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Re: Re: Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?

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• Subject: [mg54182] Re: [mg54120] Re: Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?
• From: yehuda ben-shimol <bsyehuda at gmail.com>
• Date: Sat, 12 Feb 2005 01:58:41 -0500 (EST)
• References: <cuf4rq$gk0$1@smc.vnet.net> <200502110833.DAA09153@smc.vnet.net>
• Reply-to: yehuda ben-shimol <bsyehuda at gmail.com>
• Sender: owner-wri-mathgroup at wolfram.com

I noticed that this is corrected in Math 5.1
Arg'[1+2I] does not return an evaluated value
yehuda


On Fri, 11 Feb 2005 03:33:32 -0500 (EST), Valeri Astanoff
<astanoff at yahoo.fr> wrote:
> Marie wrote:
> > How do I solve a derivative of a complex function Arg(z) or Re(z) +
> > Im(z) by definition? (f(z) - f(w))/(z-w) -> f'(w) as z->w...
> 
> Let f(z) = P + I Q be the value of a complex function.
> I remember that if P is constant and Q variable
> (or P variable and Q constant)
> then f can't be complex differentiable because the
> Cauchy-Riemann conditions can't be met.
> 
> Hence Arg, Re, Im, Abs are not complex differentiable.
> 
> Curiously, Mathematica [5.0] doesn't refuse to evaluate Arg' :
> 
> In[1]:= Arg'[1 + 2 I] // N
> 
> Out[1]= -0.4
> 
> 
> Valeri
> 
>

• References:
  • Re: Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?
    • From: "Valeri Astanoff" <astanoff@yahoo.fr>