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

[Poopsilon]

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

$$\int_{\alpha}f(z)=\int_{\alpha}\frac{z}{(z-1)(z+1)} = 2\pi i$$

Can I automatically conclude from this that $f(z)=\frac{z}{(z-1)(z+1)}$ 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)

 

[mathwonk]

yep. that's how you do it. having a primitive makes it possible to evaluate the integral by evaluating the primitive at both ends of the curve and subtracting, so you get zero if those two ends are the same. So actually you are asking it in the trivial direction. The other direction is harder.

 

[Poopsilon]

Ok you that makes complete sense, thanks mathwonk.