Let $z_j=r(cosφ_j+isinφ_j), r\in R$ for $j=1,2,3$ be different complex numbers.
If the numbers $w_1=z_1+z_2z_3$, $w_2=z_2+z_1z_3$, $w_3=z_3+z_1z_2$ are real, prove that $z_1z_2z_3=1$
I know one solution for this problem, but it just seems too complex. Wonder if there is anything simpler. (i won't post the solution that i know, at least for now)
Hopefully this is not the "too complex" method you were talking about...
$$ z_j = r(\cos \varphi_j + i \sin \varphi_j) = re^{i\varphi_j}, j=1,2,3 $$
Using the parallelogram rule for addition of complex numbers, $0 = \arg(w_1) = \arg(z_1 + z_2 z_3)=\frac1{2} (\arg z_1 + \arg z_2 z_3) = \frac1{2}(\arg z_1z_2z_3) $. This implies that $z_1 z_2 z_3 $ is a positive real number.
$$ z_1 z_2 z_3 = r^3e^{i(\varphi_1 + \varphi_2 + \varphi_3)}>0. \\ \therefore \varphi_1 + \varphi_2 + \varphi_3 = 2n\pi, n \in \mathbb{Z} \\ \because \left|e^{ix}\right| =1, \therefore z_1z_2z_3 = r^3. \\ w_1 = z_1 + \frac{r^3}{z_1}, w_2=z_2 + \frac{r^3}{z_2}, w_3=z_3+\frac{r^3}{z_3}\\ w_j = z_j + r^3z_j^{-1}=r(\cos\varphi_j + i \sin\varphi_j)+r^2(\cos \varphi_j - i \sin \varphi_j)=(r+r^2)\cos \varphi_j + (r-r^2)i \sin \varphi_j, j=1,2,3\\ \Im(w_j)=0\Rightarrow r-r^2 = 0\Rightarrow r=0,1\Rightarrow z_1z_2z_3=0,1. $$ $\Im(z)$ is the imaginary part of $z$. r can't be zero because the precondition stated that the complex numbers were different values - if it were they would be the same i.e. $z_1 = z_2 = z_3 = 0$.
$$ \therefore z_1z_2z_3 = 1 $$