I am trying exercise of Ch - 1 of Tom M Apostol Modular functions and Dirichlet series in number theory.
I am self studying and I am struck on this problem of Elliptic Functions.
Problem is --> Let S(0) denote the sum of zeroes of an elliptic function f in a period parallelogram and let S($\infty $) denote the sum of the poles in the same parallelogram. Prove that S(0) - S( $\infty $) is a period of f. [ Hint : integrate $\frac { z f'(z) } {f(z) } $ . ]
Can somebody please tell how to prove this? I am unable to think about this.
Let $\omega_1, \omega_2 \in \mathbb{C}$ be $\mathbb{R}$-linearly independent periods of $f$. Let $$\mathcal{F} =a+ \lbrace t \omega_1 + s \omega_2 : s, t \in [0,1)\rbrace$$ denote a fundamental parallelogram, where $a \in \mathbb{C}$ is chosen so that $f$ has no zeros and no poles on $\partial \mathcal{F}$. Let $z_0 \in \mathcal{F}^{\circ}$ be a zero of $f$. Then we know by complex analysis that there is an integer $m_p \geq 1$ and a holomorphic function $h$, defined and nowhere vanishing near $z_0$, such that $f(z) = h(z)(z-z_0)^{m_p}$ in a neighborhood of $z_0$. Then $$ \frac{zf'(z)}{f(z)} = z\frac{h'(z) (z-z_0)^{m_p} + h(z)m_p(z-z_0)^{m_p-1}}{h(z) (z-z_0)^{m_p}}. $$ The integral of this function, over a small circle around $z_0$, is $m_p z_0$ (by Cauchy's Theorem and direct computation). Similarly, if $z_0$ is a pole of order $n_p \geq 1$, the integral of $zf'(z)/f(z)$ over a small circle around $z_0$ will give you $-n_p z_0$. The integral of $\int_{\partial \mathcal{F}}{\frac{zf'(z)}{f(z)}dz}$ is (by the residue theorem) therefore equal to $S(0) - S(\infty)$. On the other hand, using that $f'(z)/f(z)$ is $\omega_1, \omega_2$-periodic you can easily evaluate this integral directly in terms of $\omega_1$ and $\omega_2$, by noting that two of the four line integrals are related to the others by $$ \int_{a+\omega_1}^{a+\omega_1 + \omega_2}{ \frac{zf'(z)}{f(z)}dz} = \int_{a}^{a+ \omega_2}{ \frac{(z+ \omega_2)f'(z)}{f(z)}dz}\\ \int_{a+\omega_2}^{a+\omega_2 + \omega_1}{ \frac{zf'(z)}{f(z)}dz} = \int_{a}^{a+ \omega_1}{ \frac{(z+ \omega_1)f'(z)}{f(z)}dz} $$ where we use straight-line paths everywhere