Collinearity of complex numbers and the origin

Question

If $Z_1$ , $Z_2$ and $Z_3$ are complex numbers such that $|Z_1| + |Z_2| + |Z_3| = |Z_1 + Z_2 + Z_3|$ then find the value of , $\frac{Z_1Z_2}{{Z_3}^2} + \frac{Z_2Z_3}{{Z_1}^2} + \frac{Z_1Z_3}{{Z_2}^2}$

Accordingly to the solution of the above problem , the data given the problem suggests that $0$ , ${Z_1}$ , ${Z_2}$ , ${Z_3}$ are collinear . How can we derive that from the data given in the problem ? And even if we do , how do we know for certain that the points are collinear with the origin on the real axis and not in the imaginary axis ? Then how can we proceed with the solution ? Please help.

Answer 1

If any one of the $Z_k$ is $=0$ the quantity $Q$ in question is undefined. Therefore assume $Z_k=r_ke^{i\alpha_k}$ with $r_k>0$ $\>(1\leq k\leq3)$. It is a fact of elementary geometry (the converse of the triangle inequality) that $|Z_1+Z_2+Z_3|<r_1+r_2+r_3$ unless all $\alpha_k$ are equal mod $2\pi$. It follows that $$Q={r_1r_2\over r_3^2}+{r_2r_3\over r_1^2}+{r_3r_1\over r_2^2}=3\>r_1r_2r_3\cdot{1\over3}\left({1\over r_3^3}+{1\over r_1^3}+{1\over r_2^3}\right)\geq3$$ by the AGM inequality. It follows that $Q$, when defined, assumes real values $\geq3$.