\begin{align*} {\left\lvert {w_1 - c^2 w_2} \right\rvert}^2 &= (w_1 - c^2 w_2) ( \overline{w_1} - c^2 \overline{w_2} ) \\ &= {\left\lvert {w_1} \right\rvert}^2 + c^4 {\left\lvert {w_2} \right\rvert}^2 - 2c^2 \Re(w_1 \overline{w_2}) \\ &= {\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 \overline{w_2}) \\ &= 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 \overline{w_2}) \\ &= 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 \({1\over z} = {\overline{z} \over z\overline{z}} = {\overline{z}\over {\left\lvert {z} \right\rvert}^2}\) and if this equals \(\overline{z}\), then \({\left\lvert {z} \right\rvert}^2=1\). Alternatively, \(1 = \overline{z}z = {\left\lvert {z} \right\rvert}^2\).

• 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\overline{w}} \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\overline{w}} \right\rvert} - {\left\lvert {z-w} \right\rvert}^2 &= (1-z\overline{w})(1-\overline{z} w) - (z-w)(\overline{z} - \overline{w}) \\ &= 1 - z\overline{w} - \overline{z} 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\overline{z} + z\overline{w} \\ &= 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 - \overline{z} w} \right\rvert}^2 \leq 1 &\iff 0 \leq {\left\lvert {1 - \overline{z} 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 \(\overline{w} z = 1\). This can only happen if \(z, w \in S^1\): taking the modulus yields \begin{align*} \overline{w} 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 \(z=1/\overline{w}\).

The claim is that the singularity \(1/\overline{w}\) is removable. Note that \(1\over w = \overline{w}\) on the circle, so \(1/\overline{w} = \overline{\overline{w}} = 2\), so \begin{align*} \qty{ z- \overline{w}^{-1}} \qty{w-z \over 1-\overline{w} z} &= \qty{\overline{w} z - 1 \over \overline{w}} \qty{w-z \over 1-\overline{w} z} \\ &= \overline{w}^{-1}(w-z) \\ &= w(w-z) \\ &\overset{z\to \overline{w}^{-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-\overline{w} z} } }{1-\overline{w} \qty{ \frac{w-z}{1-\overline{w} z} } } \\ &= \frac{w(1-\overline{w} z)-(w-z)}{q-\overline{w} z-\overline{w}(w-z)} \\ &= \frac{w-|w|^{2} z-w+z}{1-\overline{w} z-|w|^{2}+\overline{w} 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)(\overline{z} - \overline{w}) \\ &= {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - z\overline{w} - \overline{z} w \\ &= {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - 2\Re(\overline{w} z) \\ &\geq {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - 2{\left\lvert {\overline{w} } \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(\overline{w} z) &\leq 1 + {\left\lvert {w} \right\rvert}^2 {\left\lvert {z} \right\rvert}^2 - 2\Re(\overline{w} z) \\ \implies {\left\lvert {w-z} \right\rvert}^2 &\leq {\left\lvert {1-\overline{w}z} \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 \({\left\lvert {\overline{w} z} \right\rvert} \leq {\left\lvert {w} \right\rvert}{\left\lvert {z} \right\rvert} < 1\), so \({\left\lvert {1 - \overline{w} z} \right\rvert} > 0\).

• \(F(0) = {w-0 \over 1-0} = w\)

• \(F(w) = {w-w\over 1 - \overline{w} w} = 0\)

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

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