Cauchy's integral theorem
[[cauchy-integral-theorem|Cauchy's integral theorem]] says: if f is holomorphic on a simply connected open set U (a region with no holes), then for every closed contour γ inside U, ∫_γ f(z) dz = 0. Intuition: holomorphy is so rigid that there is a complex antiderivative locally, and integrating a derivative around a loop returns you to the start. Equivalently, you can deform the path freely — the integral only sees the homotopy class.
Cauchy's integral formula
Now the astonishing consequence. [[cauchy-integral-formula|Cauchy's integral formula]] says that for f holomorphic inside and on a positively oriented simple closed curve γ, and any point a inside, f(a) = (1 / (2πi)) ∫_γ f(z) / (z − a) dz. The value at the center is an average of boundary values. Differentiating under the integral sign even gives every derivative: f^(n)(a) = (n! / (2πi)) ∫_γ f(z) / (z − a)^(n+1) dz.
Why f(a) = (1/2pi i) integral_gamma f(z)/(z - a) dz, sketch.
By Cauchy's theorem you may shrink gamma to a tiny circle C_r of radius r
around a (the integrand is holomorphic in the annulus between them):
integral_gamma f(z)/(z-a) dz = integral_{C_r} f(z)/(z-a) dz.
Parametrize C_r: z = a + r e^{i t}, dz = i r e^{i t} dt:
integral_{C_r} f(z)/(z-a) dz
= integral_0^{2pi} f(a + r e^{i t}) / (r e^{i t}) * (i r e^{i t}) dt
= i * integral_0^{2pi} f(a + r e^{i t}) dt.
Let r -> 0. f is continuous, so f(a + r e^{i t}) -> f(a) uniformly in t:
-> i * integral_0^{2pi} f(a) dt = i * 2pi * f(a) = 2pi i * f(a).
The left side did not depend on r, so it EQUALS 2pi i * f(a).
Divide by 2pi i: f(a) = (1/2pi i) integral_gamma f(z)/(z-a) dz. QEDLiouville and the fundamental theorem of algebra
The derivative formula gives Cauchy's estimate: |f'(a)| ≤ M·R / R² = M / R on a circle of radius R where |f| ≤ M. [[liouville-theorem|Liouville's theorem]] drops out: a bounded entire function is constant. From Liouville the fundamental theorem of algebra follows in a few lines — every nonconstant polynomial has a complex root. This is the kind of leverage the integral formula gives you.
Liouville's theorem from Cauchy's estimate. Suppose f is entire and bounded: |f(z)| <= M for all z. Fix any point a. On the circle |z - a| = R, the n=1 derivative formula gives |f'(a)| = | (1/2pi i) integral f(z)/(z-a)^2 dz |. Apply the ML inequality on that circle: |f| <= M, |z-a|^2 = R^2, length = 2pi R: |f'(a)| <= (1/2pi) * ( M / R^2 ) * (2pi R) = M / R. This holds for EVERY R > 0. Let R -> infinity: |f'(a)| <= 0, so f'(a) = 0. Since a was arbitrary, f' = 0 everywhere, hence f is constant. QED Fundamental theorem of algebra (sketch). Let p(z) be a nonconstant polynomial with NO root. Then 1/p(z) is entire. As |z| -> infinity, |p(z)| -> infinity, so 1/p(z) -> 0; thus 1/p is bounded. By Liouville, 1/p is constant, so p is constant -- contradiction. Therefore p has a root. QED