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! 

Note that $$\cos(a-b) = \cos(a)\cos(b)+\sin(a)\sin(b) = \Re(x\bar{y})$$
So $$\label{eq1}\cos(a-b)+\cos(b-c)+\cos(c-a) = \Re(x\bar{y}+y\bar{z}+z\bar{x})$$
With that in mind, note that just like $x$, $y$ and $z$, the $xyz$ complex number is of modulus 1. So if we go back to the $x+y+z=xyz$ identity, we see that $$|x+y+z|=1$$
In other words, $$\begin{split}
1 & = |x+y+z|^2 \\
& = (x+y+z)(\bar x+\bar y +\bar z) \\
 & = x\bar x + x \bar y + x \bar z + y\bar x + y \bar y + y \bar z + z \bar x + z \bar y + z \bar z \\
& = 1 + x \bar y + x \bar z + 1 + y\bar x + y \bar z + 1 + z \bar x + z \bar y \\
& = 3 + (x \bar y + y \bar z + z \bar x) + (y\bar x + z \bar y + x \bar z) \\
& = 3 + 2 \Re(x \bar y + y \bar z + z \bar x)
\end{split}$$
We conclude that $$\Re(x\bar{y}+y\bar{z}+z\bar{x})+1=0$$ which yields the desired result.
