# Jacobian elliptic functions with complex modulus

Let $$\mathrm{sn}(u,k)$$ be the usual Jacobian elliptic function defined in terms of theta functions: $$\mathrm{sn}(u,k) = \frac{\theta_3(0)}{\theta_2(0)} \frac{\theta_1(z)}{\theta_4(z)}$$ where $$z = u/\theta_3^2(0)$$. I want to show to that for any $$k \in \mathbb{C}$$. The above definition makes sense. By definition the elliptic modulus $$k$$ is given by $$k = \theta_2^2(0,q)/\theta_3^2(0,q)$$ where $$q = e^{i\pi \tau}$$ is the nome and $$\tau \in \mathbb{H}$$. Thus I need to show that the map $$F: D \to \mathbb{C}$$ given by $$F(q) = \frac{\theta_2^2(0,q)}{\theta_3^2(0,q)}$$ where $$D$$ is the open unit disc is surjective. I'm not really sure how to go about this. The function in the numerator is holomorphic on its domain but I am not sure about the zeros of the function in the numerator. Is this even the right way to approach this problem? To be clear, I just want to justify that $$\mathrm{sn}(u,k)$$ is well defined for all $$k$$ rather than the usual restriction of $$0.

• Read the Wikipedia modular lambda function article. $\lambda=k^2$ Mar 4, 2020 at 12:29

You are confusing $$\theta(q)$$ with $$\vartheta(z,\tau)=\sum_n e^{2i\pi nz} e^{i\pi n^2\tau}$$ For a fixed $$\tau$$ then $$f(z,\tau)=\frac{\vartheta(z,\tau)\vartheta(z+1/2+\tau/2,\tau)}{\vartheta(z+1/2,\tau)\vartheta(z+\tau/2,\tau)}$$ is meromorphic $$1,\tau$$ periodic in $$z$$. Its poles are known from say the Jacobi triple product.

Once you know its poles you can find the non-linear differential equation it satisfies and obtain that its inverse $$g(u,\tau)$$ (such that $$f(g(u,\tau),\tau)=u$$) is an elliptic integral of the first kind.

The coefficient of the differential equation (thus of the elliptic integral) is a modular form of $$\tau$$, the next step is to express it as a polynomial in $$\vartheta(0,\tau),\vartheta(1/2,\tau),\vartheta(1/2+\tau/2,\tau)$$ using that they are weight 1/2 modular forms and that the space of modular forms of a given weight and level is finite dimensional.

This modular form is expressed in term of the modular form $$\vartheta(0;\tau)^2$$ and $$k = \frac{\vartheta(0;\tau)^2}{\vartheta(\tau/2;\tau)^2}$$ which is a modular function of $$\tau$$, the map $$\tau \to k$$ is surjective when adding the cusps of the modular curve, and from the values at the cusps we can find if it is surjective when restricted to $$\Im(\tau)> 0$$.

This is what is usually called the inversion problem. Let the Jacobi thetanulls be defined as \begin{align} \vartheta _{2}(\tau)&=\sum_{n\in\mathbb {Z}} \exp\left\{\pi i\tau\left(n+\frac{1}{2}\right)^2\right\}\notag\\ \vartheta _{3}(\tau)&=\sum_{n\in\mathbb {Z}} \exp(\pi i\tau n^2)\notag\\ \vartheta_{4}(\tau)&=\sum_{n\in\mathbb {Z}} (-1)^n\exp(\pi i\tau n^2)\notag \end{align} The above definitions make sense when imaginary part of $$\tau$$ is positive.

The inversion problem seeks to find out if the following equation $$m=k^2=\frac{\vartheta_{2}^{4}(\tau)}{\vartheta _{3}^{4}(\tau)}\tag{1}$$ has a solution $$\tau$$ as an analytic function with $$\Im(\tau) >0$$. It can be proved with some effort that if $$m$$ is any complex number not lying in $$(-\infty, 0] \cup[1,\infty)$$ then the above equation defines $$\tau$$ as an analytic function of $$m$$.

The procedure involves showing that for these values of $$m$$ the elliptic integral $$K(m) =\int_{0}^{\pi/2}\frac{dx} {\sqrt {1-m\sin^2x}}$$ is an analytic function of $$m$$ and therefore $$\tau=i\frac{K(1-m)}{K(m)}$$ is also analytic. It can be proved that $$\Im (\tau)>0, - 1<\Re(\tau)<1$$ and it satisfies equation $$(1)$$.

You can get more details with proofs in Elliptic Functions by J. V. Armitage and W. F. Eberlein, page 109, Theorem 5.2.

Once you have got $$\tau$$ you can define elliptic functions in terms of theta functions as in your post.