Closed circle integral of 1/z and branch cuts

We examine integrals of z^n over closed circular contours and highlight the important issues involving integrations of functions requiring functions with branch cuts

[Mariongr]

In 18:30 you cannot use the L' Hopital rule because the limit is π times the derivative of sin(x) at zero. We find the derivative using the Taylor series for sin (x) at x=0, and that's how we get this limit.

[MsJhonstar]

Hi, I really liked he video, thanks! I was wondering, what would happen if a was not a integer? For example 1/2, then it would depend on R by that shouldn't happen because of Cauchy's theorem, rigth? If anyone could explain me or correct me I would really appreciate it

[sick]

why are -pi and pi the same point in the complex plane?

[dancingfox]

your writing is difficult to see and thus I had moved elsewhere

[jagadashjagan]

Your handwriting was very bad to understand

[javieramado]

This is wrong. When a=0, there is no more need of a branch, because 1/z is analitic everywhere but 0.
What you should do is take the closed integral (from R 0 to -R π, then from -R π to -r π, then to rπ and tyen to Rπ) that is equal to 2πi Res(z0)=0 (because there are no residues). Then you divide it in 4 integrals. the major arc is equal to πi, because you use the parametrization z=R*exp(i phi). The othera are equal to what you want to calculate. the minor arc is equal to I*(-π) Res (0) as 0 is a simple pole, while r-->0.
because of this, I=0