My Math Forum partial f(z bar)/partial z = partial f(z bar) / partial z bar, this cannot be right?

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 September 7th, 2014, 06:15 AM #1 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 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
 September 7th, 2014, 06:25 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Is σ analytic?
September 7th, 2014, 07:00 AM   #3
Senior Member

Joined: Sep 2010
From: Oslo, Norway

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Quote:
 Originally Posted by ZardoZ Is σ analytic?
No it is not, but it is real differentiable. What are you trying to say?

Kind regards,
Marius

 September 7th, 2014, 07:22 AM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus You are taking the composition of an analytic with a non analytic function.
 September 7th, 2014, 07:39 AM #5 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 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
 September 7th, 2014, 08:03 AM #6 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus I didn't notice the "real", ok so this makes things worse.

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