\(% 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 \(k\) represent the point \(K\) in the complex plane, we must have \[ k - a = (b - a) \exp \left (- i\frac {\pi }{3}\right ), \] and hence \[ k = a + (b - a) \exp \left (- i\frac {\pi }{3}\right ). \]
Notice that \[ \omega = \exp \left (\frac {i\pi }{6}\right ) = \cos \frac {\pi }{6} + i \sin \frac {\pi }{6} = \frac {\sqrt {3}}{2} + i \frac {1}{2}, \] and hence \[ \omega ^* = \frac {\sqrt {3}}{2} - i \frac {1}{2} \]
Hence, \(g_{ab}\) is given by \begin {align*} g_{ab} & = \frac {a + b + k}{3} \\ & = \frac {a + b + a + (b - a) \exp \left (- i\frac {\pi }{3}\right )}{3} \\ & = \frac {2a + b + (b - a) \left [\cos \frac {\pi }{3} - i \sin \frac {\pi }{3}\right ]}{3} \\ & = \frac {2a + b + (b - a) \left (\frac {1}{2} - i \frac {\sqrt {3}}{2}\right )}{3} \\ & = \frac {\left (\frac {3}{2} + i \frac {\sqrt {3}}{2}\right ) a + \left (\frac {3}{2} - i \frac {\sqrt {3}}{2}\right ) b}{3} \\ & = \frac {1}{\sqrt {3}} \cdot \left [\left (\frac {\sqrt {3}}{2} + i \frac {1}{2}\right ) a + \left (\frac {\sqrt {3}}{2} - i \frac {1}{2}\right ) b\right ] \\ & = \frac {1}{\sqrt {3}} \cdot \left (\omega a + \omega ^* b\right ), \end {align*}
as desired.
\(Q_2\) is a parallelogram if and only if \begin {align*} g_{bc} - g_{ab} & = g_{cd} - g_{da} \\ \frac {1}{\sqrt {3}} \left (\omega b + \omega ^* c\right ) - \frac {1}{\sqrt {3}} \left (\omega a + \omega ^* b\right ) & = \frac {1}{\sqrt {3}} \left (\omega c + \omega ^* d\right ) - \frac {1}{\sqrt {3}} \left (\omega d + \omega ^* a\right ) \\ \omega (b - a - c + d) & = \omega ^* (d - a - c + b) \\ \left (\omega - \omega ^*\right ) \left [(b - a) - (c - d)\right ] & = 0 \\ (b - a) - (c - d) & = 0 \\ b - a & = c - d, \end {align*}
which is true if and only if \(Q_1\) is a parallelogram. All the steps above are reversible. In particular, \(\omega - \omega ^* \neq 0\) so we can divide by \(\omega - \omega ^*\) on both sides.
Notice that \begin {align*} g_{bc} - g_{ab} & = \frac {1}{\sqrt {3}} \left (\omega b + \omega ^* c\right ) - \frac {1}{\sqrt {3}} \left (\omega a + \omega ^* b\right ) \\ & = \frac {1}{\sqrt {3}} \left [\omega ^* c - \omega a + (\omega - \omega ^*) b\right ] \\ & = \frac {1}{\sqrt {3}} \left [\omega ^* c - \omega a + bi\right ], \end {align*}
and that \begin {align*} g_{ca} - g_{ab} & = \frac {1}{\sqrt {3}} \left (\omega c + \omega ^* a\right ) - \frac {1}{\sqrt {3}} \left (\omega a + \omega ^* b\right ) \\ & = \frac {1}{\sqrt {3}} \left [\omega c + (\omega ^* - \omega )a - \omega ^* b\right ] \\ & = \frac {1}{\sqrt {3}} \left [\omega c - ai - \omega ^* b\right ]. \end {align*}
Notice that \begin {align*} \frac {\omega ^*}{\omega } & = \frac {\exp \left (-\frac {i\pi }{6}\right )}{\exp \left (\frac {i\pi }{6}\right )} = \exp \left (-\frac {i\pi }{3}\right ), \\ \frac {-\omega }{-i} & = \frac {\omega }{i} = \frac {\exp \left (\frac {i\pi }{6}\right )}{\exp \left (\frac {i\pi }{2}\right )} = \exp \left (-\frac {i\pi }{3}\right ), \\ \frac {i}{-\omega ^*} & = \frac {-i}{\omega ^*} = \frac {\exp \left (-\frac {i\pi }{2}\right )}{\exp \left (-\frac {i\pi }{6}\right )} = \exp \left (-\frac {i\pi }{3}\right ), \end {align*}
and hence we can wee \[ g_{bc} - g_{ab} = \left (g_{ca} - g_{ab}\right ) \exp \left (- \frac {i\pi }{3}\right ), \] which means \(G_{BC}\) is the image of \(G_{CA}\) under rotation through \(\frac {\pi }{3}\) clockwise about \(G_{AB}\), and this shows that \(T_2\) is an equilateral triangle.