Re: Re: Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?
- To: mathgroup at smc.vnet.net
- Subject: [mg54156] Re: [mg54120] Re: Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sat, 12 Feb 2005 01:57:04 -0500 (EST)
- References: <cuf4rq$gk0$1@smc.vnet.net> <200502110833.DAA09153@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
On 11 Feb 2005, at 08:33, Valeri Astanoff 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 > > > Actually Mathematica does not return the a value for the exact derivative: Arg'[1] Arg'[1] But it does return a numerical value N[Arg'[1]] 0. Of course what it is doing is using the "directional derivative along the real axis": g[z_] := N[Limit[(Arg[z + t] - Arg[z])/t, t -> 0]] e.g. g[2 I + 3] -0.153846 N[Arg'[2 I + 3]] -0.153846 As Arg is not-differentiable this derivative will not in general agree with with directional derivatives in other directions: Chop[N[Limit[(Arg[2*I + 3 + t] - Arg[2*I + 3])/t, t -> 0, Direction -> I]]] -0.23076923076923078*I Andrzej Kozlowski Chiba, Japan http://www.akikoz.net/~andrzej/ http://www.mimuw.edu.pl/~akoz/
- References:
- Re: Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?
- From: "Valeri Astanoff" <astanoff@yahoo.fr>
- Re: Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?