the ML inequality
Often you do not need the exact value of a contour integral — you only need to know it is small, or that it cannot be larger than some amount. The ML inequality is the workhorse estimate that delivers an upper bound with almost no effort. The name is a mnemonic: M for the maximum size of the integrand on the path, L for the length of the path.
Precisely: if f is continuous on a contour C, and |f(z)| is at most M for every z on C, and L is the length of C, then the absolute value of the contour integral of f over C is at most M times L. The idea is intuitive — you are adding up many little contributions f(z) dz; each has size at most M times |dz|, and the total of the |dz| around the path is just its length L. So the worst the integral can be is M times L. To use it you usually find a clean upper bound M for |f| on the curve (not the exact maximum — any honest bound works) and multiply by L.
Its great use is showing that certain pieces of a contour contribute nothing in a limit. When you push an arc of a large semicircle out to infinity in residue calculus, you bound the integrand by some M that shrinks faster than the arc length L grows, and conclude the arc integral tends to 0. Without such an estimate the residue method could not close. The bound is rarely sharp, but a crude correct bound is exactly what you need to make a limit vanish.
On the circle |z| = 2, take f(z) = 1/(z^2 + 1). There |z^2 + 1| is at least |z|^2 - 1 = 4 - 1 = 3, so |f| is at most M = 1/3. The circle has length L = 2 pi times 2 = 4 pi. Hence the integral around it is at most (1/3)(4 pi) = 4 pi / 3 in absolute value.
A one-line bound on a contour integral: maximum of |f| times the length of the path.
The bound is on the absolute value only — it tells you nothing about the integral's argument or whether it is zero; a value can be far below M times L, and often is.