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#### Derivative of g(z) = Arg(z) and f(z) = Re(z) + Im(z)?

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...


 0
2/10/2005 8:09:30 AM
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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


 0
astanoff (103)
2/11/2005 8:45:03 AM
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/


 0
akoz (2415)
2/12/2005 7:08:46 AM
I noticed that this is corrected in Math 5.1
Arg'[1+2I] does not return an evaluated value
yehuda

On Fri, 11 Feb 2005 03:33:32 -0500 (EST), Valeri Astanoff
<astanoff@yahoo.fr> 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
>
>


 0
bsyehuda (280)
2/12/2005 7:37:05 AM
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