       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:= Arg'[1 + 2 I] // N
>
> Out= -0.4
>
>
> Valeri
>
>
>

Actually Mathematica does not return the a value for the exact
derivative:

Arg'

Arg'

But it does return a numerical value

N[Arg']

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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