# 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!

• What are $$A,B,C$$? – Dr. Sonnhard Graubner Jul 1 '17 at 15:36
• A, B, and C are just any three angles which satisfy the given condition – arya_stark Jul 1 '17 at 15:37
• Is their sum $\pi$? – Cauchy Jul 1 '17 at 15:37
• No their sum isn't π, no such restrictions – arya_stark Jul 1 '17 at 15:38
• – lab bhattacharjee Jul 1 '17 at 17:23

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 $$(\cos A+\cos B+\cos C)+i(\sin A+\sin B+\sin C)=0.$$