
Re: what has your CAS to say about this complex equality
Posted:
Oct 14, 2013 7:37 PM


On 10/14/2013 1:30 PM, clicliclic@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? > > Martin. >
Mathematica V 9.01, says
 eq = 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)
Reduce[eq, {a, b}] 
(
Re[a] < 0 && ((Im[a] < 0 && Im[b] <= ((Re[a] Re[b])/Im[a]))
 (Im[a] == 0 && Re[b] <= 0)
 (Im[a] > 0 && Im[b] >= ((Re[a] Re[b])/Im[a]))))
 (Re[a] == 0 && ((Im[a] < 0 && Im[b] <= 0)
 Im[a] == 0
 (Im[a] > 0 && Im[b] >= 0)))
 (Re[a] > 0 && ((Im[a] < 0 && Im[b] <= ((Re[a] Re[b])/Im[a]))
 (Im[a] == 0 && Re[b] >= 0)
 (Im[a] > 0 && Im[b] >= ((Re[a] Re[b])/Im[a])))
)

I am not good in Maple, so there might be better command to handle this in than solve, but this is what solve gives in Maple 17:
 eq:=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);
solve(eq,{a,b});
{ a = a,
b = RootOf(Abs(a)^3*Abs(_Za)Abs(a)*Abs(_Za)^3 +Abs(a)*Abs(_Za)*Abs(_Z)^2Abs(Abs(a)^2*_Z^2 Abs(a)^2*a^2+a^2*Abs(_Za)^2)) }

Nasser

