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September 7th, 2014, 05:15 AM   #1
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partial f(z bar)/partial z = partial f(z bar) / partial z bar, this cannot be right?

Hello there,

I let $\sigma(z) = \overline{z}$ and $g: \Omega \to \mathbb{C}$. We let $f$ be some real differentiable function such that $g(z) = f(\overline{z})$. Then I found the following, which cannot be true.

\[
\frac{\partial g(z)}{\partial z} = \frac{\partial f(\overline{z})}{\partial z} = \frac{1}{2}\left( \frac{\partial f(\overline{z})}{\partial x} -i \frac{\partial f(\overline{z})}{\partial y} \right) = \frac{1}{2}\left( \frac{\partial f}{\partial x}\frac{\partial \sigma}{\partial x} - i \frac{\partial f}{\partial y}\frac{\partial \sigma}{\partial y} \right) = \frac{1}{2}\left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right) = \frac{\partial f(\overline{z})}{\partial \overline{z}}
\]

The most counterinutative problem is that $\frac{\partial f(\overline{z})}{\partial z} = \frac{\partial f(\overline{z})}{\partial \overline{z}}$ for all $f$, which sounds like nonsense. What is wrong?

Thank you for your time.

Kind regards,
Marius
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September 7th, 2014, 05:25 AM   #2
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Is σ analytic?
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September 7th, 2014, 06:00 AM   #3
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Quote:
Originally Posted by ZardoZ View Post
Is σ analytic?
No it is not, but it is real differentiable. What are you trying to say?

Thank you for your time.

Kind regards,
Marius
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September 7th, 2014, 06:22 AM   #4
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You are taking the composition of an analytic with a non analytic function.
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September 7th, 2014, 06:39 AM   #5
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How do you know that $f$ is analytic? Its real differentiable, but not holomorphic, hence not analytic, so neiher $f$ nor $\sigma$ are analytic.

Though, why would it be a problem if one of them were not analytic whilst the other was?

Thanks.

Marius
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September 7th, 2014, 07:03 AM   #6
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I didn't notice the "real", ok so this makes things worse.
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