TITLE: Find the Möbius transformation that...
QUESTION [5 upvotes]: ...maps: 
$-1 \mapsto i,$
$\infty \mapsto 1,$
$i \mapsto 1+i$
This is driving me mad!  Should be easy but I keep getting tied up in knots.  Any help much appreciated!

REPLY [5 votes]: We can find $\omega(z)=\dfrac{az+b}{cz+d}$ using formula $$\frac{z-z_1}{z-z_2}:\frac{z_3-z_1}{z_3-z_2}=\frac{\omega-\omega_1}{\omega-\omega_2}:\frac{\omega_3-\omega_1}{\omega_3-\omega_2}$$ where $(z_1,z_2,z_3)=(-1,\infty,i)$ and $(\omega_1,\omega_2,\omega_3)=(i,1,1+i).$
In our case we have $$\frac{z+1}{1}:\frac{i+1}{1}=\frac{\omega-i}{\omega-1}:\frac{1}{i},$$ where we change $z-z_2$ with $1$ because $z_2=\infty,$ etc.
And now we find that $\omega=\omega(z)=\dfrac{z+2+i}{z+2-i},$ like previous solutions.