(Course 2) Q2 Let $\alpha , \beta$ and $\gamma$ be three complex numbers representing three different points A, B and C on a complex plane. Also, $\alpha , \beta$ and $\gamma$ satisfy
$$\begin{aligned} & ( \gamma - \alpha ) ^ { 2 } + ( \gamma - \alpha ) ( \beta - \alpha ) + ( \beta - \alpha ) ^ { 2 } = 0 \quad \cdots (1)\\ & | \beta - 2 \alpha + \gamma | = 3 \quad \cdots (2) \end{aligned}$$
We are to find the area of the triangle ABC.
Since from (1)
$$\frac { \gamma - \alpha } { \beta - \alpha } = \frac { - \mathbf { M } \pm \sqrt { \mathbf { N } } i } { \mathbf { O } } ,$$
we have
$$\left| \frac { \gamma - \alpha } { \beta - \alpha } \right| = \mathbf { P } , \quad \arg \frac { \gamma - \alpha } { \beta - \alpha } = \pm \frac { \mathbf { Q } } { \mathbf { R } } \pi ,$$
where $- \pi < \arg \frac { \gamma - \alpha } { \beta - \alpha } < \pi$. Also, since
$$\beta - 2 \alpha + \gamma = ( \beta - \alpha ) \cdot \frac { \mathbf { S } \pm \sqrt { \mathbf { T } } } { \mathbf { U } } ,$$
we have from (2) that
$$| \beta - \alpha | = \mathbf { V } .$$