The question goes like this:
In triangle $ABC$, the lengths of the sides $BC$, $CA$, $AB$ are $a$, $b$, $c$ respectively, and the radian measures of the angles at $A$, $B$, $C$ are $\alpha$, $\beta$, $\gamma$ respectively. Prove that
$$a=c\mathrm{e}^{\mathrm{i}\beta}+b\mathrm{e}^{-\mathrm{i}\gamma}.$$
My attempt at a solution:
For convenience sake, let $z^{}_1$, $z^{}_2$ and $z^{}_3$ denote the complex numbers represented by the points $A$, $B$ and $C$ respectively in an Argand diagram.
Referring to my drawing above, since $\overrightarrow{CB}$ is $\overrightarrow{CA}$ stretched by a factor of $\dfrac{a}{b}$ and rotated through $\gamma$ radians anticlockwise (about $C$), I conclude that
$$z_2-z_3=\frac{a}{b}e^{i\gamma}(z_1-z_3).$$
Using similar arguments for the directed line segment pairs $\overrightarrow{BC}, \overrightarrow{BA}$ and $\overrightarrow{AB}, \overrightarrow{AC}$, I obtain
$$z_1-z_2=\frac{c}{a}e^{i\beta}(z_3-z_2), \quad z_3-z_1=\frac{b}{c}e^{i\alpha}(z_2-z_1).$$
How do I proceed from here onwards to establish the given result?
Best Answer
Generally with problems like this, you want to pick a nice coordinate system which makes the calculations become as trivial as possible. In this case, let's pick coordinates so that $B=0$, $C$ is on the positive real axis, and $A$ is in the upper half plane. This means that as a complex number, $C$ is just $a$. Travelling from $B$ to $A$, we find that $A=ce^{i\beta}$. On the other hand, travelling instead from $C$ to $A$, we get $A=a+be^{i(\pi-\gamma)}=a-be^{-i\gamma}$. Setting these two expressions for $A$ equal gives the desired equation.