the residue theorem
Some integrals along a closed loop in the complex plane look hopeless, yet the answer depends only on a handful of special points inside the loop where the function blows up (its poles). The residue theorem says the whole loop integral is just 2 pi i times the sum of little numbers -- the residues -- one attached to each enclosed pole. It turns integration into the algebra of finding those numbers.
Precisely, if f is holomorphic on and inside a simple closed contour C except at isolated singularities z_k inside, then the contour integral of f dz around C (counterclockwise) equals 2 pi i times the sum over k of Res(f, z_k). The residue Res(f, z_k) is the coefficient of the 1/(z - z_k) term in f's Laurent series about z_k. For a simple pole, Res(f, z_k) = limit as z -> z_k of (z - z_k) f(z); for a pole of order n it is (1/(n-1)!) times the (n-1)-th derivative of (z - z_k)^n f(z), evaluated at z_k.
This is the physicist's most powerful integration tool. Closing a real-axis integral with a large semicircle in the upper half-plane (where the arc contributes nothing) evaluates otherwise-intractable real integrals -- for instance the integral of dx/(1 + x^2) from -infinity to infinity = pi, from the single pole at z = i. It underlies the evaluation of Fourier and inverse-Laplace transforms, propagators, and Green's functions, and the causal (retarded versus advanced) prescriptions come precisely from which side of a pole the contour passes.
The integral over the unit circle of dz/z = 2 pi i. The only pole is at z = 0, a simple pole with residue 1, so the answer is 2 pi i x 1 -- the single most important contour integral in all of physics.
One enclosed pole of residue 1 gives 2 pi i.
The theorem needs the singularities to be isolated poles (or removable). A branch point is not a pole -- you cannot assign it a residue, and a contour must be cut so as not to enclose it.