Series and residues · Definition

Residue theorem

complex analysisresidues

Before you start you need to know

It turns a contour integral into a finite sum of residues: the computation reduces to finding the coefficients of \(1/(z-z_k)\).

Definition

If \(f\) is holomorphic on a simple closed contour \(\gamma\) and inside it except at isolated singularities \(z_k\), its integral over \(\gamma\) is \(2\pi i\) times the sum of its residues there.

\[\oint_{\gamma} f(z)\,dz=2\pi i\sum_{k}\operatorname{Res}(f,z_k)\]

Deforming the contour toward the interior singularities (Cauchy–Goursat theorem), the integral over \(\gamma\) splits into small circles around each \(z_k\). On each one the Laurent series is integrated term by term and only \(a_{-1}\), the residue, survives; hence the factor \(2\pi i\).

Two conditions are essential: \(\gamma\) must be positively oriented (counterclockwise) and only the singularities inside \(\gamma\) count; those outside contribute nothing. Traversing the contour backwards flips the sign.

To compute residues use the right shortcut: at a simple pole, \(\operatorname{Res}(f,z_0)=\lim_{z\to z_0}(z-z_0)f(z)\); at a pole of order \(m\), \(\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}\big[(z-z_0)^m f(z)\big]\). For essential singularities read the coefficient \(a_{-1}\) off the series, as in \(e^{1/z}\). A zero residue does not mean no singularity, as in \(1/z^2\).

The theorem generalizes Cauchy's integral formula (the case of a simple pole) and is the underlying tool for evaluating improper real integrals.

Think about it

Why is \(\oint_{|z|=1} z^{-2}\,dz=0\) even though \(0\) is an interior singularity?

Because the residue of \(z^{-2}\) at \(0\) is the coefficient of \(z^{-1}\), which is \(0\). Equivalently, \(-1/z\) is a primitive on \(\mathbb{C}\setminus\{0\}\) and the closed integral vanishes.

If \(f\) is holomorphic inside and on \(\gamma\), what does the theorem say?

There are no residues to add, so the integral is \(0\): Cauchy–Goursat is recovered.

Exercises 10

  1. Compute \(\oint_{|z|=2}\frac{dz}{z-1}\), with the circle traversed counterclockwise.
  2. Compute \(\oint_{|z|=1}\frac{e^{z}}{z^{2}}\,dz\) counterclockwise.
  3. Compute \(\oint_{|z|=1}\frac{dz}{z^{2}+4}\) counterclockwise.
  4. Compute \(\oint_{|z|=2}\frac{z}{z^{2}+1}\,dz\) counterclockwise.
  5. Compute \(\oint_{|z|=2}\frac{dz}{z-1}\) traversed clockwise (negatively).
  6. Compute \(\oint_{|z|=3}\frac{dz}{z^{2}-1}\) counterclockwise.
  7. Compute \(\oint_{|z|=1}\frac{\cos z}{z^{3}}\,dz\) counterclockwise.
  8. Compute \(\oint_{|z-i|=1}\frac{dz}{z^{2}+1}\) counterclockwise.
  9. Compute \(\oint_{|z|=2}\frac{e^{z}}{z(z-1)}\,dz\) counterclockwise.
  10. Compute \(\oint_{|z|=1}e^{1/z}\,dz\) counterclockwise (the singularity at \(0\) is not a pole).

References

  • Elias M. Stein, Rami Shakarchi, Complex Analysis, 1st edition, Princeton University Press, 2003. Cap. 3, §1
  • James Ward Brown, Ruel V. Churchill, Complex Variables and Applications, 9th edition, McGraw-Hill, 2014. Cap. 6, §76
  • Lars V. Ahlfors, Complex Analysis, 3rd edition, McGraw-Hill, 1979. Cap. 4, §5.2