Holomorphic: differentiable on an open set
A function is [[holomorphic-function|holomorphic]] on an open set U if it is complex differentiable at every point of U. The word “open” matters: differentiability at a single point is almost useless, but differentiability on a whole neighborhood unlocks a cascade of theorems. A function holomorphic on all of C is called [[entire-function|entire]] — polynomials, exp(z), sin(z), and cos(z) are entire.
Integrating along a curve
A [[contour-integral|contour integral]] integrates f along a path. Parametrize a smooth curve γ by z = γ(t) for t in [a, b]. Then define ∫_γ f(z) dz = ∫_a^b f(γ(t)) γ'(t) dt, a perfectly ordinary integral of a complex-valued function of the real variable t. The factor γ'(t) is the chain-rule “dz”. The value is independent of how you parametrize (as long as you keep the direction).
Workhorse computation: integrate z^n around the unit circle.
Let gamma(t) = e^{i t}, t in [0, 2*pi]. Then gamma'(t) = i e^{i t}.
integral_gamma z^n dz
= integral_0^{2pi} (e^{i t})^n * (i e^{i t}) dt
= i * integral_0^{2pi} e^{i (n+1) t} dt.
Case n = -1:
= i * integral_0^{2pi} e^{0} dt = i * (2*pi) = 2*pi*i.
Case n != -1 (integer):
integral_0^{2pi} e^{i m t} dt = [ e^{i m t} / (i m) ]_0^{2pi}
= ( e^{i m 2pi} - 1 ) / (i m) = (1 - 1)/(i m) = 0, m = n+1 != 0.
So the integral is 0.
Result: integral over |z|=1 of z^n dz = 2*pi*i if n = -1, else 0.The ML estimate
Almost every proof in complex analysis uses one bound, the ML inequality: if |f(z)| ≤ M on γ and γ has length L, then |∫_γ f dz| ≤ M·L. It comes straight from the triangle inequality for integrals. This single estimate is what lets us say an integral is small by making either the bound M or the length L small.
ML inequality and its proof sketch.
Claim: | integral_gamma f(z) dz | <= M * L,
where M = max over gamma of |f|, and L = length(gamma).
Proof. | integral_a^b f(gamma(t)) gamma'(t) dt |
<= integral_a^b | f(gamma(t)) | * | gamma'(t) | dt (triangle ineq. for integrals)
<= M * integral_a^b | gamma'(t) | dt
= M * L, since integral_a^b |gamma'(t)| dt = arclength = L. QED
Sample use: bound integral over |z|=2 of 1/(z^2+1) dz.
On |z| = 2: |z^2 + 1| >= |z|^2 - 1 = 4 - 1 = 3, so |1/(z^2+1)| <= 1/3 = M.
Length L = 2*pi*2 = 4*pi.
Hence | integral | <= (1/3)(4*pi) = 4*pi/3. (a crude but valid bound)