Mobius transformation between those two domains I am trying to understand why there is no such mobius transformation from $A$ to $B$ .
$$B= D_1(0) / 0 , A=\{z \mid \operatorname{Im}(z)+\operatorname{Re}(z)>0\}$$
I do not understand how do I approach this kind of problem. what contradiction do I get if there will be exist one? I would also like if some one can help me how to visualize domain $A$.
 A: The half-plane $A=\{z \mid \operatorname{Im}(z)+\operatorname{Re}(z)>0\}$ is a $-\pi/4$ rotation of the upper half-plane, i.e. $A = e^{-i \pi /4} \mathbb{H}$, where $\mathbb{H} = \{z \mid Im(z) > 0\}$. Therefore, we can build ourselves a Möbius transformation $T:\mathbb{H}\rightarrow\mathbb{D}/\{0\}$, and simply apply another Möbius transormation beforehand, namely $T':A\rightarrow\mathbb{H},\; z\mapsto e^{i \pi /4}z$. Our final answer will then be $T \circ T'$, once we find our $T$.
When given $\zeta \in d\mathbb{H}$ and $w\in\mathbb{H}$, there exists precisely one biholomorph Möbius transformation that maps $\mathbb{H} \rightarrow \mathbb{D}$ with $\zeta \mapsto 1, \; w \mapsto 0$ on the extended complex plane $\mathbb{C}\,\cup\,\{\infty\}$, namely
$$
T(z) = \frac{(z - w)(\zeta - \overline{w})}{(z-\overline{w})(\zeta-w)}.
$$
Note that this Möbius transformation maps $\zeta \mapsto 1, \; w \mapsto 0, \; \overline{w} \mapsto \infty\;$ (check this yourself!).
Herein lies the contradiction. We wish that $T:\mathbb{H}\rightarrow\mathbb{D}/\{0\}$, which is equivalent to $T:\mathbb{H}\rightarrow\mathbb{D}$  with $\infty \mapsto 0$ in the extended complex plane. The "point at infinity" $\infty$ can be reached by travelling infinitely far in any direction. This implies (with a slight abuse of notation) $\overline{\infty} = \infty$ (again, please do not read into that too far).
We wish therefore to map $\infty \mapsto 0$, but also $\infty \mapsto \infty$.
