柯西積分定理
[[cauchy-integral-theorem|柯西積分定理]] 說：若 f 在單連通開集 U（無洞的區域）上 全純，則對 U 內每條閉 圍道 γ 都有 ∫_γ f(z) dz = 0。直觀上：全純極其剛性，以至於局部存在複 原函數，而把一個導數沿閉路積分回到起點便得零。等價地說，你可以自由形變路徑——積分只看同倫類。
柯西積分公式
現在來看那個驚人的推論。[[cauchy-integral-formula|柯西積分公式]] 說：若 f 在正向簡單閉曲線 γ 內部及其上全純，則對內部任一點 a 有 f(a) = (1 / (2πi)) ∫_γ f(z) / (z − a) dz。內部 一點的值竟是邊界值的某種平均。在積分號下求導，甚至給出所有階導數：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劉維爾定理與代數基本定理
導數公式給出 柯西估計：在半徑 R、其上 |f| ≤ M 的圓上，|f'(a)| ≤ M·R / R² = M / R。[[liouville-theorem|劉維爾定理]] 隨之而出：有界 整函數 必為常數。由劉維爾定理，幾行就能推出 代數基本定理——每個非常數多項式都有複根。這正是 積分公式 給你的那種槓桿力量。
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