- #1

kostas230

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- Thread starter kostas230
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- #1

kostas230

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- #2

Klungo

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- #3

raopeng

- 86

- 0

Eh a not so delicate way is:

[itex]u(x + dx, y + dy) - u(x, y)= \frac{\partial{u}}{\partial{x}}dx + \frac{\partial{u}}{\partial{y}}dy[/itex]. The same can be written for v.

Then using the Equation to write [itex]df = (\frac{\partial{u}}{\partial{x}} + i\frac{\partial{v}}{\partial{x}})(dx + idy)[/itex]. then we can see [itex](\frac{\partial{u}}{\partial{x}} + i\frac{\partial{v}}{\partial{x}})[/itex] is independent of the direction you choose, hence differentiable.

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

[itex]u(x + dx, y + dy) - u(x, y)= \frac{\partial{u}}{\partial{x}}dx + \frac{\partial{u}}{\partial{y}}dy[/itex]. The same can be written for v.

Then using the Equation to write [itex]df = (\frac{\partial{u}}{\partial{x}} + i\frac{\partial{v}}{\partial{x}})(dx + idy)[/itex]. then we can see [itex](\frac{\partial{u}}{\partial{x}} + i\frac{\partial{v}}{\partial{x}})[/itex] is independent of the direction you choose, hence differentiable.

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

Last edited:

- #4

A. Bahat

- 150

- 0

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

- #5

kostas230

- 96

- 3

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...

- #6

homeomorphic

- 1,773

- 130

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.

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