A N Niel schrieb:   
   >   
   > In article <525C3859.9FE922D5@freenet.de>,    
   > wrote:   
   >   
   > > Over which extended region of complex numbers a and b does the   
   > > relation   
   > >   
   > >   ABS(a^2*ABS(b - a)^2 + ABS(a)^2*(b^2 - a^2))   
   > >   = ABS(a)*ABS(b - a)*(ABS(a)^2 + ABS(b)^2 - ABS(b - a)^2)   
   > >   
   > > hold?   
   > >   
   >   
   > how about:  a=0 or Re(b/a) >= 0.   
      
   Splendid. This makes Mathematica's answer look downright shabby. Your   
   anonymous CAS gets full points plus!   
      
   In fact, the condition is equivalent to   
      
     ABS(a)^2 + ABS(b)^2 >= ABS(b - a)^2   
      
   and the amended equality   
      
     ABS(a^2*ABS(b - a)^2 + ABS(a)^2*(b^2 - a^2))   
     = ABS(a)*ABS(b - a)*ABS(ABS(a)^2 + ABS(b)^2 - ABS(b - a)^2)   
      
   holds for all complex a,b.   
      
   Martin.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)