In one of the books I came across a question which was as follows:
If | $z_1$ | = | $z_2$ | = | $z_3$ | = 1

$\frac{(z_1^2)}{(z_2)(z_3)}$ + $\frac{(z_2^2)}{(z_1)(z_3)}$ + $\frac{(z_3^2)}{(z_2)(z_1)}$ + 1 = 0

Find the range of |$z_1$ + $z_2$ + $z_3$|

Attempt to question

I tried to rearrange the given equation to obtain something like
$z_1^3$ + $z_2^3$ + $z_3^3$ + $z_1$$z_2$$z_3$ = 0
And then
($z_1$ + $z_2$ + $z_3$)($z_1^2$ + $z_2^2$ + $z_3^2$ - $z_1$$z_2$ - $z_2$$z_3$ - $z_3$$z_1$)= -4$z_1$$z_2$$z_3$
Then taking $\mathbf {mod} $ on both sides but I couldn't solve it further.

  • $\begingroup$ Hint. Rewrite $z_k = \text{cis}\; \alpha_k$ (where $\text{cis}\; \alpha = \cos\alpha+i\sin\alpha$), and then each fraction is of the form $\text{cis}\; (2\alpha_k-\alpha_m-\alpha_n)$, and is therefore still has unit magnitude. Are there restrictions on the three fractions? If so, what are they? How are they expressed in terms of the $\alpha_k$? $\endgroup$ – Brian Tung May 15 '17 at 18:07
  • $\begingroup$ Sorry, I couldn't get it. Please can you elaborate? $\endgroup$ – Shivam Bansal May 16 '17 at 1:39
  • $\begingroup$ Sir, I am not able to find the restrictions, please Sir help. $\endgroup$ – Shivam Bansal May 16 '17 at 15:23
  • $\begingroup$ I saw your request the first time; no need to repeat. I'm busy at the moment, and that's why I haven't had time to sit down and write out an answer. But, a second tip: Each of the fractions has unit magnitude, so they are completely defined by their argument (the angle they make with the positive real axis). There are only two degrees of freedom to these fractions; once you choose values for the first two fractions, the third is determined. So first, figure out what relation binds these three fractions together. $\endgroup$ – Brian Tung May 16 '17 at 17:40
  • $\begingroup$ Next, given the constraint you just figured out, identify how they can add up to $-1$ and satisfy that second equation. Then figure out what values of $z_k$ produce those fractions; there are multiple solutions, but I believe they are discrete and can be enumerated. Finally, take each of the solutions, add up the $z_k$ in each of them, and figure out what the magnitude of the sum is. $\endgroup$ – Brian Tung May 16 '17 at 17:42

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