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Re: simplify Arg[E^(I x)]?

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]


Bob Hanlon

On Sat, Mar 31, 2012 at 4:42 AM, Neal Becker <ndbecker2 at> wrote:
> 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

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