复可导与柯西–黎曼方程

相同的定义，更强的要求

在实数轴上，导数是差商的极限：当 x → a 时，f'(a) = lim (f(x) − f(a)) / (x − a)。对于复可导，我们写下 完全相同的公式，只是此时 z 和 a 是复数，商也是复数除以复数。玄机藏在“当 z → a 时”这句话里：在平面上，z 可以沿任意路径、从无穷多个方向逼近 a。这一个复极限必须对所有方向给出 相同的值。

逼出柯西–黎曼方程

记 f(z) = u(x, y) + i v(x, y)，其中 z = x + i y，u、v 取实值。先沿实方向逼近 a（z = a + h，h 为实数 → 0），再沿虚方向逼近（z = a + i k，k 为实数 → 0）。若 f 复可导，两者必给出相同的 f'(a)。令二者相等便得到 [[cauchy-riemann-equations|柯西–黎曼方程]]：∂u/∂x = ∂v/∂y 以及 ∂u/∂y = −∂v/∂x。它们正是“极限与方向无关”所必须付出的必要代价。

Claim: if f = u + i v is complex differentiable at a, then CR holds at a.

Let f'(a) = L (a single complex number). Then for z -> a,
   (f(z) - f(a)) / (z - a) -> L,  along EVERY path.

Path 1 (horizontal):  z = a + h,  h real, h -> 0,  so z - a = h.
   (f(a+h) - f(a)) / h
     = ( u(x+h,y) - u(x,y) )/h  +  i ( v(x+h,y) - v(x,y) )/h
     -> u_x + i v_x.
So  L = u_x + i v_x.            ...(1)

Path 2 (vertical):  z = a + i k,  k real, k -> 0,  so z - a = i k.
   (f(a+ik) - f(a)) / (i k)
     = ( u(x,y+k) - u(x,y) )/(i k)  +  i ( v(x,y+k) - v(x,y) )/(i k)
     = (1/i)( u_y + i v_y )  =  v_y - i u_y     (since 1/i = -i).
So  L = v_y - i u_y.            ...(2)

Equate real and imaginary parts of (1) and (2):
   u_x = v_y      and      v_x = -u_y.
These are the Cauchy-Riemann equations.   QED

两个逼近方向相等——方程自然落地。

一个快速验证

我们检验两个函数。平方映射 f(z) = z² 应当处处复可导；共轭映射 g(z) = z̄ 则应当失败。把每个写成 u + i v 并核对柯西–黎曼方程组，几行就能判定。

Example A:  f(z) = z^2.
  z = x + i y,  so z^2 = (x^2 - y^2) + i (2 x y).
  u = x^2 - y^2,   v = 2 x y.
  u_x = 2x,   v_y = 2x   -> u_x = v_y    OK
  u_y = -2y,  v_x = 2y   -> u_y = -v_x   OK
  CR holds everywhere; partials are continuous, so f is holomorphic on all of C.
  (And indeed f'(z) = u_x + i v_x = 2x + i 2y = 2z, as expected.)

Example B:  g(z) = conj(z) = x - i y.
  u = x,   v = -y.
  u_x = 1,   v_y = -1   -> u_x = v_y  becomes  1 = -1   FAILS.
  CR fails at every point, so g is complex differentiable NOWHERE,
  even though u and v are as smooth as could be.

z² 通过；共轭尽管极其光滑，却处处失败。