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! 

Squaring & adding we get
$$\cos^2(a+b+c)+\sin^2(a+b+c)=(\cos a+\cos b+\cos c)^2+(\sin a+\sin b+\sin c)^2$$
$$\implies1=1+1+1+2\sum_{\text{cyc}(a,b,c)}\cos(a-b)$$
Observe that we actually don't need $x+y+z=xyz$
The sufficient condition is $$ x+y+z=\cos p+i\sin p\iff|x+y+z|=1$$
