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

• To: mathgroup at smc.vnet.net
• Subject: [mg54120] Re: Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?
• From: "Valeri Astanoff" <astanoff at yahoo.fr>
• Date: Fri, 11 Feb 2005 03:33:32 -0500 (EST)
• References: <cuf4rq$gk0$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

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

• Follow-Ups:
  • Re: Re: Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
  • Re: Re: Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?
    • From: yehuda ben-shimol <bsyehuda@gmail.com>