= Solution
Take a real modulus $0<k<1$ and complementary modulus $k'=\sqrt{1-k^2}$. The <Jacobi elliptic sine> is constructed by inverting a <Schwarz-Christoffel mapping>. The main reason for the Schwarz-Christoffel <derivative> is local angle behavior: if a <prevertex> $a_j$ corresponds to a <polygon> angle $\pi\alpha_j$, straightening that corner gives $U(w)-U(a_j)\sim C_j(w-a_j)^{\alpha_j}$. Hence $U'$ has exponent $\alpha_j-1$. Along each straight boundary side its argument is constant. Dividing $U'$ by $\prod_j(w-a_j)^{\alpha_j-1}$ removes those angle changes; <Schwarz reflection> extends the quotient across the real boundary and the corner singularities. With all vertices finite, its behavior at infinity is also regular, because the sum of the <derivative> exponents is $-2$. It is therefore a nonzero constant on the sphere. This yields
$$
U'(w)=C\prod_j(w-a_j)^{\alpha_j-1}.
$$
This argument explains the exponents, the branch choices and the role of boundary reflection, rather than just writing down the formula.
For a <rectangle>, four angles are $\pi/2$, so choose prevertices $-1/k,-1,1,1/k$ and normalize the inverse map as
$$
U(w)=\int_0^w\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}.
$$
Use the branch positive on $(-1,1)$. Set
$$
K=\int_0^1\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}},\qquad K'=K(k').
$$
The two middle corner values are $U(\pm1)=\pm K$. The right side length is
$$
\int_1^{1/k}\frac{dt}{\sqrt{(t^2-1)(1-k^2t^2)}}=K',
$$
as the substitution $t=(1-k'^2s^2)^{-1/2}$ shows. The selected branch makes the increment along that side $iK'$. Thus the other corner values are $K+iK'$ and $-K+iK'$. The two boundary intervals through infinity have combined horizontal length $2K$; the substitution $t=1/(ks)$ gives length $K$ from $1/k$ to infinity. In particular $U(\infty)=iK'$. As in the <square> mapping, the boundary goes once around a <rectangle>, and the <argument principle> gives a <conformal> <bijection>
$$
U:\mathbb H\longrightarrow\{-K<\operatorname{Re}z<K,\quad0<\operatorname{Im}z<K'\}.
$$
Its inverse $s(z)=\operatorname{sn}(z,k)$ maps that <rectangle> onto the <complex upper half-plane>, and near zero has $s(0)=0$, $s'(0)=1$. Differentiating the inverse relation gives
$$
(s')^2=(1-s^2)(1-k^2s^2),\qquad s''=-(1+k^2)s+2k^2s^3.
$$
These identities continue meromorphically.
All four <rectangle> edges have real images, so <Schwarz reflection> extends $s$ across them. Reflection in the bottom edge gives $s(\bar z)=\overline{s(z)}$. Reflection in the right edge, combined with the first reflection, gives $s(2K-z)=s(z)$. The even inverse-integral <derivative> gives $s(-z)=-s(z)$, and hence
$$
s(z+2K)=-s(z),\qquad s(z+4K)=s(z).
$$
Reflection in the top edge and then the bottom gives $s(z+2iK')=s(z)$. The resulting <period lattice of the Jacobi elliptic sine> is
$$
\boxed{\Lambda=4K\mathbb Z+2iK'\mathbb Z}.
$$
To see the <poles>, use $U'(w)\sim-1/(kw^2)$ near the boundary point infinity. Then $U(w)-iK'\sim1/(kw)$, so the inverse has a <simple pole> at $iK'$ with <residue> $1/k$. The real half-period shift gives another <pole> at $2K+iK'$ with opposite <residue>. Reflection tiles the plane with copies of the <rectangle>, so these are exactly two <simple poles> per displayed fundamental cell. They also show there is no additional period: a period must permute these two <pole> classes; exchanging them would shift by $2K$ modulo the displayed lattice, which reverses the function's sign and <residues> rather than preserving it.
For any finite $w$, integrate $s'(z)/(s(z)-w)$ around a <fundamental parallelogram> avoiding its zeros and <poles>. The opposite edges cancel by periodicity, so the number of zeros of $s-w$ equals the number of <poles>, namely two. Therefore \b[every value has two inverse images modulo periods, counted with multiplicity]. The value infinity likewise has the two <simple poles> as preimages. Usually the finite-value preimages are distinct; the involution $z\mapsto2K-z$ interchanges them. At $w=\pm1,\pm1/k$, a critical point supplies one double inverse image instead. For example $s(K)=1$, $s'(K)=0$, $s''(K)=-(1-k^2)\ne0$. Thus the printed “exactly two” requires the standard multiplicity convention; it would be false if interpreted as two distinct solutions at these <branch values of a holomorphic map>.
Now choose $z_0=K$ and put $f(z)=s(z+K)$. Since $s(2K-z)=s(z)$, $f$ is even, elliptic for $\Lambda$, and has degree two. The <Weierstrass elliptic function> for the same lattice is even and has one <double pole> per cell, hence also degree two. Its generic fibers are precisely $\{z,-z\}$. Consequently an even $f$ is constant on those fibers and descends to a <meromorphic function> $R$ of $\wp$. At the four fixed classes of $z\mapsto-z$, the even local Laurent expansion makes this descent <meromorphic> in the squared local coordinate. A <meromorphic function> on the <Riemann sphere> is rational, so $f=R(\wp)$. Generic map degrees multiply:
$$
2=\deg f=\deg R\,\deg\wp=2\deg R.
$$
Thus $R$ has degree one and the <degree-two even elliptic function> conclusion is
$$
\boxed{\operatorname{sn}(z+K,k)=\frac{a\wp(z)+b}{c\wp(z)+d},\qquad ad-bc\ne0}.
$$
The <determinant> condition follows because $f$ is nonconstant. This proves the existence requested, with the explicit allowable shift $z_0=K$.
Back to article page
