Dassinia said:Hello,
I don't get why using the fact that ∫dz/z = 2*pi*i accros the circle
This integral gives:
1/3∫(1/(z-2)-1/(z-1/2))dz = 1/3(-2*pi*i) across the circle
Thanks !
The Cauchy theorem for integrals, also known as the Cauchy integral theorem, states that if a function is analytic in a simply connected region, then the integral of that function along any closed path within that region is equal to 0.
The Cauchy theorem can be used to evaluate integrals by breaking down complex integrals into simpler ones that can be easily solved using techniques such as the residue theorem or Cauchy's integral formula.
The Cauchy theorem is an important tool in complex analysis as it allows for the evaluation of complex integrals and provides a connection between analytic functions and their integrals. It also has applications in various fields such as physics, engineering, and economics.
Yes, the Cauchy theorem can still be applied to integrals with singularities as long as the singularities lie outside of the contour of integration. In such cases, the residues of the singularities can be used to evaluate the integral.
One limitation of the Cauchy theorem is that it only applies to simply connected regions, meaning that there are no holes or gaps in the region. It also only applies to functions that are analytic within the region of integration.