# Proving a result regarding period of elliptic functions

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 $$\frac { z f'(z) } {f(z) }$$ . ]

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
• everything in your answer is clear to me except that how did you computed integral of $\frac {z h'(z) } { h(z) }$ . 2 nd integral in same line can be computed by Cauchy Theorem. My another doubt is how to use this to prove that S(0) - S($\infty$) is period of f and I know f'(z) /f(z) is periodic with above mentioned periods but where it's integral would be used. – Tim Dec 21 '19 at 17:04
• when you changed the integral limit ( in last paragraph 1st integral) from a+ $\omega_1$ + $\omega_2$ and a+ $\omega_1$ to a+ $\omega_2$ and a why and how did you changed z to z +$\omega_2$ in integrand? Also I am sorry but I am still not able to understand 1. How to compute $\frac { z h'(z) } { h(z) }$ ? 2. How to use it to prove S(0) - S($\infty$) is a period of Parellogram? . I am just a beginner . Can you please explain these doubts of mine? – Tim Dec 23 '19 at 10:32
• The integral of $zh'(z)/h(z)$ is over a circle is zero because that function is holomorphic (this is known as Cauchy's theorem). For the change of variables, note that both $f$ and $f'$ are periodic with respect to the lattice generated by $\omega_1$ and $\omega_2$. (Sorry for the delay ) – m.s Dec 25 '19 at 15:31