Given the Möbius Transformation $T(z)=\frac{2+iz}{z+i}$ we have to find $T(\mathbb{D})$ where $\mathbb{D}$ is the unit disk on $\mathbb{C}_\infty$. I wanna know if what I did is correct:
Clearly, $-i \in \partial\mathbb{D}$ and then $T(-i)=\infty \in \partial T(\mathbb{D})$. This implies that $\partial T(\mathbb{D})$ is a line and thus $T(\mathbb{D})$ is a half-plane. Since $0$ and $\infty$ are symmetric with respect to $\partial\mathbb{D}$, $T(0)=-2i$ and $T(\infty)=i$ must be symmetric with respect to $\partial T(\mathbb{D})$. This implies that $\partial T(\mathbb{D})=${$z\in\mathbb{C}_\infty : $ Im$(z)=-1/2$ or $z=\infty$} and since $0 \in \mathbb{D}$ we have $T(0)=-2i \in T(\mathbb{D})$ and finally it must happen that
$T(\mathbb{D})=${$z\in\mathbb{C}_\infty : $ Im$(z)<-1/2$ or $z=\infty$}.
I think it is correct but I'm not totally sure if $z \in \partial \mathbb{D}$ implies that $T(z)\in\partial T(\mathbb{D})$, which I used at the beginning.
Thanks in advance.
Your method is fine ... for a quick sanity check ... try the method below.
Paramaterise the boundary using \begin{eqnarray*} z= \cos \theta + i \sin \theta. \end{eqnarray*} Thus \begin{eqnarray*} T & =& \frac{ 2- \sin \theta + i\cos \theta}{ \cos \theta + i (1+ \sin \theta)} \times \frac{ \cos \theta - i(1+ \sin \theta) } { \cos \theta - i(1+ \sin \theta) } \\ & =& \cdots = \frac{3 \cos \theta }{2(1+\sin \theta)} - \frac{1}{2}i. \\ \end{eqnarray*} And your conclusion thet the image is the said lower half plane is correct.