<s>[INST] If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$ 
If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $$\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$$

Here's how I tried it 
$$x+y+z=xyz $$
So, by De Moivre's Theorem,
$$(\cos a + \cos b + \cos c)  + i(\sin a + \sin b + \sin c) = \cos(a+b+c)  + i \sin(a+b+c) 
$$
Equating real and imaginary parts, 
$$\cos a + \cos b + \cos c= \cos(a+b+c)$$
and similarly for sine. Now,
$$(a-b)  + (b-c)  + (c-a)  =0$$
What to do now?  Please help. And please use De Moivre Theorem! 
 [/INST] (This is a followup on Stefan Lafon's answer, too long for a comment.)
Using that $\,|x|=1 \iff \bar x = \dfrac{1}{x}\,$ and $\,\cos(a-b)= \operatorname{Re}\left(\dfrac{x}{y}\right)=\operatorname{Re}(x \bar y) = \operatorname{Re}(\bar x y)\,$, it follows that:
$$
\begin{align}
\cos(a-b) + \cos(b-c) + \cos(c-a) &= \cos(a-b) + \cos(a-c) + \cos(b-c) \\
 &= \operatorname{Re}\left(x\bar y+ x \bar z + y \bar z\right) \\
 &= \operatorname{Re}\left(x\left(\bar y + \bar z\right) + y \bar z\right) \\
 &= \operatorname{Re}\left(x\left(\bar x \bar y \bar z - \bar x\right) + y \bar z\right) \\
 &= \operatorname{Re}\left(\bar y \bar z + y\bar z -1 \right) \\
 &= \operatorname{Re}\left(\left(y + \bar y\right) \bar z -1 \right) \\
 &= -1 + 2 \operatorname{Re}(y) \operatorname{Re}(z)
\end{align}
$$
For the latter RHS to be $\,-1\,$, the second term must be zero, so one of $\,y,z\,$ must be a purely imaginary number of modulus $\,1\,$ i.e. $\,\pm i\,$. With some more legwork, it follows that the solution set of the given constraints is $\,\{(x,-x,z)\mid x=\pm i, |z| = 1\}\,$ or permutations thereof.
</s>