%e %i log(- 1)
Correct result is %e*%pi/2
Logged In: YES
I think this is very likely caused by the fact that maxima
is log(-1) = %i*%pi/2. The correct answer is -%i*%pi/2,
because argument of the log is always in the third quadrant
(x < 0, y < 0).
I think it's because maxima thinks limit log(x) = log(limit
x) in this case since the limit of the arg is a finite number.
This is fixed. See the discussion in