- #1
Poopsilon
- 294
- 1
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:
[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)
[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)