Contour

Work in progress. This note is still being written and incomplete.

A contour is a curve CC in the complex plane constructed by joining finitely many smooth curves end to end.

Riemann Sum

Let CC be a contour with initial point AA and terminal point BB.

A partition of CC is a finite set of points P={z0,z1,,zn}P = \{z_0, z_1, \ldots, z_n\}, occurring in order along CC, with z0=Az_0 = A and zn=Bz_n = B.

Δzk=zkzk1,k=1,2,,n\Delta z_k = z_k - z_{k-1}, \qquad k = 1, 2, \ldots, n

ckc_k is any point on the part of CC between zk1z_{k-1} and zkz_k.

The Riemann sum for this partition, for a function ff is

k=1nf(ck)Δzk\sum_{k=1}^n f(c_k)\Delta z_k

Contour Integral

The contour integral of ff over CC is

Cf(z)dz=limP0k=1nf(ck)Δzk\int_C f(z)\,\text{d}z = \lim_{\|P\| \to 0} \sum_{k=1}^n f(c_k)\Delta z_k

provided the limit exists and is independent of the choice of partition and points ckc_k.

P\|P\| is the maximum length of the subarcs determined by PP.

Evaluation via Parametrization

Let f(z)f(z) be continuous on a set containing the contour CC, and let z=z(t)z = z(t), atba \le t \le b, be a piecewise smooth parametrization of CC.

Cf(z)dz=abf(z(t))z(t)dt\int_C f(z)\,\text{d}z = \int_a^b f(z(t))z'(t)\,\text{d}t

When z(t)z(t) is only piecewise smooth, the right side is evaluated piecewise.

Integral Triangle Inequality

Let f(t)=u(t)+iv(t)f(t) = u(t) + iv(t) be continuous on [a,b][a,b].

abf(t)dtabf(t)dt\left|\int_a^b f(t)\,\text{d}t\right| \le \int_a^b |f(t)|\,\text{d}t

ML Inequality

Let ff be continuous on a contour CC, with f(z)M|f(z)| \le M for all zz on CC, and let LL be the length of CC.

Cf(z)dzML\left|\int_C f(z)\,\text{d}z\right| \le ML

Examples

Line Integral of zz from 00 to 1+i1+i

Let CC be the line segment from 00 to 1+i1+i, parametrized by z(t)=t(1+i)z(t) = t(1+i), 0t10 \le t \le 1. z(t)=1+iz'(t) = 1+i.

Czdz=01t(1+i)(1+i)dt=(1+i)201tdt\int_C z\,\text{d}z = \int_0^1 t(1+i)(1+i)\,\text{d}t = (1+i)^2\int_0^1 t\,\text{d}t

(1+i)2=2i(1+i)^2 = 2i and 01tdt=12\displaystyle\int_0^1 t\,\text{d}t = \dfrac{1}{2}

Czdz=2i12=i\int_C z\,\text{d}z = 2i \cdot \frac{1}{2} = i
Written by September 13, 2026 3 min read
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