Let $E=\mathbb{C}/\Lambda$ be a complex elliptic curve where $\Lambda=\omega_1\mathbb{Z}\oplus\omega_2\mathbb{Z}$

Let $f$ be a nonconstant elliptic function with respect to $\Lambda$.

Let $P=\{x_1\omega_1+x_2\omega_2:x_1,x_2\in[0,1]\}$ be the parallelogram with opposing boundary identified appropriately.

There exists some constant $t$ such that $f$ has no zeros and poles on $t+\partial P$.

Now I want to compute the integral $$ \frac{1}{2\pi i}\int_{t+\partial P}\frac{zf'(z)}{f(z)}d{z} $$

I need to show that it is $$ \sum_{x\in E}\nu_x(f)x $$ where $\nu_x(f)$ is the order of $f$ at $x$, which means $f(z)=(z-x)^{\nu_x(f)}g(z)$ where $g(x)\neq 0$.

I can find that it is $$ \omega_1\int_{t}^{t+\omega_2}\frac{f'(z)}{f(z)}dz-\omega_2\int_{t}^{t+\omega_1}\frac{f'(z)}{f(z)}dz $$

and each integral is an integer because $f$ is the same at the two endpoints. But I cannot get the desired formula.

Any help?


Doesn't this follow at once from (Cauchy's) Argument Principle? We know that

$$\frac{1}{2\pi i}\int_Cg(z)\frac{f'(z)}{f(z)}dz=\sum_{a} g(a)-\sum_{b}g(b)$$

where the firt sum is over the zeros $\,a\,$ of $\,f\,$ in the domain enclsoed by $\,C\,$ and the second over the poles of $\,f\,$ in the same region (of course, $\,g\,$ analytic on $\,C\,$ and inside it).

  • $\begingroup$ Oh! I know the argument principle with $g(z)=1$, but not familiar with this formula. It is right, thank you!. $\endgroup$ – hxhxhx88 Jan 7 '13 at 5:15

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