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.

  • $\begingroup$ For $Z_1=Z_2=Z_3=0$ the value is $\infty$. $\endgroup$ – Dietrich Burde Jan 31 '18 at 17:53
  • $\begingroup$ @Dietrich Could you please elaborate on what you mean and how it could help in the solution ? $\endgroup$ – Aditi Jan 31 '18 at 17:55
  • $\begingroup$ @Dietrich I’m sorry I’m not able to understand what you’re trying to say . Do you mean that all of the complex numbers will lie on the real axis ? $\endgroup$ – Aditi Jan 31 '18 at 17:57
  • $\begingroup$ @Dietrich which of the following alternatives seems most appropriate to you for an answer and why ?a) real no. b) purely imaginary c) of modulus 1 d) none of these $\endgroup$ – Aditi Jan 31 '18 at 18:00

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 $\geq3$.


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