# [Help]Cauchy-Riemann Equations Proof

1. Sep 29, 2012

### kostas230

Can someone give me a proof of the of Cauchy-Riemann equations? I understand how a differentiable complex function f(x,y)=u(x,y)+iv(x,y) satisfies the Cauchy-Riemann equations. How do we prove that if a complex function satisfies the Cauchy-Riemann equations, then it's differentiable. I didn't quite get the proof of Churchill's book, and I'm looking for a more elegant thought. Thanks in advance :)

2. Sep 29, 2012

### Klungo

Well. I'm only familiar with the proof in the: if f is differentiable, then f satisfies cauchy riemann equations. The converse, Is actually a bit more complicated.

3. Sep 30, 2012

### raopeng

Eh a not so delicate way is:
$u(x + dx, y + dy) - u(x, y)= \frac{\partial{u}}{\partial{x}}dx + \frac{\partial{u}}{\partial{y}}dy$. The same can be written for v.
Then using the Equation to write $df = (\frac{\partial{u}}{\partial{x}} + i\frac{\partial{v}}{\partial{x}})(dx + idy)$. then we can see $(\frac{\partial{u}}{\partial{x}} + i\frac{\partial{v}}{\partial{x}})$ is independent of the direction you choose, hence differentiable.

Don't know if it is completely correct...

Last edited: Sep 30, 2012
4. Sep 30, 2012

### A. Bahat

This as stated is actually not true. We also need to assume that the partial derivatives are continuous.

5. Sep 30, 2012

### kostas230

OK, let's assume that its partial derivatives are continuous. How do we prove that it's differentiable? I was thinking a straightforward proof using the definition of the derivative, but I don't think it's gonna give me any good result...

6. Oct 1, 2012

### homeomorphic

Well, the big idea here is that a linear transformation acts on R^2, just like multiplication by a complex number does if it satisfies the Cauchy-Riemann equations. So, basically, it's condition for a linear transformation to be a dilation (i.e. a rotation composed with rescaling). That's probably the main thing to understand. You're just applying this fact to the differential of a function from R^2 to R^2.

Going from complex differentiability to this condition isn't too bad. The converse has some technical details to prove rigorously. If you want more intuition, read Visual Complex Analysis.

I Help with expression $F(it)-F(-it)$ in the Abel-Plana form Aug 7, 2017