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Re: Re: Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?
*To*: mathgroup at smc.vnet.net
*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
>
>
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