Riemmann Surface integral

Bop

Dec 2009
48
0
Hello, here's the problem:

Let be C a curve in Riemmann Surface log(z) that goes from z=1 (in the log(1)=0 sheet) to z=1 turning around z=0 'n' times. Calculate:


\(\displaystyle \int_C log(z) dz\)


I have tried to calculate in this way:

\(\displaystyle log(z)=log(r)+it\)

\(\displaystyle z(t)=r(cos(t)+isin(t))\) so \(\displaystyle d(z(t))=r(-sin(t)+icos(t))dt\)

thus:

\(\displaystyle \int_C log(z) dz=\int^{2n\pi}_0 logz(t)z'(t)dt=\int^{2n\pi}_0(log(r)+it)r(-sin(t)+icos(t))dt=2\pi i rn\)

As r=1: (I know I could have simplified before)

\(\displaystyle \int_C log(z) dz=2\pi i n\)


What do you think? Is it right?


Thank you.
 

Bop

Dec 2009
48
0
I'm not sure about the 'n'..

Thank you.
 

shawsend

MHF Hall of Honor
Aug 2008
903
379
\(\displaystyle \int_{C(n)} \log(z)dz=2n\pi i\)

Everytime you go one time around, the integral is \(\displaystyle 2\pi i\). You could show this piece-wise for each circuit, or I would analytically extend the antiderivative and write:

\(\displaystyle \int_{C(n)}\log(z)dz=\left(z\log(z)-z\right)\biggr|_1^1=(2n\pi i-1)-(0-1)=2n\pi i\)

it's meant to look controversial just to get people thinking :)

I do believe an argument for it's validity could be made however.
 
Last edited:
  • Like
Reactions: Bop