Given the domain ℂ\[-1,1] and the function, [itex]f(z)=\frac{z}{(z-1)(z+1)}[/itex], defined on this domain, the Residue Theorem shows that for [itex]\alpha[/itex] a positive parametrization of the circle of radius two centered at the origin, that:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int_{\alpha}f(z)=\int_{\alpha}\frac{z}{(z-1)(z+1)} = 2\pi i[/tex]

Can I automatically conclude from this that [itex]f(z)=\frac{z}{(z-1)(z+1)}[/itex] does not have a primitive in ℂ\[-1,1]?

I already know it's true the other way, so I'm suspecting that these two statements in the title are equivalent.

(Note, this is the contrapositive of the title)

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# F has a primitive on D ⊂ ℂ ⇒ ∫f = 0 along any closed curve in D?

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