oint_Cf(z)\,dz = 2\pi i \sum_k\operatorname{Res}(f;z_k). } \]Checking Strategies
- Determine singularity type before selecting a residue formula.
- For a simple pole, verify that the denominator has a simple zero.
- For a higher-order pole, double-check both the derivative order and the factorial.
- In residue-theorem problems, mark enclosed versus exterior singularities before computing.
- Add residues before multiplying by \(2\pi i\).
- Check orientation.
- If a Laurent series is available, confirm that the residue equals the coefficient of \((z-z_0)^{-1}\).
Connections to Earlier Mathematics
- Laurent series: residues are the \(a_{-1}\) coefficients.
- Cauchy's theorem: removing singularities leaves analytic regions where contour integrals vanish.
- Cauchy's integral formula: a simple residue calculation generalizes the same \(2\pi i\) mechanism.
- Partial fractions: rational functions often reveal residues term by term.
- Real analysis: contour methods can evaluate difficult real improper integrals.
- Trigonometry: \(z=e^{i\theta}\) converts periodic real integrals into unit-circle contour integrals.
What to Remember
If
\[ f(z) = \sum_{n=-\infty}^{\infty} a_n(z-z_0)^n, \]then
\[ \boxed{ \operatorname{Res}(f;z_0)=a_{-1}. } \]At a simple pole,
\[ \boxed{ \operatorname{Res}(f;z_0) = \lim_{z\to z_0}(z-z_0)f(z). } \]At a pole of order \(m\),
\[ \boxed{ \operatorname{Res}(f;z_0) = \frac1{(m-1)!} \lim_{z\to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right]. } \]And for enclosed isolated singularities \(z_k\),
\[ \boxed{ \oint_Cf(z)\,dz = 2\pi i \sum_k \operatorname{Res}(f;z_k). } \]The essential idea is: the global contour integral is assembled from local singularity data.
Where We Go Next
Residues show how local singular behavior controls global contour integrals. The final lesson in this section, Conformal Mapping and Applications, returns to the geometric side of analytic functions and studies how analytic maps transform regions, preserve angles, simplify boundary-value problems, and support applications in physics, engineering, and applied mathematics.