Evaluate $|z_1 + z_2 + z_3|$ on unit circle given $ \frac{ z_1^2}{z_2 z_3} + \frac{ z_2^2}{z_1 z_3} + \frac{ z_3^2}{z_1 z_2} +1 =0$

Question

given three complex numbers $(z_1 , z_2 , z_3)$ lying on the unit circle and related by the equation $ \frac{ z_1^2}{z_2 z_3} + \frac{ z_2^2}{z_1 z_3} + \frac{ z_3^2}{z_1 z_2} +1 =0$, find the sum of all possible values of $|z_1 + z_2 + z_3|$

I have no idea how to solve it systematically but with some guessing I found that $(1,1,-1) $ is a solution triplet for this constraint. And, if $(1,1,-1)$ is a solution then by the symmetry of how the equation is, it means that $(-1,1,1)$ and $(1,-1,1)$ are also solutions.

But, I'm not sure how to rule out/ find more solutions. Ideally, the answer I"m looking for is a general method to approach these kind of questions.

Answer 1

Let us define the angles $\alpha$ and $\beta$ as \begin{eqnarray} \alpha &=& \mathrm{Arg}\left[\frac{z_2}{z_1}\right] \, , \\ \beta &=& \mathrm{Arg}\left[\frac{z_3}{z_1}\right] \, . \end{eqnarray} Both the condition and the norm $|z_1 + z_2 + z_3|$ can be written in terms of only $\alpha$ and $\beta$ since \begin{eqnarray} 0 &=& \frac{z^2_1}{z_2 z_3} + \frac{z^2_2}{z_3 z_1} + \frac{z^2_3}{z_1 z_2} + 1 \, \\ &=& \frac{1}{\frac{z_2}{z_1} \frac{z_3}{z_1}} + \frac{\left(\frac{z_2}{z_1}\right)^2}{\frac{z_3}{z_1}} + \frac{\left(\frac{z_3}{z_1}\right)^2}{\frac{z_2}{z_1}} + 1 \, , \end{eqnarray} and \begin{equation} |z_1 + z_2 + z_3| = |z_1|\left|1 + \frac{z_2}{z_1} + \frac{z_3}{z_1}\right| = \left|1 + \frac{z_2}{z_1} + \frac{z_3}{z_1}\right| \, . \end{equation} This means that for each pair of values of $\alpha$ and $\beta$ that solve the condition above, $z_1$ (or, equivalently, its $\mathrm{Arg}$) still remains an unconstrained degree of freedom, which is probably considered a redundancy when computing the sum that the exercise asks for.

An additional source or redundancy is the fact that $\alpha$ and $\beta$ that solve the condition above are interchangeable, given the symmetry of the expression.

If we focus on only the ranges $0 \leq \alpha < 2 \pi$ and $0 \leq \beta < 2 \pi$, the pairs of values that solve the condition are \begin{equation} \left( \pi, \pi \right), \left( 0, \pi \right), \left( \frac{\pi}{3} , \frac{2\pi}{3} \right) , \left( \frac{\pi}{3} , \frac{5\pi}{3} \right), \left( \frac{4\pi}{3} , \frac{5\pi}{3} \right) \, , \end{equation} where, again, each can be either $\alpha$ or $\beta$. The corresponding values of $|z_1 + z_2 + z_3|$ are \begin{equation} 1, 1, 2, 2, 2. \end{equation} From here it's not completely clear how much redundancy should we consider in order to compute the sum, but the fundamental components are there. If we are expected to sum each possible value of the norm once then we get $1+2 = 3$.

Answer 2

Let $\frac{z_1}{z_2}=e^{i \alpha}$, $\frac{z_2}{z_3}= e^{i \beta} $ and $\frac{z_3}{z_1}= e^{i \gamma} $. WLOG, $\alpha +\beta +\gamma=2\pi$, $\alpha,\beta,\gamma\ge 0$. Then, from the given

$$e^{i(\alpha-\gamma)}+ e^{i(\beta-\alpha)}+ e^{i(\gamma-\beta)}+1=0$$

or

$$\sin (\alpha-\gamma) + \sin(\beta-\alpha)+ \sin(\gamma-\beta)=0$$ $$\cos (\alpha-\gamma) + \cos(\beta-\alpha)+ \cos(\gamma-\beta)+1=0$$

Factorize $$\sin \frac{\alpha-\gamma}2 \sin\frac{\beta-\alpha}2\sin\frac{\gamma-\beta}2=0$$ $$\cos \frac{\alpha-\gamma}2 \cos\frac{\beta-\alpha}2\cos\frac{\gamma-\beta}2=0$$

which, along with $\alpha +\beta +\gamma=2\pi$, yield the solutions for $(\alpha,\beta,\gamma)$

$$(\pi,0,\pi),(\pi,\pi,0) ,(0,\pi,\pi), (\frac\pi3,\frac\pi3,\frac{4\pi}3), (\frac\pi3,\frac{4\pi}3, \frac\pi3), (\frac{4\pi}3, \frac\pi3,\frac\pi3) $$

As expected, they are symmetric with respect to $\alpha,\beta,\gamma$, and it suffices to evaluate below with $(\pi,0,\pi)$ and $(\frac\pi3,\frac{4\pi}3, \frac\pi3)$ only to obtain

\begin{equation} |z_1 + z_2 + z_3| = |1+e^{-i\alpha}+e^{i \gamma}|= 1,2 \end{equation}

Answer 3

Given $$\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1\Rightarrow z^3_{1}+z_{2}^3+z_{3}^3=-z_{1}z_{2}z_{3}$$

Now Using $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

So $$\left(z_{1}+z_{2}+z_{3}\right)\left[z^2_{1}+z^2_{2}+z^2_{3}-z_{1}z_{2}-z_{2}z_{3}-z_{3}z_{1}\right] = -4z_{1}z_{2}z_{3}$$

So $$(z_{1}+z_{2}+z_{3})\cdot ((z_{1}+z_{2}+z_{3})^2-3(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1})) = -4z_{1}z_{2}z_{3}.......(1)$$

$(z_{1}+z_{2}+z_{3})\cdot \bigg((z_{1}+z_{2}+z_{3})^2-3z_1z_2z_3\bigg(\dfrac{1}{z_1}+\dfrac{1}{z_2}+\dfrac{1}{z_3}\bigg)\bigg) = -4z_{1}z_{2}z_{3}.....(2)$

Using $\bar z_{1}=\dfrac{1}{z_1}, \bar z_{2}=\dfrac{1}{z_2}, \bar z_{3}=\dfrac{1}{z_3}$

Equation $(2)$ turns to $(z_{1}+z_{2}+z_{3})\cdot \bigg((z_{1}+z_{2}+z_{3})^2-3z_1z_2z_3\big(\overline{z_1+z_2+z_3}\big)\bigg) = -4z_{1}z_{2}z_{3}$

$\implies (z_1+z_2+z_3)^3-3z_1z_2z_3(z_1+z_2+z_3)\overline{(z_1+z_2+z_3)}=- 4z_1z_2z_3$

$\implies (z_1+z_2+z_3)^3-3z_1z_2z_3|(z_1+z_2+z_3)|^2=- 4z_1z_2z_3 $

$\implies (z_1+z_2+z_3)^3=3z_1z_2z_3|(z_1+z_2+z_3)|^2- 4z_1z_2z_3$

$\implies (z_1+z_2+z_3)^3=z_1z_2z_3\bigg(3|(z_1+z_2+z_3)|^2- 4\bigg)$

Now take modulus on both side

$|z_1+z_2+z_3|^3=|z_1||z_2||z_3|\bigg|3|z_1+z_2+z_3|^2-4\bigg|.......(3)$

Now let $|z_1+z_2+z_3|=t$

So equation $(3)$ turns into

$t^3=|3t^2-4|.....(4)$

Case $1$: $t^2\geq 4/3$ Then equation $(4)$ transform to $t^3=3t^2-4$

So roots of above equation are $t=-1$ and $t=2$

Case $2$: $t^2< 4/3$ Then equation $(4)$ transform to $t^3=-3t^2+4$

So roots of above equation are $t=-2$ and $t=1$

Discard the negative values of $t$ because $t\geq 0$

Hence $|z_1+z_2+z_3|=1$ and $|z_1+z_2+z_3|=2$