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/

**References**:**Problem with the Derivative of a Arg-function***From:*klishko@mail.ru (Alex Klishko)