show that $\cos(\theta_{2}-\theta_{3})+\cos(\theta_{3}-\theta_{1})+\cos(\theta_{1}-\theta_{3})+1=0$ If $$a = \cos\theta_{1} + i\sin\theta_{1}, \\ b = \cos\theta_{2} + i\sin\theta_{2}, \\ c = \cos\theta_{3} + i\sin\theta_{3}$$ and $a+b+c=abc$, 
then show that 

$$\cos(\theta_{2}-\theta_{3})+\cos(\theta_{3}-\theta_{1})+\cos(\theta_{1}-\theta_{2})+1=0$$

I am confused with where to start. I did try to proceed with the De Moivre's Theorem but was not able to get hold of the required equation. 
Kindly provide a hint. 
Thanks in advance !

If we divide $a+b+c = abc$ by $a$ on both sides, we obtain
$$ 1 + \frac{b}{a} + \frac{c}{a} = bc. $$
Similarly, dividing by $b$ and $c$ yields
$$ \frac{a}{b} + 1 + \frac{c}{b} = ac $$
$$ \frac{a}{c} + \frac{b}{c} + 1 = ab. $$
Adding the three equations yields
$$\frac{a}{b} + \frac{b}{a} + \frac{a}{c} + \frac{c}{a} + \frac{b}{c} + \frac{c}{b} + 3 = ab+ac+bc. $$
Note that $\frac{a}{b} + \frac{b}{a} = 2\cos(\theta_2-\theta_1)$ (check this if you are not convinced), and similar equalities follow for the other two pairs. We thus have
$$ 2(\cos(\theta_2-\theta_1)+\cos(\theta_3-\theta_1)+\cos(\theta_3-\theta_2)+1) = ab+ac+bc-1.$$
It now suffices to show that $ab+ac+bc-1=0$. Since the LHS of the above equation is real, it suffices to show that the real part of $ab+ac+bc$ is $1$.
For any $z\in\mathbb{C}$ such that $|z|=1$, we have that $\text{Re}(z) = \text{Re}(z^{-1})$. Hence $\text{Re}(ab) = \text{Re}(1/ab)$ since $|ab|=|a||b|=1$, and similarly with the other two terms, so
$$\text{Re}(ab+ac+bc) = \text{Re}\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right) = \text{Re}\left(\frac{a+b+c}{abc}\right) = \text{Re}(1) = 1$$
as desired.
