Möbius transformation with infinity in both the $w$-plane and $z$-plane.

Question

I want to find the Möbius transformation which brings $f(0)=\infty$, $ f(\infty)=0$, and $f(5)=i$. If I use the formula

\begin{equation} \frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)} \end{equation}

then I get a cancellation of the $w$-variable. That would give just another complex number, $z=5i$. But what does this mean?

Thanks

Forget that formula for a second. Which Möbius transformations map $0$ to $\infty$ and $\infty$ to $0$? — Martin R, Dec 21 '21 at 12:00
A Möbius transformation is a function of the form $T(z) = \frac{az+b}{cz+d}$ with complex constants $a, b, c, d$. — Martin R, Dec 21 '21 at 12:08
I thought that |w|>0 has an infinite number of points on the w plane, so infinity would be there too. I consider your example. So $T(0)=\frac{b}{d}=\infty $ and $T(\infty)=\frac{a}{c}=0$, then solve the two equations — Luthier415Hz, Dec 21 '21 at 12:09
This gives, along with the third condition: a=c=0, $d\infty=di$. But this is no equation, and does not give any transformation? — Luthier415Hz, Dec 21 '21 at 12:14
@ViolaPlayer There are several elementary mobius transformations you can examine by setting most of the constants $a,b,c,d$ to zero. See if you can find one of those that takes zero to infinity. — CyclotomicField, Dec 21 '21 at 13:40
You can use the ansatz $f(z)=a/z$ because you know $1/z$ swaps $0$ and $\infty$. — anon, Dec 23 '21 at 18:35

score 1 · Accepted Answer · answered Dec 22 '21 at 21:08

The formula $$ \frac{(f(z)-w_1)(w_2-w_3)}{(f(z)-w_3)(w_2-w_1)}=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)} $$ can only be applied if all $z_j$ and $w_j$ are complex numbers, i.e. not $\infty$. The general formula is $$ (f(z), w_1, w_2, w_3) = (z, z_1, z_2, z_3) $$ where $(z, z_1, z_2, z_3)$ is the cross-ratio of the numbers $z, z_1, z_2, z_3$, that is the value of $z$ under the Möbius transformation which maps $z_1, z_2, z_3$ to $0, 1, \infty$, respectively.

If $z, z_1, z_2, z_3$ are all (distinct) complex numbers then their cross-ratio is $$ (z, z_1, z_2, z_3) = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)} \, . $$ If one of the $z_j$ is equal to $\infty$ then the corresponding formula can be obtained by a limiting process. As an example, for $z_1 \to \infty$ we get $$ (z, \infty, z_2, z_3) = \frac{z_2-z_3}{z-z_3} \, . $$ Similarly, $$ (z, z_1, \infty, z_3) = \frac{z-z_1}{z-z_3} $$ and $$ (z, z_1, z_2, \infty) = \frac{z-z_1}{z_2-z_1} \, . $$

In your case we need the solution for $$ (f(z), \infty, 0, i) = (z, 0, \infty, 5) \, . $$ Using the above formulae this translates to $$ \frac{0-i}{f(z)-i} = \frac{z-0}{z-5} \iff \boxed{f(z) = \frac{5i}{z}} \, . $$

Of course one could have obtained that result with an educated guess. A Möbius transformation exchanges the points $0$ and $\infty$ if it is of the form $T(z) = a/z$ for some non-zero constant $a$, and the condition $f(5)=i$ gives $a=5i$.

I retried your method, and I get only to $z/(z-5)$. But from there to $5i/z$ I don't know how you do it. How did you do it, if this is all equal to $w$? In fact, I got $w=/frac{5i}{z}+i+1$, which is the same as you got, just shifted. — Luthier415Hz, Dec 27 '21 at 11:16
@ViolaPlayer: Try it again. If you solve the equation $\frac{0-i}{w-i} = \frac{z-0}{z-5} $ for $w$ then you'll get $w = \frac{5i}{z}$. — Martin R, Dec 27 '21 at 15:02
Got it, used erroneously the identity $-i=1/i$, but I got $w=-\frac{5i}{z}$ — Luthier415Hz, Dec 27 '21 at 15:25

Möbius transformation with infinity in both the $w$-plane and $z$-plane.

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