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Re: When is Exp[z]==Exp[w]??
*To*: mathgroup at smc.vnet.net
*Subject*: [mg113657] Re: When is Exp[z]==Exp[w]??
*From*: Murray Eisenberg <murray at math.umass.edu>
*Date*: Sat, 6 Nov 2010 05:00:39 -0500 (EST)
OK, but it's still strange to have to add that.
And that also leaves the peculiar term Log[E^z], which is also totally
redundant (even though Log[E^z] need not equal z, but still it will
differe from z by an integer multiple of 2 Pi I).
On 11/5/2010 7:06 AM, Bob Hanlon wrote:
>
> You can "force" it by being just as redundant.
>
> Simplify[Reduce[Exp[z] == Exp[w], {z, w}], E^z != 0]
>
> Element[C[1], Integers]&&
> w == 2*I*Pi*C[1] + Log[E^z]
>
>
> Bob Hanlon
>
> ---- Murray Eisenberg<murray at math.umass.edu> wrote:
>
> =============
> Mathematica 7.0.1 gives (as InputForm of the result):
>
> Reduce[Exp[z]==Exp[w],{z,w}]
> Element[C[1], Integers]&& E^z != 0&& w == (2*I)*Pi*C[1] + Log[E^z]
>
> How can Mathematica be forced to simplify this to what is the fact,
> namely, the following?
>
> Element[C[1], Integers]&& w == (2*I)*Pi*C[1] + z
>
> (At the very least, certainly the expression E^z != 0 is redundant.)
>
--
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
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