On Dec 5, 5:31 am, Dan Drake <dr...@kaist.edu> wrote:
> I keep wondering whether Sage is making a mistake, or I'm not
> understanding complex analysis. I'm a little afraid to learn the answer.
> :)
>
> Take f(z) = (z-I)*(z-1)^2/(z-(-1/2-I/3)). It's analytic everywhere
> except at -1/2-I/3, where it has a simple pole. So, if I integrate over
> a circle centered at 0 of radius, say, 2, the answer should be
>
> 2*pi*I*(residue of f at -1/2 - I/3),
>
> which is pi*(181/27 + 19*I/36). However, when I try to do the contour
> integral, I get:
>
> sage: integrate(f(2*exp(I*t)) * 2*I*exp(I*t), (t, 0, 2*pi))
> 0
>
> even though the contour encloses the pole. It works if I center the
> circle around the pole:
>
> sage: integrate(f(-1/2-I/3 + exp(I*t)) * I*exp(I*t), (t, 0, 2*pi))
> (19/36*I + 181/27)*pi
>
> and also if I integrate over the square with vertices 1+i, 1-i, -1-i,
> -1+i. What's wrong with the circle at the origin?
>
> Note that Mathematica gets this right, although you need to ask for full
> simplification: with f[z_] = (z-I)*(z-1)^2/(z-(-1/2-I/3)), you get
>
> In[5]:= Integrate[f[2*Exp[I*t]] * 2*I*Exp[I*t], {t, 0, 2*Pi}]//FullSimplify
>
> 181 19 I
> Out[5]= (--- + ----) Pi
> 27 36
>
> Any ideas?
>
> Dan
>
> --
> --- Dan Drake
> ----- http://mathsci.kaist.ac.kr/~drake
> -------
>
> signature.asc
> < 1KViewDownload
Note that pari has a good approximate solution:
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: 2*n(pi)*I*pari('intcirc(z=0,2,f(z))')
21.0603063073982 + 1.65806278939461*I
sage: CC(pi*(181/27 + 19*I/36))
21.0603063073982 + 1.65806278939461*I
Andrzej Chrzeszczyk
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