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Re: How to use Mathematica find the solution of an

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• Subject: [mg130939] Re: How to use Mathematica find the solution of an
• From: "Eduardo M. A. M. Mendes" <emammendes at gmail.com>
• Date: Mon, 27 May 2013 04:24:24 -0400 (EDT)
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Many thanks for the reply.

If we consider

1/((I w)(1+I w)^2)

we have

-angle(I w)-2*angle(1+I w) = -180

-90-2*angle(1+ I w)=-180

-2*angle(1+i w)=-90
angle(1+iw)=45

therefore w = 1

and probably something similar for w=-1.

I feel that I do not know how Mathematica deals with the argument of a complex number. 

To be honest I would like to know what is going on under the hood of PhaseMargins.

Many thanks

Ed



On May 24, 2013, at 7:23 AM, Bob Hanlon <hanlonr357 at gmail.com> wrote:

>
> The equation does not appear to have a solution except as a limit (from
> below) but then the solution is either -1 or 1.
>
>
> eqn = Arg[-(I/((1 + I w)^2 w))] == -Pi;
>
>
> eqn // Simplify
>
>
> False
>
>
> eqn /. {{w -> -1}, {w -> 1}}
>
>
> {False, False}
>
>
> Limit[Arg[-(I/((1 + I w)^2 w))],
>  w -> -1, Direction -> 1] == -Pi
>
>
> True
>
>
> Limit[Arg[-(I/((1 + I w)^2 w))],
>  w -> 1, Direction -> 1] == -Pi
>
>
> True
>
>
>
> Bob Hanlon
>
>
>
>
> On Thu, May 23, 2013 at 4:04 AM, Eduardo M. A. M. Mendes <
> emammendes at gmail.com> wrote:
>
>> Hello
>>
>> I need to solve the following equation:
>>
>> Arg[-(I/((1+I \[Omega])^2 \[Omega]))]==-\[Pi]
>>
>> I have tried Solve (empty output), Reduce (it gives some results but not
>> the answer Omega=1) and FindRoot (it gives Omega=1 but it is a
>> numerical search).   Is there a way to get the solution not using a
>> numerical search?
>>
>> Many thanks
>>
>> Ed
>>
>> PS.   I need to solve several equation of the same kind.
>>
>>

• Follow-Ups:
  • Re: How to use Mathematica find the solution of an
    • From: Murray Eisenberg <murray@math.umass.edu>

• References:
  • How to use Mathematica find the solution of an equation when Arg is part of it?
    • From: "Eduardo M. A. M. Mendes" <emammendes@gmail.com>
  • Re: How to use Mathematica find the solution of an
    • From: Bob Hanlon <hanlonr357@gmail.com>