the method of steepest descent
Suppose you must estimate an integral of the form (integral over a contour C of e^(N g(z)) dz) when the number N is large. The integrand e^(N g(z)) is wildly lopsided: where the real part of g is largest the integrand is enormous, and everywhere else it is comparatively tiny. So almost the entire value of the integral comes from a small neighborhood of the point where the real part of g peaks. The method of steepest descent is a disciplined way to find that peak and read off the integral's size from the shape of the bump around it.
Because g is holomorphic, you are free to deform the contour C (Cauchy's theorem) without changing the integral. The trick is to slide C so that it passes through a saddle point z_0 — a point where g'(z_0) = 0 — and runs along the direction in which the real part of g falls off fastest on each side. Along that steepest-descent path the imaginary part of g stays constant, so there is no oscillation to spoil the estimate: e^(N g(z)) just rises to a single hump at z_0 and decays. Near z_0 you expand g(z) approximately as g(z_0) + (1/2) g''(z_0) (z - z_0)^2, a Gaussian bump, and integrate it. The leading result is (integral) approximately e^(N g(z_0)) times the square root of (2 pi / (N times |g''(z_0)|)), with a phase factor fixed by the direction of the descent path.
This is one of the great payoffs of complex analysis, genuinely different from the residue calculus: residues capture poles, while steepest descent captures the large-N growth of smooth integrals. It delivers the leading asymptotics of special functions and physical quantities — Stirling's formula for the factorial, the large-argument behaviour of Bessel and Airy functions, the saddle-point approximation in statistical mechanics and the WKB approximation in quantum mechanics. An honest caveat: it gives an asymptotic expansion, valid as N -> infinity, not an exact answer, and the error terms must be tracked if you want rigor.
To find Stirling's formula write n! = (integral from 0 to infinity of t^n e^(-t) dt) and substitute t = n s, giving n! = n^(n+1) (integral of e^(n (log s - s)) ds). Here g(s) = log s - s has g'(s) = 1/s - 1 = 0 at s = 1, the saddle, with g''(1) = -1. Steepest descent gives n! approximately e^(n(log 1 - 1)) times n^(n+1) times the square root of (2 pi / n), which tidies up to n! approximately (square root of 2 pi n) times (n/e)^n.
Stirling's formula falls straight out of the saddle at s = 1.
The method requires you to actually deform the contour onto the steepest-descent path; if the original contour cannot be moved there without crossing a singularity, you must account for the poles you cross (their residues) on the way.