Suppose we are focusing a Möbius Map that sends the upper half plane $H^+:=\{z:\Im(z)>0\}$ to the unit circle $B:=\{z:|z|<1\}$ bijectively. My textbook says if such Möbius Map sends a complex number, say $z_0$, to $0$ then it must send the conjugate of $z_0$ to $\infty$.
Why is this the case? I get that its conjugate must get sent to a point that lies outside of unit circle but why infinity point?
A geometrical argument is that Möbius transformations preserve symmetry: $z_0$ and $\bar z_0$ are symmetric with respect to the real line. $0$ and $\infty$ are symmetry respect to the unit circle.
Putting this into formulae: $S_1(z) = \bar z$ is the reflection at the real axis, and $S_2(z) = 1/\bar z$ is the reflection at the unit circle. If $T$ is a Möbius transformations mapping the upper halfplane onto the unit disk then $S_2 \circ T \circ S_1$ is another Möbius transformations mapping the upper halfplane onto the unit disk, and these mappings coincide on the real axis. It follows that these mappings are identical, so that $T \circ S_1 =S_2 \circ T $, i.e. $$ T(\bar z) = \left( \overline{T( z)}\right)^{-1} $$ for all $z$. In particular does $T(z_0) = 0$ imply that $T(\bar z_0) = \infty$.