1. [SOLVED] Mobius Transformation

Consider a Mobius transformation

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

Determine the constants a, b, c, and d, (all $\epsilon \mathbb{C}$), such that the

transformation maps
0 $\rightarrow$ i, −i $\rightarrow$ 1, and −1 $\rightarrow$ 0.

How would I work this out?

2. What's mapped to $i$? Looks like you forgot to type it.

3. Originally Posted by chiph588@
What's mapped to $i$? Looks like you forgot to type it.
sorry 0 is mapped to i

4. Originally Posted by ronaldo_07
Consider a Mobius transformation

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

Determine the constants a, b, c, and d, (all $\epsilon \mathbb{C}$), such that the

transformation maps
0 $\rightarrow$ i, −i $\rightarrow$ 1, and −1 $\rightarrow$ 0.

How would I work this out?
The way to do this is to cross products equal to each other.

$(z,0,-i,-1) = (w,i,1,0)$

$\frac{z-0}{z-(-i)} \frac{-1-(-i)}{-1-0} = \frac{w-i}{w-1} \frac{0-1}{0-i}$

$-\frac{z(-1+i)}{z+i}= -i \frac{w-i}{w-1}$

Solving for $w$, we get $w = \frac{-iz-i}{z-1}$

$a = -i$

$b = -i$

$c = 1$

$d = -1$

5. Setting the cross products equal, I get

(z-(-i))/z-(-1) x 0-(-1)/0-(-i) = (w-1)/(w-0) x (i-0)/(i-1)

(z+i)/(iz+i) = (iw-i)/(iw-w)

and getting w by itself

w = (z+1)/(iz-i) a = 1, b = 1, c = i, d = -i

6. Multiply numerator and denominator by $-i$ and you'll get my answer.