# Does $\frac 1{z^2+1}$ have a primitive on $\mathbb C-\{i,-i\}$?

Please explain why the above is False. (I do not understand what the hint is trying to say either: the Cauchy Integral Theorem that I know states that If $f$ is analytic on a simply-connected domain containing the loop L, then the contour integral of $f$ along L is zero.)

• If $f(z)=1/(z^2+1)$ is exact, then the contour integral of $f$ about any closed curve vanishes. Try to evaluate one of these contour integrals explicitly (specifically, a small circle around $i$ or $-i$). – awwalker May 7 '13 at 21:08

If the function has a primitive, the integral over any closed curve is $0$. Use the residue theorem or Cauchy's integral theorem to see that the suggested integral is not $0$.
If your function is a primitive, i.e. has anti derivative, then if integrate around any loop you should get zero, by pretty much the fundamental theorem of calculus ($\int_\gamma f' = f(z_0) - f(z_0)$ for any 'starting' point $z_0 \in \gamma$).
Now if you hear integrate around loops, think residues...since you could take a tiny loop around any singularity, that would mean $\frac{1}{1+ z^2}$ didn't have residues at its singularities, $\pm i$.