How does that work with quaternions H or octonions O ? You can't just switch one coefficient. (Or you can, but then you're just operating in a single complex subspace of H or O, so it's really the same thing.) --Dan On 2013-04-26, at 5:50 AM, Fred lunnon wrote:
Etc. for quaternions and octonions, where available ... WFL
On 4/26/13, Warut Roonguthai <warut822@gmail.com> wrote:
I don't know who discovered this, but I know that Pythagorean triples can be obtained from the fact that |z^2| = |z|^2. This is similar to the way you can prove that the product of numbers representable as sums of two squares is also representable as a sum of two squares by using the fact that |z1 * z2| = |z1| * |z2|.
On Fri, Apr 26, 2013 at 3:04 PM, Bill Gosper <billgosper@gmail.com> wrote:
In[570]:= (1+2*I)/(2+1*I) Out[570]= 4/5+(3 I)/5
In[571]:= (1+3*I)/(3+1*I) Out[571]= 3/5+(4 I)/5
In[572]:= (2+3*I)/(3+2*I) Out[572]= 12/13+(5 I)/13
In[573]:= (1+4*I)/(4+1*I) Out[573]= 8/17+(15 I)/17
In[586]:= Simplify/@ComplexExpand[(a+I*b)/(b+I*a)] Out[586]= (2 a b)/(a^2+b^2)-(I (a^2-b^2))/(a^2+b^2) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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