\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Let the complex number representing \(R(P)\) be \(z'\). Therefore, \begin {align*} z' - a & = \exp (i\theta ) (z - a), \\ z' & = z \exp (i\theta ) + a (1 - \exp (i\theta )), \end {align*}
as desired.
Let the complex number representing \(SR(P)\) be \(z''\). Therefore, \begin {align*} z'' - b & = \exp (i\phi ) (z' - b), \\ z'' & = z' \exp (i\phi ) + b(1 - \exp (i\phi )), \\ z'' & = \left [z \exp (i\theta ) + a (1 - \exp (i\theta ))\right ] \exp (i\phi ) + b(1 - \exp (i\phi )), \\ z'' & = z\exp (i\left (\theta + \phi \right )) + a (1 - \exp (i \theta )) \exp (i \phi ) + b (1 - \exp (i \phi )). \\ \end {align*}
This will be an anti-clockwise rotation around \(c\) over an angle of \((\theta + \phi )\), where \[ c \left [1 - \exp (i (\theta + \phi ))\right ] = a \exp (i \phi ) - a\exp (i \left (\theta + \phi \right )) + b - b \exp (i \phi ),\\ \]
If \(\theta + \phi = 2n\pi \) for some integer \(n \in \ZZ \), \(1 - \exp (i (\theta + \phi )) = 0\), therefore \(c\) cannot be determined.
Multiplying both sides by \(\exp \left (-\frac {i(\theta + \phi )}{2}\right )\), we have \begin {align*} & \phantom {=} c \left [\exp \left (-\frac {i(\theta + \phi )}{2}\right ) - \exp \left (\frac {i (\theta + \phi )}{2}\right )\right ] \\ & = a \left [\exp \left (\frac {i(\phi - \theta )}{2}\right ) - \exp \left (\frac {i(\theta + \phi )}{2}\right )\right ] + b\left [\exp \left (-\frac {i(\theta + \phi )}{2}\right ) - \exp \left (\frac {i(\phi - \theta )}{2}\right )\right ], \end {align*}
and hence \begin {align*} -2ci\sin \left (\frac {\theta + \phi }{2}\right ) & = - 2 ai \exp \left (\frac {i\phi }{2}\right ) \sin \left (\frac {\theta }{2}\right ) - 2 bi \exp \left (-\frac {i\theta }{2}\right ) \sin \left (\frac {\phi }{2}\right ), \\ c \sin \left (\frac {\theta + \phi }{2}\right ) & = a \exp \left (\frac {i\phi }{2}\right ) \sin \left (\frac {\theta }{2}\right ) + b \exp \left (-\frac {i\theta }{2}\right ) \sin \left (\frac {\phi }{2}\right ). \end {align*}
If \(\theta + \phi = 2\pi \), we will have \(z'' = z + a \exp (i\phi ) - a + b (1 - \exp (i\phi )) = z + (b-a) (1 - \exp (i\phi ))\), which is a translation by \((b-a) (1 - \exp (i\phi ))\).
If \(RS = SR\), then we have \begin {align*} a (1 - \exp (i \theta )) \exp (i \phi ) + b (1 - \exp (i \phi )) & = b (1 - \exp (i \phi )) \exp (i \theta ) + a (1 - \exp (i \theta )), \\ a (-1 + \exp (i \phi ) + \exp (i \theta ) - \exp (i (\theta + \phi ))) & = b (-1 + \exp (i \phi ) + \exp (i \theta ) - \exp (i (\theta + \phi ))), \\ (a-b) (1 - \exp (i \phi )) (1 - \exp (i \theta )) & = 0. \end {align*}
Therefore, \(a = b\), or \(\phi = 2 n \pi \), or \(\theta = 2 n \pi \), for some integer \(n \in \ZZ \).