# Riemmann Surface integral

#### Bop

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

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

Thank you.

#### shawsend

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

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:
• Bop