so I am doing an assignment on triangles and complex numbers, but I am stuck in the very first question. I am not asking for the solution, I would just like a hint or some ideas on what I need to look at to get me started. Thanks!
Consider a triangle $ABC$ in the plane with angles $A$, $B$, $C$ and sides $a$, $b$, $c$ with the side of length a opposite the angle $A$, etc. Suppose that the angles appear in the order $A$, $B$, $C$ as one traverses the vertices in an anticlockwise direction. Let the 3 vertices be complex numbers $\alpha$, $\beta$, $\gamma$ with $\alpha$ corresponding to the vertex with angle A, etc. Therefore, a triangle $(\alpha, \beta, \gamma)$ is a point in a 3-dimensional complex plane.
Question: Show that for the triangle $ABC$ we have $$(be^{iA}-c)\alpha - be^{iA}\beta+c\gamma=0$$
In the following, I assume you know that the complex plane is (also) a two-dimensional (real) vector space, and what $e^{ix}$ is.
Considering an arbitrary triangle with vertices, angles and sides as you have given them, what is $\alpha-\beta$? What is $\gamma-\alpha$? Can you rearrange the terms in the equation so that these differences appear? Once you have simplified it to something like $X+Y=0$, it is clear what $Y$ must be. Why is this indeed the case?