Hello, On Tue, Apr 27, 2010 at 03:16:00AM +0200, Burcin Erocal wrote:
For conjugation, power objects just compute the conjugate of the basis and the exponent, and construct a new power object from these.
Proof: (a^b)* = exp(b log(a))* = exp((b log(a))*) = exp(b* (log(a))*) (log(a))* = ( log |a| + i arg(a))* = log |a| - i arg(a) = log |a*| + i arg(a*) = log(a*) So (a^b)* = exp(b* log(a*)) = (a*)^(b*)
conjugate(sqrt(-3)); sqrt(-3)
You are trying to compute the value of the function on the branch cut (which is ill defined), so you get the nonsense result. Please note: it's a midnight now (here in Ukraine), and I had a busy day, so the above might be a total nonsense. Feel free to point out mistakes (if any). Best regards, Alexei