algebra precalculus - Solve using complex analysis?

$\cos(A-B) + \cos(B-C) + \cos(C-A) = \frac{-3}{2}$
We need to prove that $\cos A + \cos B + \cos C = \sin A + \sin B + \sin C = 0$

I was wondering if it's possible to prove this result by showing that the real and imaginary parts of $z = \cos A + \cos B + \cos C$ are equal to zero, somehow invoking Vieta's or De Moivre's theorem if required. I tried, starting with $\cos(B-C)$ and other cyclic terms but couldn't really get anywhere.

Any other method is also appreciated.

Thanks a lot!

Answer

Let $u$, $v$, $w$ be $e^{iA}$, $e^{iB}$ and $e^{iC}$.

Then
$$(u+v+w)(u^{-1}+v^{-1}+w^{-1})
=2\cos(A-B)+2\cos(B-C)+2\cos(C-A)+3.$$
Your condition is equivalent to $(u+v+w)(u^{-1}+v^{-1}+w^{-1})=0$.
These brackets are complex conjugates, so $u+v+w=0$
and this means
$$(\cos A+\cos B+\cos C)+i(\sin A+\sin B+\sin C)=0.$$