\(% 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}}\)
Since \(A, Q, C\) lie on a straight line, \(\vect {AQ} = \lambda \vect {AC}\) for some \(\lambda \in \RR \). This means \[ q - a = \lambda (c - a), \] and hence \[ \frac {q - a}{c - a} = \lambda \in \RR , \] as required.
Hence, we must have \[ \frac {q - a}{c - a} = \left (\frac {q - a}{c - a}\right )^* = \frac {q^* - a^*}{c^* - a^*}. \]
Cross-multiplying the terms out give \[ (c - a)(q^* - a^*) = (c^* - a^*)(q - a) \] exactly as desired.
Substituting in \(a^* = 1 / a\) and \(c^* = 1 / c\), we have \[ (c - a) \left (q^* - \frac {1}{a}\right ) = \left (\frac {1}{c} - \frac {1}{a}\right )(q - a), \] and expanding the brackets gives \[ cq^* - aq^* - \frac {c}{a} + 1 = \frac {q}{c} - \frac {a}{c} - \frac {q}{a} + 1, \] and hence \[ cq^* - aq^* - \frac {c}{a} = \frac {q}{c} - \frac {a}{c} - \frac {q}{a}. \]
Multiplying by \(ac\) on both sides gives us \[ a c^2 q^* - a^2 c q^* - c^2 = aq - a^2 - cq, \] and hence \[ ac (c - a) q^* = (a - c)q - (a^2 - c^2) = (a - c)q - (a - c)(a + c). \]
We can divide through \((a - c)\) on both sides since \(a \neq c\). Hence, \[ 0 = q - (a + c) + ac q^*, \] and hence \[ acq^* + q = a + c, \] as desired.
By part 1, we must have \[ acq^* + q = a + c, bdq^* + q = b + d. \]
Since \(q = q\), we have \[ acq^* - (a + c) = bdq^* - (b + d), \] and rearranging gives \[ (ac - bd) q^* = (a + c) - (b + d), \] exactly as desired.
We also have \(q^* = q^*\), and hence \[ \frac {a + c - q}{ac} = \frac {b + d - q}{bd}, \] which gives \[ (bd)(a + c - q) = (ac)(b + d - q), \] and rearranging gives \[ (ac - bd)q = ac(b + d) - bd(a + c). \]
Summing this with previously, we have \[ (ac - bd)(q + q^*) = (a + c) - (b + d) + ac(b + d) - bd(a + c). \]
We notice that \begin {align*} (a + c) - (b + d) + ac(b + d) - bd(a + c) & = a + c - b - d + abc + acd - abd - bcd \\ & = a - b + acd - bcd + c - d + abc - abd \\ & = (a - b)(1 + cd) + (c - d)(1 + ab), \end {align*}
and hence \[ (ac - bd)(q + q^*) = (a - b)(1 + cd) + (c - d)(1 + ab), \] exactly as desired.
By part 1, we must have \[ p + abp^* = a + b. \]
Since \(p\) is real, \(p = p^*\), and hence \[ (1 + ab)p = a + b, \] as desired.
Similarly, we must have \[ (1 + cd)q = c + d, \] and putting this back into the result from part 2, we have \[ (ac - bd)(q + q^*) = \frac {(a - b)(c + d)}{p} + \frac {(c - d)(a + b)}{p}, \] and hence since \(ac - bd \neq 0\), we have \begin {align*} p(q + q^*) & = \frac {(a - b)(c + d) + (c - d)(a + b)}{ac - bd} \\ & = \frac {ac + ad - bc - bd + ac + bc - ad - bd}{ac - bd} \\ & = \frac {2ac - 2bd}{ac - bd} \\ & = 2, \end {align*}
as desired.