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Cauchy-Riemann --> Laplace equations

If a function satisfies the Cauchy-Riemann equations, then it also satisfies the Laplace equation.

(1) Proof/derivation?

(2) Is the reverse also true? (I'm guessing not..)
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InteractiveMind
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3 Solutions

Billing EngineerCommented:
This looks like homework... sorry, you know the rules, InteractiveMind...
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Author Commented:
haha, I can see why you might think that, but I can assure you it's not.
I've finished school now, and won't be starting university for another couple of months (plus, it's the summer! ;)).

I'm just curious.
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Billing EngineerCommented:
http://en.wikipedia.org/wiki/Laplace's_equation
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Commented:
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Author Commented:
The Mathworld site provides a simple proof that if a complex function satisfies the Cauchy-Riemann equations, then it also satisfies the Laplace equation.

And it would appear that the reverse proof results in some additional constants -- so I think the answer to (2) will be a "no".

However, I do have a question about the derivation provided on the Mathworld page: it seems to make two assumptions:

i) That u+iv is second differentiable; and
ii) that D/(DxDy) = D/(DyDx) (D for Delta).

How to justify these?
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Commented:
If you have the time and patience to read the following forum threads, it is an interesting discussion on this subject:
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