Contour integral

A contour integral is the integral of a complex function along an oriented curve in $\mathbb{C}$. The complex-analytic analog of the vector line integral in vector calculus.

Definition

A contour is a piecewise-smooth oriented curve in $\mathbb{C}$, parameterized by $z(t)$, $a\le t\le b$, with $z^{\prime }(t)$ piecewise continuous. For a continuous function $f$ along $C$,

${\int }_{C}f(z)dz={\int }_a^{b}f(z(t))z^{\prime }(t)dt.$

Same shape as the vector line integral: substitute the parameterization, multiply by the velocity (now complex), integrate over the parameter.

Properties

• Linear in $f$: $\int (af+bg)dz=a\int fdz+b\int gdz$.
• Orientation-sensitive: ${\int }_{-C}fdz=-{\int }_{C}fdz$.
• Decomposition: ${\int }_{C_1+C_2}fdz={\int }_{C_1}fdz+{\int }_{C_2}fdz$.
• Reparameterization-invariant within the same orientation.

Closed contours

When $C$ is a closed contour (starts and ends at the same point), write ${\oint }_{C}f(z)dz$. By convention, closed contours are traversed counterclockwise (positive orientation) unless stated otherwise.

Worked examples

Example 1: $\oint \bar{z}dz$ where $C$ is the unit circle, counterclockwise.

Parameterize $z(t)=e^{it}$, $0\le t\le 2\pi$. $z^{\prime }(t)=i{e}^{it}$. $\overline{z(t)}=e^{-it}$.

$\oint \bar{z}dz={\int }_0^{2\pi }{e}^{-it}\cdot i{e}^{it}dt=i{\int }_0^{2\pi }1dt=2\pi i.$

Note: $\bar{z}$ is not analytic. The integral didn’t have to vanish, and didn’t.

Example 2: $\oint zdz$ where $C$ is the unit circle.

$z(t)=e^{it}$, $z^{\prime }(t)=i{e}^{it}$, integrand $z\cdot z^{\prime }=i{e}^{2it}$.

${\int }_0^{2\pi }i{e}^{2it}dt=i\cdot \frac{1}{2i}[e^{2it}{]}_0^{2\pi }=\frac{1}{2}(1-1)=0.$

This time the integral was zero. $z$ is analytic — and (as Cauchy’s theorem will say) every closed integral of an analytic function over a contour in a simply connected region is zero.

Example 3: $\oint dz/z$ where $C$ is the unit circle.

$z=e^{it}$, $dz=i{e}^{it}dt$, $1/z=e^{-it}$. Integrand: $e^{-it}\cdot i{e}^{it}=i$.

${\int }_0^{2\pi }idt=2\pi i.$

The integral was $2\pi i$. $1/z$ is analytic on $\mathbb{C}\setminus \{0\}$ but has a singularity at the origin which the contour encloses. The $2\pi i$ is the residue contribution that becomes systematic in the Residue theorem.

The fundamental integral

Generalizing example 3, on the unit circle:

${\oint }_{|z|=1}{z}^{n}dz=\{\begin{array}{ll}2\pi i& n=-1\\ 0& \text{otherwise.}\end{array}$

For $n\ge 0$: $z^n$ is analytic on the disk, Cauchy’s theorem gives $0$.

For $n\le -2$: parameterize and the integral is ${\int }_0^{2\pi }{e}^{i(n+1)t}\cdot idt$, which vanishes because $n+1\ne 0$.

For $n=-1$: the integral is ${\int }_0^{2\pi }idt=2\pi i$.

Internalize this. Only the $1/z$ term contributes to a circular integral around the origin. This single fact is the kernel of the residue theorem.

Connection to vector line integral

Unpacking $f(z)dz=(u+iv)(dx+idy)=(udx-vdy)+i(vdx+udy)$:

$\oint f(z)dz=\oint (udx-vdy)+i\oint (vdx+udy).$

The first piece is the circulation of the 2D vector field $⟨u,-v⟩$; the second is the flux of $⟨u,-v⟩$ (or equivalently the circulation of $⟨v,u⟩$). A contour integral is two line integrals packaged into one complex number.

When $f$ is analytic, the Cauchy-Riemann equations make $⟨u,-v⟩$ both conservative and source-free, so both line integrals vanish by Green’s theorem — giving Cauchy’s theorem. The major theorems of complex analysis are restatements of vector calculus identities in complex notation.

In context

The contour integral is the foundation for:

• Cauchy’s theorem: $\oint f=0$ for analytic $f$.
• Cauchy integral formula: expresses $f$ inside a contour from boundary values.
• Residue theorem: $\oint f=2\pi i\sum \mathrm{Res}$.

These collapse complex contour integrals to algebra.