proof of the Cauchy-Riemann equations


Existence of complex derivative implies the Cauchy-Riemann equations.

Suppose that the complex derivativeMathworldPlanetmath

f(z)=limζ0f(z+ζ)-f(z)ζ (1)

exists for some z. This means that for all ϵ>0, there exists a ρ>0, such that for all complex ζ with |ζ|<ρ, we have

|f(z)-f(z+ζ)-f(z)ζ|<ϵ.

Henceforth, set

f=u+iv,z=x+iy.

If ζ is real, then the above limit reduces to a partial derivativeMathworldPlanetmath in x, i.e.

f(z)=fx=ux+ivx,

Taking the limit with an imaginary ζ we deduce that

f(z)=-ify=-iuy+vy.

Therefore

fx=-ify,

and breaking this relation up into its real and imaginary parts gives the Cauchy-Riemann equationsMathworldPlanetmath.

The Cauchy-Riemann equations imply the existence of a complex derivative.

Suppose that the Cauchy-Riemann equations

ux=vy,uy=-vx,

hold for a fixed (x,y)2, and that all the partial derivatives are continuous at (x,y) as well. The continuity implies that all directional derivativesPlanetmathPlanetmath exist as well. In other words, for ξ,η and ρ=ξ2+η2 we have

u(x+ξ,y+η)-u(x,y)-(ξux+ηuy)ρ0,as ρ0,

with a similar relation holding for v(x,y). Combining the two scalar relations into a vector relation we obtain

ρ-1(u(x+ξ,y+η)v(x+ξ,y+η))-(u(x,y)v(x,y))-(uxuyvxvy)(ξη)0,as ρ0.

Note that the Cauchy-Riemann equations imply that the matrix-vector product above is equivalent to the product of two complex numbersMathworldPlanetmathPlanetmath, namely

(ux+ivx)(ξ+iη).

Setting

f(z) = u(x,y)+iv(x,y),
f(z) = ux+ivx
ζ = ξ+iη

we can therefore rewrite the above limit relation as

|f(z+ζ)-f(z)-f(z)ζζ|0,as ρ0,

which is the complex limit definition of f(z) shown in (1).

Title proof of the Cauchy-Riemann equations
Canonical name ProofOfTheCauchyRiemannEquations
Date of creation 2013-03-22 12:55:39
Last modified on 2013-03-22 12:55:39
Owner rmilson (146)
Last modified by rmilson (146)
Numerical id 6
Author rmilson (146)
Entry type Proof
Classification msc 30E99
Defines complex derivative