Find a Mobius transformation that maps $i,-i,1$ to $0,2,\infty$ respectively.

Question

Find a Mobius transformation that maps $i,-i,1$ to $0,2,\infty$ respectively.

$$w(z)=\frac{az+b}{cz+d}$$

I think I don't have enough clear elements here, and I don't know which numbers to plugin and why for $a,b,c$ and $d$. This one that maps to $\infty$ I have no idea, how to work with. Can someone help with this. I feel like I have no clear direction, and that too much is unclear and unknown.

Answer 1

The first condition is $$ f(i)=\frac{ai+b}{ci+d}=0. $$ This implies $ai+b$ equals what? Next, $$ f(-i)=\frac{b-ai}{d-ci}=2, $$ so $b-ai=2d-2ci$. Finally, $$ f(1)=\frac{a+b}{c+d}=\infty, $$ so $c+d=0$. We can now conclude $c=-d$. As noted in the comments, we have freedom in the variables since we have less restrictions than free variables, so lets say $c=1$, $d=-1$. Can you finish from here?