contour integral
A contour integral adds up the values of a complex function as you walk along a curve in the plane, weighting each value by the tiny complex step you take. Imagine threading a path through a landscape and tallying the function as you go — except that here both the function value and the step you take are complex numbers, so the multiplications carry rotation and scaling information, not just magnitude. The direction you traverse the curve matters, and reversing it flips the sign.
Formally, given a piecewise-smooth curve (a contour) parametrized by z(t) for t in [a, b], the integral of f along it is the integral from a to b of f(z(t)) z'(t) dt — a perfectly ordinary one-variable integral of a complex-valued integrand. The result is a complex number, independent of how you parametrize the curve, depending only on the oriented path. A basic estimate, the ML-inequality, bounds the modulus of the integral by (maximum of |f| on the curve) times (length of the curve).
What makes contour integrals magical is their interaction with holomorphicity. For holomorphic integrands the value can be wildly insensitive to the exact path — Cauchy's theorem says a closed-loop integral vanishes, and deformation arguments let you slide contours around freely as long as you do not cross a singularity. This is the engine behind residue calculus, which evaluates hard real integrals by detouring through the complex plane.
Integrate 1/z counterclockwise once around the unit circle z(t) = e^{it}, t from 0 to 2pi. Then z'(t) = i e^{it} and f(z(t)) z'(t) = (1/e^{it})(i e^{it}) = i, so the integral is the integral of i dt from 0 to 2pi = 2pi i. This nonzero answer signals the singularity of 1/z at the origin enclosed by the loop.
The fundamental computation: integral of 1/z around the unit circle is 2pi i.