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From: Pianiel on 17 Dec 2009 07:22
Dear All,

With Mathematica 7.0.1, the following expression;

Simplify[ Im[E^(-2 I a) b], Assumptions -> Element[a, Reals] &&
Element[b, Reals]]

gives

-b Sin[2 a]

But:

Simplify[ Im[E^(-2 I a) 1/b], Assumptions -> Element[a, Reals] &&
Element[b, Reals]]

gives

Im[E^(-2 I a)/b]

So we see that the Im[] is not simplified. Why?

Any idea how to help mathematica to handle the simplification shown in
the second example. It seems to be so similar to the first kind...

Thanks in advance

Pianiel

From: DC on 18 Dec 2009 06:24
You might try

In[3]:= ComplexExpand[Im[E^(-2 I a) 1/b], TargetFunctions -> {Re, Im}]

Out[3]= -(Sin[2 a]/b)

-Francesco

On 12/17/2009 12:22 PM, Pianiel wrote:
> Dear All,
>
> With Mathematica 7.0.1, the following expression;
>
> Simplify[ Im[E^(-2 I a) b], Assumptions -> Element[a, Reals]&&
> Element[b, Reals]]
>
> gives
>
> -b Sin[2 a]
>
> But:
>
> Simplify[ Im[E^(-2 I a) 1/b], Assumptions -> Element[a, Reals]&&
> Element[b, Reals]]
>
> gives
>
> Im[E^(-2 I a)/b]
>
> So we see that the Im[] is not simplified. Why?
>
> Any idea how to help mathematica to handle the simplification shown in
> the second example. It seems to be so similar to the first kind...
>
> Thanks in advance
>
> Pianiel
>

From: Bob Hanlon on 19 Dec 2009 06:25

Simplify[Im[E^(-2 I a) b],
Element[{a, b}, Reals]]

(-b)*Sin[2*a]

In the second case, also specify that b is not zero

Simplify[Im[E^(-2 I a) 1/b],
Element[{a, b}, Reals] && b != 0]

-(Sin[2*a]/b)


Bob Hanlon

---- DC <b.gatessucks(a)gmail.com> wrote:

=============
You might try

In[3]:= ComplexExpand[Im[E^(-2 I a) 1/b], TargetFunctions -> {Re, Im}]

Out[3]= -(Sin[2 a]/b)

-Francesco

On 12/17/2009 12:22 PM, Pianiel wrote:
> Dear All,
>
> With Mathematica 7.0.1, the following expression;
>
> Simplify[ Im[E^(-2 I a) b], Assumptions -> Element[a, Reals]&&
> Element[b, Reals]]
>
> gives
>
> -b Sin[2 a]
>
> But:
>
> Simplify[ Im[E^(-2 I a) 1/b], Assumptions -> Element[a, Reals]&&
> Element[b, Reals]]
>
> gives
>
> Im[E^(-2 I a)/b]
>
> So we see that the Im[] is not simplified. Why?
>
> Any idea how to help mathematica to handle the simplification shown in
> the second example. It seems to be so similar to the first kind...
>
> Thanks in advance
>
> Pianiel
>

From: Peter Breitfeld on 19 Dec 2009 06:26

Simplify doesn't like the b in the denominator (I don't know why), but

Simplify[Im[Exp[-2 I a] 1/b], {a, b} \[Element] Reals && b != 0]

will work. But usually it's easier to use ComplexExpand instead of
Simplify, which treats all variables as Reals.


Pianiel wrote:

> Dear All,
>
> With Mathematica 7.0.1, the following expression;
>
> Simplify[ Im[E^(-2 I a) b], Assumptions -> Element[a, Reals] &&
> Element[b, Reals]]
>
> gives
>
> -b Sin[2 a]
>
> But:
>
> Simplify[ Im[E^(-2 I a) 1/b], Assumptions -> Element[a, Reals] &&
> Element[b, Reals]]
>
> gives
>
> Im[E^(-2 I a)/b]
>
> So we see that the Im[] is not simplified. Why?
>
> Any idea how to help mathematica to handle the simplification shown in
> the second example. It seems to be so similar to the first kind...
>
> Thanks in advance
>
> Pianiel
>

--
_________________________________________________________________
Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de

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