柯西积分定理
[[cauchy-integral-theorem|柯西积分定理]] 说：若 f 在单连通开集 U（无洞的区域）上 全纯，则对 U 内每条闭 围道 γ 都有 ∫_γ f(z) dz = 0。直观上：全纯极其刚性，以至于局部存在复 原函数，而把一个导数沿闭路积分回到起点便得零。等价地说，你可以自由形变路径——积分只看同伦类。
柯西积分公式
现在来看那个惊人的推论。[[cauchy-integral-formula|柯西积分公式]] 说：若 f 在正向简单闭曲线 γ 内部及其上全纯，则对内部任一点 a 有 f(a) = (1 / (2πi)) ∫_γ f(z) / (z − a) dz。内部 一点的值竟是边界值的某种平均。在积分号下求导，甚至给出所有阶导数：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. QED刘维尔定理与代数基本定理
导数公式给出 柯西估计：在半径 R、其上 |f| ≤ M 的圆上，|f'(a)| ≤ M·R / R² = M / R。[[liouville-theorem|刘维尔定理]] 随之而出：有界 整函数 必为常数。由刘维尔定理，几行就能推出 代数基本定理——每个非常数多项式都有复根。这正是 积分公式 给你的那种杠杆力量。
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