# Deriving equations regarding triangles in the complex plane

I am having real trouble wrapping my head around this question. Any help would be much appreciated.

Consider a triangle ABC in the plane with angles $A,B,C$ and side lengths $a,b,c$ with the side of length $a$ opposite the angle $A$, etc, as usual. Suppose also that the angles appear in the order $A,B,C$ as one traverses the vertices in an anticlockwise direction.

Regard the three vertices as complex numbers $\alpha$,$\beta$,$\gamma$ with $\alpha$ corresponding to the vertex with angle $A$, etc. So a triangle corresponds to a point ($\alpha,\beta,\gamma$) in $3$-dimensional complex space. Two triangles are "similar" if they have the same angles. Let us say they are "directly similar" if they are similar and the angles are in the same cyclic order.

1) Show that for the triangle $ABC$ we have $(\beta e^{iA}-c)\alpha -b e^{iA}\beta + c \gamma=0$.

2) Show that the same equation holds for all triangles with vertices $\alpha,\beta,\gamma$ directly similar to $ABC$.

3) Find an analogous equation that holds for triangles that are similar, but not directly similar, to $ABC$.

1) Vectors AB and AC in this triangle can be represented as the complex numbers $\beta-\alpha$ and $\gamma-\alpha$ respectively. We know that the angle between these two vectors is A, and that to go from AB to AC we must rotate counterclockwise. We also know that the ratio of the magnitudes of AB and AC is $c:b$.
Thus we can write the equation $$b(\beta-\alpha)e^{iA}=c(\gamma-\alpha)$$ $$b\beta e^{iA}-b\alpha e^{iA}+c\alpha-c\gamma=0$$ This differs slightly from what you wrote, but I am fairly sure that mine is correct, so I suggest you check that answer.
The transformation that we want to maintain similarity is $$T(\alpha,\beta,\gamma)=(u\alpha+v,u\beta,u\gamma+v)$$ Where $u$ and $v$ are complex numbers. The $u$ is for dilation, and the $v$ is transformation.
Since b and c are the same (the triangle is similar), plugging these back into the equation we get, $$b(u\beta+v)e^{iA}-b(u\alpha+v) e^{iA}+c(u\alpha+v)-c(u\gamma+v)=0$$ Note that the $v$'s cancel out and then we can divide by $u$ to get our original equation.
3) For triangles which are simlar but not directly similar, the angle between AB and AC is either clockwise OR counterclockwise, so we can write the equation $$b(\beta-\alpha)e^{\pm iA}=c(\gamma-\alpha)$$ $$b\beta e^{\pm iA}-b\alpha e^{\pm iA}+c\alpha-c\gamma=0$$ Which will hold for all similar triangles.