0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Tim Dec 21 '19 at 17:04
  • $\begingroup$ 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? $\endgroup$ – Tim Dec 23 '19 at 10:32
  • $\begingroup$ can you please explain the doubts which I have asked above!! I am a beginner in analytic number theory and I am self studying it without any help from professor. $\endgroup$ – Tim Dec 24 '19 at 9:19
  • $\begingroup$ can you please answer my doubts b, otherwise I must put a bounty on it or Ask on mathoverflow!! $\endgroup$ – Tim Dec 24 '19 at 20:05
  • $\begingroup$ 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 ) $\endgroup$ – m.s Dec 25 '19 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.