       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, Integers] &&
w == 2*I*Pi*C + 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, Integers] && E^z != 0 && w == (2*I)*Pi*C + Log[E^z]

How can Mathematica be forced to simplify this to what is the fact,
namely, the following?

Element[C, Integers]&& w == (2*I)*Pi*C + 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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