1. ## Riemmann Surface integral

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:

$\int_C log(z) dz$

I have tried to calculate in this way:

$log(z)=log(r)+it$

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

thus:

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

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

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

What do you think? Is it right?

Thank you.

2. I'm not sure about the 'n'..

Thank you.

3. $\int_{C(n)} \log(z)dz=2n\pi i$

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

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

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

I do believe an argument for it's validity could be made however.