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MathGroup Archive 2005

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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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