Re: When is Exp[z]==Exp[w]??
- To: mathgroup at smc.vnet.net
- Subject: [mg113656] Re: When is Exp[z]==Exp[w]??
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sat, 6 Nov 2010 05:00:28 -0500 (EST)
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