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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*: [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/
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