On Dec 6, 3:26 pm, kcrisman <kcris...@gmail.com> wrote:
> > The problem is that the integral should not depend on the center of
> > the circle
> > containing the pole. It looks like maxima bug (?)
>
> I've reported this
> athttps://sourceforge.net/tracker/?group_id=4933&atid=104933
>
> Dan, if you want to open a ticket, just be sure to refer to that.
It is nice (but slightly dangereous) that:
sage: pari('f(z) = (z-I)*(z-1)^2/(z-(-1/2-I/3))')
(z)->(z-I)*(z-1)^2/(z-(-1/2-I/3))
sage: pari('intcirc(z=-1/2-I/3,1,f(z))')==pari('intcirc(z=0,2,f(z))')
True
(Since the calculation is numerical, the equality is modulo pari
default precision)
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