# [Help]Cauchy-Riemann Equations Proof

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 :)

## Answers and Replies

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.

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...

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A. Bahat
How do we prove that if a complex function satisfies the Cauchy-Riemann equations, then it's differentiable.
This as stated is actually not true. We also need to assume that the partial derivatives are continuous.

kostas230
This as stated is actually not true. We also need to assume that the partial derivatives are continuous.

My bad. Sorry :shy:
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...

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.