Re: Problem with the Derivative of a Arg-function

• To: mathgroup at smc.vnet.net
• Subject: [mg48512] Re: [mg48482] Problem with the Derivative of a Arg-function
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Fri, 4 Jun 2004 04:49:27 -0400 (EDT)
• References: <200406020821.EAA15504@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

On 2 Jun 2004, at 17:21, Alex Klishko wrote:

> Hello group,
>
> Let us define a simple complex function Exp[-j*x], j is a square root
> of -1.
> Fase of this function is computed by Arg -function.
> If x is a real number the fase is equal -x and it's derivative is -1.
> If x is a complex number the fase is equal Re[-x].
> But if we would write:
>
> f[x_] = Arg[E^(-j*x\)];  N[D[f[x], x]]\  /. \ x -> 5+j*3
>
> we would get a complex number not equal -1.
>
> In case of a real x, this problem may be solved by
> ComplexExpand-function:
>
> f[x_] = ComplexExpand[Arg[E^(-j*x\)]];  N[D[f[x], x]]\  /. \ x -> 5
>
> we would get -1.
>
> In case of a complex x, Matematica gives a wrong solution.
>
> Would you correct me if I make a mistake.
>
> Wishes
> Alex Klishko
>
>
>
I am not sure what you mean by "fase"? Is that just Arg[E^(-I*x)]? Are
you saying that this is -Re[x] for all complex x? Well, that certianly
is not true, even for real ones!
For example

Arg[E^(-I*20)]

-20+6 Pi

or more famously:

E^(I*Pi)
-1

Arg[-1]
Pi

But

FullSimplify[ComplexExpand[Arg[E^((-I)*x)],
TargetFunctions -> {Re, Im}], -Pi/2 < x < Pi/2]

-x

Maybe I have misunderstood something?

Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/

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