simplify Arg[E^(I x)]?

[Neal Becker]

Why doesn't this simplify?

Fullsimplify[Arg[E^(I x)], x \[Element] Reals]

I expect to get 'x'.

Is there some way I can get this to simplify?

[Bob Hanlon]

Because it is only correct for a small domain

Plot[Arg[E^(I x)], {x, -2 Pi, 2 Pi}]

Arg[E^(I x)] /. {{x -> -Pi}, {x -> Pi}}

{Pi, Pi}

FullSimplify[Arg[E^(I x)], -Pi < x <= Pi]

x


Bob Hanlon

Post by Neal Becker
Why doesn't this simplify?
Fullsimplify[Arg[E^(I x)], x \[Element] Reals]
I expect to get 'x'.
Is there some way I can get this to simplify?

[Andrzej Kozlowski]

The first and obvious problem is that it should be FullSimplify and not
Fullsimplify. But fixing that won't help since your expected answer is
incorrect. All you need to do is to look up Mathematica's help for Arg
and you will see:

The result from Arg[z] is always between -\[Pi] and +\[Pi]

But then, it is immediately obvious that if you take an x not in this
range than the answer is not going to be x, for example:

Arg[Exp[3 Pi/2 I]]

-(\[Pi]/2)

Also, this tells you what you have to do to get the answer you are
looking for:

FullSimplify[Arg[Exp[I x]], -Pi < x <= Pi]

x


Andrzej Kozlowski

Post by Neal Becker
Why doesn't this simplify?
Fullsimplify[Arg[E^(I x)], x \[Element] Reals]
I expect to get 'x'.
Is there some way I can get this to simplify?

[Neal Becker]

So can I specify I just want the principle value?

First, I presume you just made a typo in your message -- it should be
FullSimplify, with an upper-case S, of course.
Second, one reason the expression doesn't simplify to x is that such a
Arg[Exp[(2 Pi + Pi/2) I]]
has value Pi/2 rather than 2 Pi + Pi/2.

Post by Neal Becker
Why doesn't this simplify?
Fullsimplify[Arg[E^(I x)], x \[Element] Reals]
I expect to get 'x'.
Is there some way I can get this to simplify?

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Amherst, MA 01003-9305

[Anthony Hodsdon]

The following seems to work:

In[24]:= FullSimplify[Arg[E^((2 Pi n + x) I)], x > -Pi && x <= Pi &&
Element[n, Integers]]
Out[24]= x

Does that help?

--Anthony

-----Original Message-----
From: Neal Becker [mailto:***@gmail.com]
Sent: Saturday, March 31, 2012 1:42 AM
Subject: simplify Arg[E^(I x)]?

Why doesn't this simplify?

Fullsimplify[Arg[E^(I x)], x \[Element] Reals]

I expect to get 'x'.

Is there some way I can get this to simplify?

[Murray Eisenberg]

First, I presume you just made a typo in your message -- it should be
FullSimplify, with an upper-case S, of course.

Second, one reason the expression doesn't simplify to x is that such a
result would be wrong! For example:

Arg[Exp[(2 Pi + Pi/2) I]]

has value Pi/2 rather than 2 Pi + Pi/2.

Post by Neal Becker
Why doesn't this simplify?
Fullsimplify[Arg[E^(I x)], x \[Element] Reals]
I expect to get 'x'.
Is there some way I can get this to simplify?

[David Reiss]

For a given value of x, Arg[Exp[I x]] is not necessarily equal to x:

Plot[Arg[Exp[I x]], {x, 0, 4 Pi}]

So, to get the answer you want, you need to restrict x as in:

FullSimplify[Arg[E^(I x)], {x > -\[Pi], x <= \[Pi]}]

--David

Post by Neal Becker
Why doesn't this simplify?
Fullsimplify[Arg[E^(I x)], x \[Element] Reals]
I expect to get 'x'.
Is there some way I can get this to simplify?

[Neal Becker]

FullSimplify[Arg[E^((x) I)], x > -Pi && x <= Pi]

This seems to do what I wanted. Thanks!

Post by Anthony Hodsdon
In[24]:= FullSimplify[Arg[E^((2 Pi n + x) I)], x > -Pi && x <= Pi &&
Element[n, Integers]]
Out[24]= x
Does that help?
--Anthony
-----Original Message-----
Sent: Saturday, March 31, 2012 1:42 AM
Subject: simplify Arg[E^(I x)]?
Why doesn't this simplify?
Fullsimplify[Arg[E^(I x)], x \[Element] Reals]
I expect to get 'x'.
Is there some way I can get this to simplify?