Re: a^n*b^n != (a*b)^n
- To: mathgroup@smc.vnet.net
- Subject: [mg12298] Re: a^n*b^n != (a*b)^n
- From: Jeff Callahan <jcallaha@uiuc.edu>
- Date: Thu, 7 May 1998 18:51:35 -0400
- Organization: University of Illinois at Urbana-Champaign
- References: <6imi04$gvl@smc.vnet.net>
Michael Milirud wrote: > > I always considered it trivial that a^n*b^n == (a*b)^n when a,b are > complex and n is real. However: this is most definitely NOT trivial. it holds for natural numbers n, where a and b are complex, but not necessarily with other n. the problem enters because there are many solutions to the equation x^n = k for any particular k. thus some ambiguity exists in taking a (note that i say _a_ and not _the_) square root of -4 or a sqare root of -6, as below. this ambiguity is what explains your problem. for example, had you chosen Sqrt[-4] == -2i (it is easy to verify (-2i)^2 == -4), your line of 'equalities' would have been true. > 6 == Sqrt[36] == Sqrt[-4*-9] == Sqrt[-4]*Sqrt[-9] == 2i*3i == 6*i^2 == > -6 hope this helps clear things up at least a bit... jdc.