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>