Zeros of Jacobi elliptic functions For the past month or so, I have been studying Jacobi elliptic functions $sn$, $cn$ and $dn$. However I stumbled upon a problem and it goes as following:
I am trying to find $z \in \mathbb{C}$, such that $sn(z) = sin({\psi}) = 0$, where $\psi$ is the inverse of the integral:
$$z(\psi) = \int_0 ^\psi \frac{1}{\sqrt{1 - k^2 \sin^2(x)}} dx .$$
By definig constant $K$ as:
$$K = K(k) = \int_0 ^\frac{\pi}{2} \frac{1}{\sqrt{1 - k^2 \sin^2(z)}} dz $$ I was able to prove by induction that:
$$2K = \int_{\pi(k-1)}^{\pi k}\frac{1}{\sqrt{1 - k^2 \sin^2(z)}} dz $$ for $k \in \mathbb{Z}.$ To find zeros of $sn$ I did the following:
$$sn(z) = \sin(\psi) = 0 \iff \psi = n\pi \iff z = \int_0 ^{n\pi} \frac{1}{\sqrt{1 - k^2 \sin^2(z)}} dz $$ for $n \in \mathbb{Z}$. I rewrote the last integral as:
$$z = \sum_{k=1}^n \int_{\pi(k-1)}^{\pi k}\frac{1}{\sqrt{1 - k^2 \sin^2(z)}}dz = \sum_{k=1}^n 2K = 2nK.$$
Now it appears as is I found all zeros of $sn(z)$ for $z\in \mathbb{C}$, however there are also zeros of the form $2iK’$, where $K’$ is:
$$ K’ = K(k’) = 2K = \int_0^\frac{\pi}{2}\frac{1}{\sqrt{1 - k’^2 \sin^2(z)}} dz $$ where $k’ = \sqrt{1 - k^2}.$ Now I obviously made mistake somewhere however I have no idea where. I would appreciate any help or tips.

You should look into Jacobi's Imaginary Transformation.
It is
$$\sin\varphi=i\tan\psi$$ under this,
$$\int\frac{d\varphi}{\sqrt{1-k^2\sin^2\varphi}}=
i\int\frac{d\psi}{\sqrt{1-k^{\prime 2}\sin^2\psi}}$$
This allows you to prove that the periods of $sn(z)$ are $2K$ and $4K^{\prime}i$. Modulo these periods you have all the roots.
