Only three *of* each of those there a+bi complexes?
Today's "image special" is even a better deal.
How about (3+7i) of those there a+bi complex numbers? If you take 3 of those there a+bi complexes, you can have the 7i(mage) of them, for free ... today only!
But read the directions, carefully. For, 7i of each of those there a+bi complexes is actually 7 of their rotational images ... those there (-b+ai) complexes ... and the combination will vector you out somewhere between 3(a+bi) and 7(-b+ai).
Enjoy your trip! ((:-)>
- -------------------------------------------------- From: "Joe Niederberger" <niederberger@comcast.net> Sent: Sunday, September 09, 2012 12:47 AM To: <math-teach@mathforum.org> Subject: Re: Non-Euclidean Arithmetic
> Hmmm.... > I'll take 3 *of* those there complex numbers. > That explains it??? > > Joe N