Find a Mobius transformation $f$ that maps $\mathbb{H}=\{z \in \mathbb{C}:Im(z) >0\}$ bijectively to ball $B(0,2)$ such that $f(i)=1, f(1)=-2 \ ?$ Could anyone advise me on this problem: Find a Möbius transformation $f$ that maps $\mathbb{H}=\{z \in \mathbb{C}:\text{Im}(z) >0\}$ bijectively to ball $B(0,2)$ such that $f(i)=1, f(1)=-2 \ ?$
Can I consider $\frac{1}{2}f,$ so $\frac{1}{2}f$ is conformal between $\mathbb{H} $ to $D(0,1)$ and $f(z)=2e^{it}\dfrac{z-\alpha}{z-\overline{\alpha}}, \alpha \in \mathbb{H}.$
This means $f(i)=1=2e^{it}\dfrac{i-\alpha}{i-\overline{\alpha}},  \ f(1)=-2=2e^{it}\dfrac{1-\alpha}{1-\overline{\alpha}} \ $
$\implies \text{Im}\left(\dfrac{\dfrac{1-\alpha}{1-\overline{\alpha}}}{\dfrac{i-\alpha}{i-\overline{\alpha}}}\right)=0 \implies |\alpha|=\sqrt{2} \ ?$
 A: There's a standard technique to find a Möbius transformation mapping $z_1\rightarrow w_1$, $z_2\rightarrow w_2$, and $z_3\rightarrow w_3$, namely to use the cross-ratio:
$$\frac{(z-z_1) (z_2-z_3)}{(z-z_3)(z_2-z_1)}.$$
This maps $z_1\rightarrow0$, $z_2\rightarrow 1$, and $z_3\rightarrow \infty$.  If we now set
$$\frac{(z-z_1) (z_2-z_3)}{(z-z_3)(z_2-z_1)}=\frac{(w-w_1) (w_2-w_3)}{(w-w_3)(w_2-w_1)}$$
and solve for $w$, we get the desired Möbius transformation.  
We can now apply this taking $i\rightarrow1$, $1\rightarrow-2$, and $-1\rightarrow2$ to obtain your desired Möbius transformation.  The result is
$$T(z) = \frac{(2+4 i)-(4+2 i) z}{(2+i)-(1+2 i) z}.$$
It's easy to check that $T$ maps $i\rightarrow1$, $1\rightarrow-2$, $-1\rightarrow2$, and $0\rightarrow(8+6i)/5$.  The point behind that last one is that, since $|(8+6i)/5|=2$, we know that three real points map to points with absolute value $2$ so that the real line maps to the circle with radius 2.
Here's the image of the rectangle $[-8,8]\times[0,8]$ under $T$ with the points $i$ and $T(i)$ marked respectively.

