Can the Mapping $F$ Always Interchange Two Points in a Unit Disk?

[Chris L T521]

Here's this week's problem.

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Problem: (a) Let $z,\,w$ be two complex numbers such that $\overline{z}w\neq 1$. Prove that\[\left|\frac{w-z}{1-\overline{w}z}\right|<1\quad\text{if $\left|z\right|<1$ and $\left|w\right|<1$,}\]
and also that
\[\left|\frac{w-z}{1-\overline{w}z}\right|=1\quad\text{if $\left|z\right|=1$ or $\left|w\right|=1$.}\]

(b) Prove that for a fixed $w$ in the unit disk $\mathbb{D}$, the mapping\[F:z\mapsto \frac{w-z}{1-\overline{w}z}\]
satisfies the following conditions:

(i) $F$ maps the unit disc to itself (that is, $F:\mathbb{D}\rightarrow\mathbb{D}$), and is holomorphic.
(ii) $F$ interchanges $0$ and $w$, namely $F(0)=w$ and $F(w)=0$.
(iii) $\left|F(z)\right|=1$ if $\left|z\right|=1$.
(iv) $F:\mathbb{D}\rightarrow\mathbb{D}$ is bijective. [Hint: Calculate $F\circ F$.]
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[Chris L T521]

There were no takers this week. Here's my solution.

Spoiler

(a) Proof: Consider the following:\[\begin{aligned}
\left|\frac{w-z}{1-\overline{w}z}\right|^2 &= \frac{w-z}{1-\overline{w}z}\frac{\overline{w}-\overline{z}}{1-w\overline{z}}\\ &= \frac{\overline{w}w-\overline{w}z-w\overline{z}+\overline{z}z}{1-\overline{w}z-w\overline{z}+\overline{w}w\overline{z}z}\\ &= \frac{(1-\overline{w} z-w\overline{z}+\overline{w}w\overline{z}z) + ( \overline{z} z+\overline{w}w-1-\overline{w}w\overline{z}z)}{1-\overline{w}z-w\overline{z}+\overline{w}w\overline{z}z}\\ &= 1-\frac{(1-\left|z\right|^2)+(\left|w\right|^2\left|z\right|^2-\left|w\right|^2)}{\left|1-\overline{w}z\right|^2}\\ &= 1-\frac{(1-\left|z\right|^2)(1-\left|w\right|^2)}{\left|1-\overline{w}z\right|^2}.
\end{aligned}\]
Now, if $\left|z\right|<1$ and $\left|w\right|<1$, then
\[1-\frac{(1-\left|z\right|^2)(1-\left|w\right|^2)}{\left|1-\overline{w}z\right|^2}<1-\frac{(1-1)(1-1)}{\left|1-\overline{w}z\right|^2}=1\]
since $\overline{z}w\neq 1$. If $\left|z\right|=1$ (or $\left|w\right|=1$), then
\[1-\frac{(1-\left|z\right|^2)(1-\left|w\right|^2)}{\left|1-\overline{w}z\right|^2}=1-\frac{(1-1)(1-\left|w\right|^2)}{\left|1-\overline{w}z\right|^2}=1.\]
Thus,
\[\left|\frac{w-z}{1-\overline{w}z}\right|^2<1 \implies \left|\frac{w-z}{1-\overline{w}z}\right|<1\]
and
\[\left|\frac{w-z}{1-\overline{w}z}\right|^2=1 \implies \left|\frac{w-z}{1-\overline{w}z}\right|=1.\]
This completes the proof. Q.E.D.

(b):
(i) Proof: Let $z\in\mathbb{D}\implies\left|z\right|<1$. Since $w\in\mathbb{D}$ is fixed (and $\left|w\right|<1$), it is clear that for each $z$, $F(z) = \dfrac{w-z}{1-\overline{w}z}\in\mathbb{D}$ since $\left|F(z)\right|<1$ by part (a). Therefore $F(\mathbb{D})\subseteq\mathbb{D}$. Now, since $F(z)$ is rational and $\left|1-\overline{w}z\right|>1-\left|w\right|\left|z\right|>0$ for each $z\in\mathbb{D}$, it follows that $F$ is holomorphic at each $z\in\mathbb{D}$. Q.E.D.

(ii) Proof: Clearly, $F(0)=\dfrac{w}{1}=w\quad\text{and}\quad F(w)=\dfrac{0}{1-\left|w\right|^2}=0$. Q.E.D.

(ii) Proof: This is a direct consequence of part (a) since $\left|w\right|\neq 1$ and $\left|z\right|=1$. Q.E.D.

(iv) Proof: It follows that\[\begin{aligned}
F\circ F(z) &= \frac{w-\dfrac{w-z}{1-\overline{w}z}}{1-\overline{w}\dfrac{w-z}{1-\overline{w}z}}\\ &= \frac{\dfrac{w-\left|w\right|^2z-w+z}{1-\overline{w}z}}{\dfrac{1-\overline{w}z-\left|w\right|^2+\overline{w}z}{1-\overline{w}z}}\\ &= \frac{z(1-\left|w\right|^2)}{1-\left|w\right|^2}\\ &= z\quad(\text{since $w\in\mathbb{D}\implies\left|w\right|<1$}.)
\end{aligned}\]
Thus, $F$ is bijective. Q.E.D.