Finding bilinear transformation which maps $|z|=1$ on to $|w|=1$

Question

How can I show that every bilinear transformation which maps $|z|=1$ on to $|w|=1$ must be of the form $$w=K\frac{z-\alpha}{\overline{\alpha}z-1}$$ where $|K|=1$?

Please help me. Thanks.

Answer 1

If we conjugate a transformation $f$ which fixes the circle $C$ with some other transformation $\phi$, then the conjugate transformation $g=\phi\circ f\circ \phi^{-1}$ fixes $\phi(C)$.

So then, let's apply a transformation that takes the circle to something convenient - say, the real line. The transformation $\phi(z)=i\frac{1+z}{1-z}$ will do it. (The ratio there is pure imaginary because of the right angle between $-1,z,$ and $1$.) With this choice, $\phi^{-1}(w)=\frac{w-i}{w+i}$

OK, now what transformations $g(z)=\frac{az+b}{cz+d}$ fix the real axis? Precisely those with all coefficients real - or at least, since we can freely multiply all the coefficients by a constant without changing the transformation, all coefficients real multiples of the same thing. Choose a representation with all coefficients real for convenience.

So then, $\frac{az+b}{cz+d}=g(z)=\phi\circ f\circ \phi^{-1}(z)$. Then \begin{align*}f(z)&=\phi^{-1}\circ g\circ \phi(z)\\ &= \phi^{-1}\circ g\left(i\frac{1+z}{1-z}\right)\\ &= \phi^{-1}\left(\frac{ia\frac{1+z}{1-z}+b}{ic\frac{1+z}{1-z}+d}\right) = \phi^{-1}\left(\frac{ia(1+z)+b(1-z)}{ic(1+z)+d(1-z)}\right)\\ &= \frac{\frac{ia(1+z)+b(1-z)}{ic(1+z)+d(1-z)}-i}{\frac{ia(1+z)+b(1-z)}{ic(1+z)+d(1-z)}+i} = \frac{ia(1+z)+b(1-z)-i(ic(1+z)+d(1-z))}{ia(1+z)+b(1-z)+i(ic(1+z)+d(1-z))}\\ f(z) &= \frac{(ia+b+c-id)+(ia-b+c+id)z}{(ia+b-c+id)+(ia-b-c-id)z}\end{align*} The coefficients of $1$ in the numerator and $z$ in the denominator have the same imaginary part and opposite real parts, as do the coefficients of $z$ in the numerator and $1$ in the denominator. In order to get something that looks like the form we want, we normalize - divide the numerator by $ia-b+c-id$ and the denominator by its conjugate $-(ia+b-c+id)$. That leaves us with $$f(z)=\frac{c-b-ia-id}{c-b+ia+id}\cdot\frac{z-\frac{-b-c-ia+id}{c-b+ia+id}}{\frac{b+c-ia+id}{b-c+ia+id}z-1}$$ The factor out front is a ratio of conjugates, so it has size $1$. That's our $K$. The stuff inside still needs some work. Those look like conjugates, except that it's the real part flipped instead of the imaginary part - so let's multiply numerator and denominator of each inner fraction by $i$. The fraction in the numerator becomes $\frac{a-d-ib-ic}{-a-d-ib+ic}$, while the fraction in the denominator becomes $\frac{a-d+ib+ic}{-a-d+ib-ic}$. Those are clearly conjugates of each other, and we have the final form $f(z) = K\frac{z-\alpha}{\overline{\alpha}z-1}$ with $|K|=1$. Done.

One thing to note here: If $|\alpha|<1$, then the transformation sends the unit disk to itself - these are most often the cases we're interested in. If $|\alpha|=1$, then the transformation degenerates to a constant map $f(z)=K$. If $|\alpha|>1$, the transformation interchanges the disk with the exterior, including sending the point $\frac1{\overline{\alpha}}$ to $\infty$.