1
$\begingroup$

enter image description here

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.)

$\endgroup$
  • $\begingroup$ 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 $-i$). $\endgroup$ – awwalker May 7 '13 at 21:08
4
$\begingroup$

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 $0$.

$\endgroup$
3
$\begingroup$

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, $\pm i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.