Let $D$ be the interior of a closed parallelogram in the complex plane with one of its vertices at the origin, and $f(z)$ be a conformal mapping of $D$ onto a half-plane $H$ ($f$ is a bijection). Is $f$ the restriction to $D$ of an elliptic function $F$ ?
If $D$ is the interior of a closed rectangle, Yes, $f$ is the restriction to $D$ of an elliptic function $F$. What about parallelogram?
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No, this does not work for any non-rectangle. For notational simplicity, let $0$ be a vertex of $\overline{D}$ and let $f(0) = 0$. Let $\theta$ be the angle of $\overline{D}$ at $0$ and let $f(z) = f_m z^m + f_{m+1} z^{m+1} + \cdots$. Near $0$, $f$ maps the cone of angle $\theta$ to a half space, so we must have $m \theta = \pi$. Thus, $\theta = \tfrac{\pi}{m}$ for some positive integer $m$.
But now, applying the same logic at an adjacent vertex of the parallelogram, we must have $\pi - \theta = \tfrac{\pi}{n}$ for some integer. So $\tfrac{\pi}{m} + \tfrac{\pi}{n} = \pi$ and we must have $m = n = 2$. In other words, our parallelogram is a rectangle.
Recall that non-constant elliptic functions $F: \mathbb C \to \overline{\mathbb C}$ must admit a pole. Indeed, if $F$ was entire the parallelogram $D$ is the fundamental domain of $F$, continuous functions map bounded sets to bounded sets, so $F(\mathbb C) = F(D)$ is bounded, which by Lioville we know implies $F$ is constant. Arguing along similar lines, we can show that $F$ is surjective.
If $f : D \to \mathbb H$ is a conformal map, by Caratheodory's theorem it extends to a homeomorphism map $f: \overline{D} \to \overline{\mathbb H}$, so it cannot be the restriction of an elliptic function with fundamental domain $D$ since it has no poles and does not map onto $\mathbb C$.
Now if you are wondering if it is possible that $f$ is the restriction of an elliptic function $F$ with fundamental domain $D'$ satisfying $D \subset D'$, which is perfectly possible, this runs into the more general question of analytic continuation, and I don't think there is enough information to settle it.