Which group is given by $\begin{pmatrix} e^{i\theta}&x\\0&e^{-i\theta}\end{pmatrix}$? I encounter the group $$\biggr\{\begin{pmatrix}e^{i\theta}&a\\0&e^{-i\theta}
\end{pmatrix}\colon\theta\in\mathbb{R},a\in\mathbb{C}\biggr\}.$$
To which more studied group is this group isomorphic? It looks like a rotation and a translation, so something like a semidirect product?
 A: Yes, we have the obvious subgroups of translations $T_a=\begin{pmatrix}1&a\\0&1\end{pmatrix}$ (isomorphic to $\Bbb C)$, and the subgroup of rotations $R_\theta=\begin{pmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{pmatrix}$ (isomorphic to $S^1$).
Note that $R_\theta T_a R_\theta^{-1}=T_{e^{wi\theta}a}$, making the translation subgroup normal and 
$$ 1\to\Bbb C\to G\to S^1\to 1$$
split.
A: Your group $G$ is the unique double cover of the group of positive Euclidean isometries $\operatorname{Isom}^+(\mathbb R^2)$.
Indeed: $G$ acts on $\mathbb P^1(\mathbb C)$ by homographies and leaves $\infty$ invariant. The action restricts to $\mathbb C$, and is faithful modulo $\pm 1$. The image of the permutation representation contains all translations $z \mapsto z + a$ and all rotations $z \mapsto e^{2i \theta} z$. Thus, under the isomorphism $\mathbb C \cong \mathbb R^2$, the image is precisely the group of direct Euclidean isometries of $\mathbb R^2$: $G / \{\pm 1\} \cong \operatorname{Isom}^+(\mathbb R^2)$.
The Lie group $\operatorname{Isom}^+(\mathbb R^2)$ is homeomorphic to $\mathbb R^2 \times S^1$, so that it has a unique (connected, Galois) double cover.
