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.