$$\begin{align*} {\left\lvert {w_1 - c^2 w_2} \right\rvert}^2 &= (w_1 - c^2 w_2) ( \mkern 1.5mu\overline{\mkern-1.5muw_1\mkern-1.5mu}\mkern 1.5mu - c^2 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu ) \\ &= {\left\lvert {w_1} \right\rvert}^2 + c^4 {\left\lvert {w_2} \right\rvert}^2 - 2c^2 \Re(w_1 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu) \\ &= {\color{green} c^2 {\left\lvert {w_2} \right\rvert}^2 } + c^4 {\left\lvert {w_2} \right\rvert}^2 - 2c^2 \Re(w_1 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu) \\ &= c^2 {\left\lvert {w_2} \right\rvert}^2 + {\color{green} c^2 {\left\lvert {w_1} \right\rvert}^2 } - 2c^2 \Re(w_1 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu) \\ &= c^2 {\left\lvert {w_1 - w_2} \right\rvert} ,\end{align*}$$ where we’ve applied the assumption ${\left\lvert {w_1} \right\rvert} = c{\left\lvert {w_2} \right\rvert}$ twice.

Using part 1: $$\begin{align*} w_1\coloneqq z-z_1, w_2 \coloneqq z-z_2 \implies {\left\lvert {w_1} \right\rvert} &= c{\left\lvert {w_2} \right\rvert} \\ \implies {\left\lvert {w_1 - c^2 w_2} \right\rvert} &= c {\left\lvert {w_1 - w_2} \right\rvert} \\ \implies {\left\lvert { z-z_1 - c^2 (z-z_2) } \right\rvert} &= {\left\lvert {(z-z_1) - (z-z_2)} \right\rvert} \\ \implies {\left\lvert {(1-c^2) z - z_3} \right\rvert} &= {\left\lvert { z_2 - z_1 } \right\rvert} \\ \implies {\left\lvert {z-z_4} \right\rvert} &= r ,\end{align*}$$ where the $z_i$ and $r$ are all constant, so this is the equation of a circle.

• A circle of radius 1 about $z=1$.

• A circle, using that Apollonius circles are characterized as the locus of distances whose ratios to some fixed points $A, B$ are constant. To actually compute this: $$\begin{align*} {\left\lvert {z-1} \right\rvert}^2 &= 4{\left\lvert {z-2} \right\rvert}^2 \\ \implies (x-1)^2 + y^2 - 4\qty{(x-2)^2 + y^2 } &=0 \\ -3x^2 + 14x - 3y^2 - 15 &= 0 && \star\\ \implies x^2 - {14\over 3}x + y^2 + 5 &= 0 \\ \implies (x- {14\over 2\cdot 3})^2 - {14\over 2\cdot 3}^2 + y^2 + 5 &= 0 \\ \implies (x-{14\over 6})^2 + y^2 = \qty{2\over 3}^2 ,\end{align*}$$ which is a circle of radius $2/3$ with center $\qty{{14\over 6}, 0}$. To avoid the calculation, use $$\begin{align*} Ax^2 + Bxy + Cy + \cdots = 0,\quad A=1, B=0, C=1 \implies \Delta \coloneqq B^2 - 4AC < 0 ,\end{align*}$$ which is an ellipse, and since $A=C$ it is in fact a circle.

• $S^1$, using that ${1\over z} = {\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \over z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} = {\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\over {\left\lvert {z} \right\rvert}^2}$ and if this equals $\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$, then ${\left\lvert {z} \right\rvert}^2=1$. Alternatively, $1 = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5muz = {\left\lvert {z} \right\rvert}^2$.

• Vertical line through $z=3$.

• Horizontal line through $z=ia$.

• Region to the right of the vertical line through $z=a$.

• Exterior of a circle: same calculation is (2), replacing $=0$ with $<0$. Note that the line marked $\star$ involves dividing by a negative, so this flips the sign, and we get $\cdots > \qty{2\over 3}^2$ at the end.

The geometry at hand:

By the inscribed angle theorem, this locus is an arc of a circle whose center $O$ is the point for which the angle $aOb$ is $2\theta$:

$\implies$: Write the vertices as $z_1, z_2, z_3$ and the sides as

• $s_1 \coloneqq z_2-z_1$
• $s_2 \coloneqq z_3 - z_2$
• $s_1 \coloneqq z_1 -z_3$

Note that $s_i = \pm \zeta_3 s_{i-1}$, dividing yields $$\begin{align*} {s_2 \over s_3} &= {s_1\over s_2} \\ &\iff s_2^2 - s_1 s_3 = 0 \\ &\iff \left(z_{2}-z_{3}\right)^{2}-\left(z_{2}-z_{1}\right)\left(z_{1}-z_{3}\right)=0 \\ &\iff \left(z_{2}^{2}+z_{3}^{2}-2 z_{2} z_{3}\right)-\left(z_{2} z_{1}-z_{2} z_{3}-z_{1}^{2}+z_{1} z_{3}\right)=0 \\ &\iff z_{1}^{2}+z_{2}^{2}+z_{3}^{2}-\left(z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\right)=0 .\end{align*}$$

$\impliedby$: We still have $s_i = \theta_i s_{i-1}$ for some angles $\theta_i$ We have

and $$\begin{align*} {s_1\over s_2} &= {\theta_1 \over \theta_2} \cdot {s_3\over s_1} \\ {s_2\over s_3} &= {\theta_2 \over \theta_3} \cdot {s_1\over s_2} \\ {s_3\over s_1} &= {\theta_3 \over \theta_1} \cdot {s_2\over s_3} .\end{align*}$$

Running the above calculation backward yields $s_2/s_3 = s_1/s_2$, and by the 2nd equality above, this forces $\theta_2 = \theta_3$. Similar arguments show $\theta_1=\theta_2 = \theta_3$ which forces $s_1=s_2 = s_3$.

First note that by the maximum modulus principal, it suffices to consider sups on the boundary, i.e. $$\begin{align*} M_R = \sup_{{\left\lvert {z} \right\rvert} = R}{\left\lvert {f(z)} \right\rvert}, \qquad N_R = \sup_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {f'(z)} \right\rvert} .\end{align*}$$

The first estimate: stuck!

The second estimate: suppose $z_0 \in {\mathbb{D}}_R(0)$, then any $D_R(z_0)$ is contained in $D_{2R}(0)$, So for any such $z_0$, apply Cauchy’s integral formula centered at $z_0$: $$\begin{align*} f^{(1)}(z_0) &= {1\over 2\pi i }\oint_{{{\partial}}{\mathbb{D}}_{R}(z_0)} {f(\xi)\over (\xi-z_0)^2 }\,d\xi\\ \implies {\left\lvert { f^{(1)}(z_0)} \right\rvert} &\leq {1\over 2\pi} \oint_{{{\partial}}{\mathbb{D}}_{R}(z_0)} {\left\lvert {f(\xi)\over (\xi-z_0)^2 } \right\rvert}\,d\xi\\ &= {1\over 2\pi} \oint_{{{\partial}}{\mathbb{D}}_R(z_0)} { {\left\lvert {f(\xi)} \right\rvert} \over {\left\lvert {\xi-z_0} \right\rvert}^2 } \,d\xi\\ &= {1\over 2\pi} \oint_{{{\partial}}{\mathbb{D}}_R(z_0)} { {\left\lvert {f(\xi)} \right\rvert} \over R^2 } \,d\xi\\ &\leq {1\over 2\pi} R^{-2} \oint_{{{\partial}}{\mathbb{D}}_R(z_0)} { \sup_{z\in {{\partial}}{\mathbb{D}}_{R}(z_0) } {\left\lvert {f(z)} \right\rvert} } \,d\xi\\ &= {1\over 2\pi} R^{-2} \sup_{{{\partial}}{\mathbb{D}}_R(z_0) } {\left\lvert {f(z)} \right\rvert} \cdot 2\pi R \\ &= R^{-1}\sup_{{{\partial}}{\mathbb{D}}_R(z_0) } {\left\lvert {f(z)} \right\rvert} \\ &\leq R^{-1}M_{2R} ,\end{align*}$$ where we’ve used in the last step that ${\mathbb{D}}_R(z_0) \subseteq {\mathbb{D}}_{2R}(0)$, and sups can only get larger when taken over larger sets. Since this was an arbitrary $z_0\in {\mathbb{D}}_R(0)$, this holds for all $z$ with ${\left\lvert {z} \right\rvert} \leq R$. Since taking sups preserves inequalities, we have $$\begin{align*} {\left\lvert {f'(z_0)} \right\rvert} \leq R^{-1}M_{2R}\, \forall {\left\lvert {z} \right\rvert} \leq R \implies N_R\coloneqq\sup_{{\left\lvert {z} \right\rvert} \leq R}{\left\lvert {f'(z)} \right\rvert} \leq R^{-1}M_{2R} .\end{align*}$$

Part 1: For ease of notation, let $z=z_1$ and $w=z_2$ We want to show $$\begin{align*} {\left\lvert {1- z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \right\rvert} - {\left\lvert {z-w} \right\rvert}^2 = \qty{{1 - {\left\lvert {z} \right\rvert}^2 }} \qty{{1 - {\left\lvert {w} \right\rvert}^2}} .\end{align*}$$

So write $$\begin{align*} {\left\lvert {1- z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \right\rvert} - {\left\lvert {z-w} \right\rvert}^2 &= (1-z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu)(1-\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w) - (z-w)(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu) \\ &= 1 - z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w - {\left\lvert {z} \right\rvert}^2{\left\lvert {w} \right\rvert}^2 - {\left\lvert {z} \right\rvert}^2 - {\left\lvert {w} \right\rvert}^2 + w\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu \\ &= 1 - {\left\lvert {z} \right\rvert}^2{\left\lvert {w} \right\rvert}^2 - {\left\lvert {z} \right\rvert}^2 - {\left\lvert {w} \right\rvert}^2 \\ &=(1-{\left\lvert {z} \right\rvert}^2)(1-{\left\lvert {w} \right\rvert}^2) .\end{align*}$$

Part 2 and 3: $$\begin{align*} {\left\lvert {z-w \over 1 - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w} \right\rvert}^2 \leq 1 &\iff 0 \leq {\left\lvert {1 - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w} \right\rvert}^2 - {\left\lvert {z-w} \right\rvert}^2 \\ &\iff 0 \leq (1 - {\left\lvert {z} \right\rvert}^2 )(1 - {\left\lvert {w} \right\rvert}^2) ,\end{align*}$$ where we’ve used part 1. But this is clearly true when ${\left\lvert {z} \right\rvert}, {\left\lvert {w} \right\rvert} < 1$, so the RHS is positive. Moreover if ${\left\lvert {z} \right\rvert} = {\left\lvert {w} \right\rvert} = 1$, the RHS is zero, yielding equalities at every step instead.

Part 1: See Spring 2021.1 above.

Part 2, holomorphicity: This is clearly meromorphic, as it’s a rational function, and has a singularity only at $z$ such that $\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z = 1$. This can only happen if $z, w \in S^1$: taking the modulus yields $$\begin{align*} \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z = 1 \implies {\left\lvert {w} \right\rvert}^2{\left\lvert {z} \right\rvert}^2 = 1 ,\end{align*}$$ and moreover since ${\left\lvert {w} \right\rvert}^2 \leq 1$ and ${\left\lvert {z} \right\rvert}^2\leq 1$, the only way this product can be one is when ${\left\lvert {w} \right\rvert}^2 = {\left\lvert {z} \right\rvert}^2 = 1$. This also forces $z=1/\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu$.

The claim is that the singularity $1/\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu$ is removable. Note that $1\over w = \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu$ on the circle, so $1/\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu = \mkern 1.5mu\overline{\mkern-1.5mu\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu\mkern-1.5mu}\mkern 1.5mu = 2$, so $$\begin{align*} \qty{ z- \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu^{-1}} \qty{w-z \over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} &= \qty{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z - 1 \over \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \qty{w-z \over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \\ &= \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu^{-1}(w-z) \\ &= w(w-z) \\ &\overset{z\to \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu^{-1}=w }\to 0 .\end{align*}$$

Part 2, being a bijection: This follows from finding an explicit inverse, using that $F^2 = \operatorname{id}$: $$\begin{align*} F(F(z)) &= \frac{w- \qty{ \frac{w-z}{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} } }{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu \qty{ \frac{w-z}{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} } } \\ &= \frac{w(1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z)-(w-z)}{q-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu(w-z)} \\ &= \frac{w-|w|^{2} z-w+z}{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z-|w|^{2}+\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \\ &= \frac{z\left(1-|w|^{2}\right)}{1-|w|^{2}} \\ &= z .\end{align*}$$

Part 2, being an involution: A direct check shows that $F(w) = 0$, since the numerator vanishes, and $F(0) = {w - 0 \over 1 - 0} = w$.

Part 3, preserving the circle: Follows from the estimate in part 1.

$$\begin{align*} {\left\lvert {z-w} \right\rvert}^2 &= (z-w)(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu) \\ &= {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w \\ &= {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - 2\Re(\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) \\ &\geq {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - 2{\left\lvert {\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu } \right\rvert}{\left\lvert {z} \right\rvert} \\ &\geq \qty{ {\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert} }^2 ,\end{align*}$$ and taking square roots introduces an absolute value on the final term. Here we’ve used the basic estimate $$\begin{align*} \Re(z) \leq {\left\lvert {z} \right\rvert} \implies -\Re(z) \geq -{\left\lvert {z} \right\rvert} .\end{align*}$$

$$\begin{align*} \prod_{1\leq k \leq n-1} \sin\qty{k\pi\over n} &= \prod_{1\leq k \leq n-1} {\omega_n^k - \omega_n^{-k} \over 2i} \\ &= \qty{1\over 2i}^{n-1} \prod_{1\leq k \leq n-1} \omega_n^{k} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2i}^{n-1} \prod_{1\leq k \leq n-1} \exp\qty{i\pi k\over n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2i}^{n-1} \exp\qty{ {i\pi\over n} \displaystyle\sum_{1\leq k \leq n-1} k } \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2i}^{n-1} e^{i\pi n(n-1)\over 2n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2}^{n-1}\qty{1\over i}^{n-1} \qty{ e^{i\pi \over 2} }^{n-1} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= 2^{1-n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= 2^{1-n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{k}} \\ &= 2^{1-n} { \Phi_n(z) \over z-1}\Big|_{z=1} \\ &= 2^{1-n} \qty{ \sum_{0\leq k \leq n-1} z^k}\Big|_{z=1} \\ &= n2^{1-n} .\end{align*}$$

$$\begin{align*} \prod_{1\leq j\leq n-1} \sin \left(\frac{j \pi}{n}\right) &=\prod_{1\leq j\leq n-1} \frac{1}{2 i} \qty{\zeta_n^{j\over 2} - \zeta_n^{- {j \over 2} }} \\ &=\left(\frac{1}{2 i}\right)^{n-1} \prod_{1\leq j\leq n-1} \zeta_n^{j\over 2} \prod_{1\leq j \leq n-1} \qty{1 - \zeta_n^{-j}} \\ &=\left(\frac{1}{2 i}\right)^{n-1} \prod_{1\leq j\leq n-1} \exp\qty{ij\pi \over n} \prod_{1\leq j \leq n-1} \qty{1 - \zeta_n^{-j}} \\ &=\left(\frac{1}{2 i}\right)^{n-1} \exp \left(\sum_{1\leq j\leq n-1} \frac{i j \pi}{n}\right) \Phi_{n}(1)\\ &=\left(\frac{1}{2 i}\right)^{n-1} \exp \left(\frac{(n-1) i \pi}{2}\right) \Phi_{n}(1)\\ &=\left(\frac{1}{2 i}\right)^{n-1}\left(e^{i \pi / 2}\right)^{n-1} \Phi_{n}(1)\\ &=\left(\frac{1}{2 i}\right)^{n-1} i^{n-1} \Phi_{n}(1)\\ &=\left(\frac{1}{2}\right)^{n-1} \Phi_{n}(1)\\ &=\frac{n}{2^{n-1}} .\end{align*}$$

$$\begin{align*} 0 &\leq (1 - {\left\lvert {w} \right\rvert}^2)(1-{\left\lvert {z} \right\rvert}^2) \\ \implies {\left\lvert {w} \right\rvert}^2 + {\left\lvert {z} \right\rvert}^2 &\leq 1 + {\left\lvert {w} \right\rvert}^2 {\left\lvert {z} \right\rvert}^2 \\ \implies {\left\lvert {w} \right\rvert}^2 + {\left\lvert {z} \right\rvert}^2 - 2\Re(\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) &\leq 1 + {\left\lvert {w} \right\rvert}^2 {\left\lvert {z} \right\rvert}^2 - 2\Re(\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) \\ \implies {\left\lvert {w-z} \right\rvert}^2 &\leq {\left\lvert {1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5muz} \right\rvert}^2 .\end{align*}$$ Note that if either ${\left\lvert {w} \right\rvert}^2 = 1$ or ${\left\lvert {z} \right\rvert}^2 = 1$ then the first line is an equality, yielding equality in the final line.

• That $F: {\mathbb{D}}\to {\mathbb{D}}$: follows from the inequality, since ${\left\lvert {z} \right\rvert}, {\left\lvert {w} \right\rvert} < 1$ for $z,w\in {\mathbb{D}}$. Holomorphicity: follows from the fact that rational expressions of holomorphic functions are holomorphic away from where the denominators vanish.

• Then just note that ${\left\lvert {\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \right\rvert} \lleq {\left\lvert {w} \right\rvert}{\left\lvert {z} \right\rvert} < 1$, so ${\left\lvert {1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \right\rvert} > 0$.

• $F(0) = {w-0 \over 1-0} = w$

• $F(w) = {w-w\over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu w} = 0$

• ${\left\lvert {z} \right\rvert} = 1$ yields equality in part 1.

Other notes: $F$ is bijective on ${\mathbb{D}}$: $$\begin{align*} F(F(z)) &= {w - \qty{w-z\over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu\qty{w-z\over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} } \\ &= {w(1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) - (w-z) \over (1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu (w-z)} \\ &= {z-{\left\lvert {w} \right\rvert}^2 z \over 1 - {\left\lvert {w} \right\rvert}^2 }\\ &= z .\end{align*}$$