2017.3.2 Question 2

1. Let the complex number representing $R(P)$ be $z^{\prime }$. Therefore,

   $$\begin{array}{llll}\hfill z^{\prime }-a& =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃})(z-a),& \hfill & \\ \hfill z^{\prime }& =z\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃})+a(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃})),& \hfill & \end{array}$$

   as desired.

2. Let the complex number representing $\mathit{SR}(P)$ be $z^{″}$. Therefore,

   $$\begin{array}{llll}\hfill z^{″}-b& =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })(z^{\prime }-b),& \hfill & \\ \hfill z^{″}& =z^{\prime }\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })+b(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })),& \hfill & \\ \hfill z^{″}& =\left[z\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃})+a(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃}))\right]\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })+b(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })),& \hfill & \\ \hfill z^{″}& =z\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(i\left(𝜃+\phi \right))+a(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃}))\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })+b(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })).& \hfill & \\ \hfill & & \hfill \end{array}$$

   This will be an anti-clockwise rotation around $c$ over an angle of $(𝜃+\phi )$, where

   $$c\left[1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(i(𝜃+\phi ))\right]=a\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })-a\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(i\left(𝜃+\phi \right))+b-b\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi }),$$

   If $𝜃+\phi =2\mathit{n\pi }$ for some integer $n\in \mathbb{Z}$, $1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(i(𝜃+\phi ))=0$, therefore $c$ cannot be determined.

   Multiplying both sides by $\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{i(𝜃+\phi )}{2}\right)$, we have

   $$\begin{array}{llll}\hfill & c\left[\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{i(𝜃+\phi )}{2}\right)-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{i(𝜃+\phi )}{2}\right)\right]& \hfill & \\ \hfill & =a\left[\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{i(\phi -𝜃)}{2}\right)-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{i(𝜃+\phi )}{2}\right)\right]+b\left[\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{i(𝜃+\phi )}{2}\right)-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{i(\phi -𝜃)}{2}\right)\right],& \hfill & \end{array}$$

   and hence

   $$\begin{array}{llll}\hfill -2\mathit{ci}\text{sin}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{𝜃+\phi }{2}\right)& =-2\mathit{ai}\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\mathit{i\phi }}{2}\right)\text{sin}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{𝜃}{2}\right)-2\mathit{bi}\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{\mathit{i𝜃}}{2}\right)\text{sin}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\phi }{2}\right),& \hfill & \\ \hfill c\text{sin}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{𝜃+\phi }{2}\right)& =a\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\mathit{i\phi }}{2}\right)\text{sin}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{𝜃}{2}\right)+b\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{\mathit{i𝜃}}{2}\right)\text{sin}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\phi }{2}\right).& \hfill & \end{array}$$

   If $𝜃+\phi =2\pi$, we will have $z^{″}=z+a\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })-a+b(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi }))=z+(b-a)(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi }))$, which is a translation by $(b-a)(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi }))$.

3. If $\mathit{RS}=\mathit{SR}$, then we have

   $$\begin{array}{llll}\hfill a(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃}))\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })+b(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi }))& =b(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi }))\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃})+a(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃})),& \hfill & \\ \hfill a(-1+\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })+\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃})-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(i(𝜃+\phi )))& =b(-1+\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi })+\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃})-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(i(𝜃+\phi ))),& \hfill & \\ \hfill (a-b)(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i\phi }))(1-\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{i𝜃}))& =0.& \hfill & \end{array}$$

   Therefore, $a=b$, or $\phi =2\mathit{n\pi }$, or $𝜃=2\mathit{n\pi }$, for some integer $n\in \mathbb{Z}$.