 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, 05:15 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, 05:25 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Is σ analytic? September 7th, 2014, 06:00 AM   #3
Senior Member

Joined: Sep 2010
From: Oslo, Norway

Posts: 162
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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? Thank you for your time.

Kind regards,
Marius September 7th, 2014, 06:22 AM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 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, 06:39 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, 07:03 AM #6 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 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. Tags bar, bar or partial, partial Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post sagicoh Real Analysis 3 December 29th, 2012 09:28 AM peripatein Calculus 1 November 10th, 2012 08:23 AM chapsticks Calculus 5 March 10th, 2012 05:57 PM ZardoZ Real Analysis 4 August 5th, 2011 07:41 AM riotsandravess Calculus 1 February 13th, 2011 04:36 PM

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