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 $0<k<1$.
 A: 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 $\Im(\tau)> 0$.
A: 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. 
