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 !
 A: Let $\bar{z}$ denote the complex conjugate of $z$.  Multiplying $a+b+c = abc$ by $\bar{a}$, we get
    \begin{align*}
 1 + \bar{a}b + \bar{a}c = bc
 \end{align*}
    Similarly,
    \begin{align*}
 1+\bar{b}c+\bar{b}a &= ca\\
 1+\bar{c}a + \bar{c}b & = ab
 \end{align*}
    Adding we get,
    \begin{align*}
 3 + (\bar{a}b +a\bar{b}+ \bar{b}c +b\bar{c}+ \bar{c}a + c\bar{a}) = ab+bc+ca
 \end{align*}
    Also, 
    from $a+b+c = abc$, we get $\bar{a} +\bar{b}+\bar{c} = \bar{a}\bar{b}\bar{c}$ and hence dividing throughout by $\bar{a}\bar{b}\bar{c}$ we get
    \begin{align*}
 bc+ca+ab = 1
 \end{align*}
    Thus,
    \begin{align*}
 (\bar{a}b +a\bar{b}+ \bar{b}c +b\bar{c}+ \bar{c}a + c\bar{a}) = -2
 \end{align*}
    and hence 
    \begin{align*}
 \cos(\theta_2-\theta_3) + \cos(\theta_3 - \theta_1) +  \cos(\theta_1 - \theta_2) + 1 = 0 
\end{align*}
A: Not adding much, but slightly shorter:
$$ 1 = (abc) (\overline{abc})= (a+b+c)(\bar{a}+\bar{b}+\bar{c}) = 3 + \left[ (a\bar{b}+b\bar{a}) + (b\bar{c}+c\bar{b})+ (c\bar{a}+a\bar{c})\right] $$
So 
$$ 0 = 2+ 2  \left( \cos (\theta_2-\theta_1) + \cos (\theta_3-\theta_2)+\cos (\theta_1-\theta_3) \right) $$
A: 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.
