The same definition, a much stronger demand
On the real line, the derivative is the limit of a difference quotient: f'(a) = lim (f(x) − f(a)) / (x − a) as x → a. For complex differentiability we write the exact same formula, with z and a now complex numbers and the quotient a complex number divided by a complex number. The catch is hidden in the words “as z → a”: in the plane, z can approach a from infinitely many directions, along any path. The single complex limit must come out the same value for all of them.
Forcing the Cauchy–Riemann equations
Write f(z) = u(x, y) + i v(x, y), where z = x + i y and u, v are real-valued. Approach a along the real direction (z = a + h, h real → 0) and then along the imaginary direction (z = a + i k, k real → 0). If f is complex differentiable, both give the same f'(a). Equating them produces the [[cauchy-riemann-equations|Cauchy–Riemann equations]]: ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x. They are exactly the necessary price of one limit that does not depend on direction.
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. QEDA quick verification
Let's test two functions. The squaring map f(z) = z² should be complex differentiable everywhere; the conjugation map g(z) = z̄ should fail. Writing each as u + i v and checking the Cauchy–Riemann system settles both in a line or two.
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.