# Find the Möbius transformation mapping $(i, 0, \infty)$ to $(0, \infty, -i)$, in precisely that order

I did this:

Assuming my Möbius transformation is some $\omega$ in terms of $z$, I want to work out a formula that gives me:

• 1) $\omega = 0$ when $z = i$
• 2) $\omega = \infty$ when $z = 0$
• 3) $\omega = -i$ when $z = \infty$

So I did this

1) Happens when $\omega = z - i$

2) Happens when $\omega = \dfrac{1}{z}$

3) Don't know when this happens

So, by combining the top two functions, I get my Möbius transformation to be $\omega = \dfrac{z - i}{z}$.

This was wrong. The answer is $\omega = \dfrac{-iz - i}{z}$ and it gets this in a really wierd way which I don't understand. The answers say:

Omitting the terms containing $\infty$ in

$$\dfrac{z - z_1}{z - z_3} \dfrac{z_2 - z_3}{z_2 - z_1} = \dfrac{\omega - \omega _1}{\omega - \omega _3} \dfrac{\omega _2 - \omega _3}{\omega _2 - \omega _1}$$

We get

$$\dfrac{z - i}{-i} = \dfrac{\omega}{\omega + i}$$

Can someone explain the answer to me, or show me another way of doing these types of questions please?

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You should have $z-i$ on the numerator in order to send $i$ to $0$. Then multiply by $-i$ in order to send $\infty$ to $-i$. Then determine the constant on the denominator so that $0$ goes to infinity. You get $$w=\frac{ -i (z-i) } {z}$$
How does multiplying it by $-i$ send $\infty$ to $-i$? –  Kaish Dec 13 '12 at 19:01
$\infty$ always goes to $\frac{a}{c}$ –  PAD Dec 13 '12 at 19:04
I don't get it. Right now I have $\frac{a}{c} = 1$, so why do you multiply it by $-i$ –  Kaish Dec 13 '12 at 19:09
I choose $a=-i$ and $c=1$. –  PAD Dec 13 '12 at 19:11
Because I want $\infty$ to go to $-i$. When you have $\frac{az+b}{cz+d}$ and you take the limit of $z$ to $\infty$ you get $\frac{a}{c}$. –  PAD Dec 13 '12 at 19:14