Show that Complex-differentiable functions are always Real-diff.

[Maybe_Memorie]

Homework Statement

"Show that Complex-differentiable functions are always Real-differetiable."

I missed the lecture where the proof of this was given and I'm trying to figure it out.

Homework Equations

Cauchy-Riemann, u_x = v_y, u_y = -v_x

The Attempt at a Solution

Suppose a function f(z) = u + iv is C-differentiable, so it satisfies the Cauchy-Riemann equations everywhere. So u_x = v_y, u_y = -v_x at all points.

A function f(x,y) is R-differentiable if f(x,y) = f(x₀,y₀) + f(x₀, y₀)(x-x₀)(y-y₀) + O(Δx,Δy)
and O is such that O(Δx,Δy)/(||(Δx,Δy)||) -> 0 as ||(Δx,Δy)|| -> 0.

Firstly, my notes are a bit sketchy so it's possible my definition of R-diff is wrong.

I'm not really sure how to use the C-R equations with this definition either. It's not as if I can just substitute some stuff into get the required result...

Any hints/tips/suggestions?


Edit: Oh wait, I just realized I can apply to definition to R-diff to both u and v, since they're both functions of x and y. Will report back with where this leads.

Edit 2: That didn't really go anywhere. I realized I should be using the C-R equations to show that f satisfies the definition of R-diff.

I'm kind of lost.

 

[vanhees71]

Write down what it means that $f$ is differentiable as a complex function in the same way as you did for differentiability as a function $f:\mathbb{R}^2 \rightarrow \mathbb{C}$. Then express the differentiability criterion in the sense of a complex function in terms of the real and imaginary parts.

 

[Maybe_Memorie]

Okay, for f to be differentiable as a complex function means that the limit
f'(z₀) = lim, z->z₀ [f(z) - f(z₀)/(z - z₀) exists.

This has led me to believe that the question is basically a fancy way of asking for the derivation of the C-R equations, or at least the first half of it.

Is that correct? I can express z, z₀ in terms of x,y etc, and then just apply the conditions of R-differentiability to the real and imaginary parts, and I'll eventually just end up with the Cauchy-Riemann equations.