If $\frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then $|z_{1}+z_{2}+z_{3}|$

Question

If $z_{1},z_{2},z_{3}$ are three complex number such that $|z_{1}| = |z_{2}| = |z_{3}| = 1$

and $\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then possible values of $|z_{1}+z_{2}+z_{3}|$

$\bf{My\; Try::}$ 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 $$(z_{1}+z_{2}+z_{3})\cdot ((z_{1}+z_{2}+z_{3})^2-3(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1})) = -4z_{1}z_{2}z_{3}.......(1)$$

Now Let $k=z_{1}+z_{2}+z_{3}.$ Taking Conjugate, we get $\displaystyle k = \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}} = \frac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}z_{2}z_{3}}$

So we get $z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=kz_{1}z_{2}z_{3}$

Put into $(1)\;,$ We get $$(z_{1}+z_{2}+z_{3})\cdot \left[(z_{1}+z_{2}+z_{3})^2-3kz_{1}z_{2}z_{3}\right] = -4z_{1}z_{2}z_{3}$$

Now How can I solve after that, Help Required, Thanks

Answer 1

(You made a mistake when taking conjugates: You have $\bar k=\ldots\ $.)

Denote by $s_1$, $s_2$, $s_3$ the elementary symmetric functions of the $z_i$. It is easy to see that we may assume $s_3=1$. Then the $z_i$ are the solutions of the third degree equation $$z^3-s_1z^2+s_2z-1=0\ .\tag{1}$$ The solutions of the equation $z^3-\bar s_1z^2+\bar s_2z-1=0$ are then the $\bar z_i$, and the solutions of $1-\bar s_1w+\bar s_2 w^2- w^3=0$, resp. $$w^3-\bar s_2w^2+\bar s_1 w-1=0\tag{2}$$ are the $\bar z_i^{-1}=z_i$ again. Comparing $(1)$ and $(2)$ we see that necessarily $s_2=\bar s_1$.

Now it is well known that the power sum $p_3:=\sum z_i^3$ can be expressed in terms of the $s_j$ as $$p_3=s_1^3-3s_1s_2+3s_3\ .$$ The given condition then amounts to $s_1^3-3s_1\bar s_1+3=-1$, or $$s_1^3=3|s_1|^2-4\ .$$ Writing $s_1=re^{i\phi}$ with $r\geq0$ we see that necessarily $e^{3i\phi}\in\{1, -1\}$, so that either $r^3-3r^2+4=0$, or $r^3+3r^2-4=0$, which then leads to $r\in\{1,2\}$.

Both these values can be realized: Letting $z_1=z_2=-1$, $z_3=1$ the given condition is satisfied with $|z_1+z_2+z_3|=1$, and letting $z_1=1$, $z_2=e^{i\pi/3}$, $z_3=e^{-i\pi/3}$ leads to $|z_1+z_2+z_3|=2$.

Answer 2

Here is a 3rd solution (inspired by yours, @Christian Blatter, but using different variables).

Let $$Z_1=\frac{z^2_{1}}{z_{2}z_{3}}, \ Z_2=\frac{z^2_{2}}{z_{3}z_{1}}, Z_3=\frac{z^2_{3}}{z_{1}z_{2}} \ \ \ (*)$$

Let us remark that all $Z_k \in \mathbb{U}$ (unit circle).

Thus the initial condition is replaced by $Z_1+Z_2+Z_3=-1 \ \ \ (1)$

Another immediate relationship is $Z_1Z_2Z_3=1 \ \ \ (2)$

At last, is we set $w=Z_1Z_2+Z_2Z_3+Z_3Z_1$, we have

$$w=\frac{z_{1}z_{2}}{z^2_{3}}+\frac{z_{3}z_{1}}{z^2_{2}}+\frac{z_{2}z_{3}}{z^2_{1}}$$

Using the rule $u \in\mathbb{U} \Longrightarrow \bar{w}=\dfrac{1}{w}$ :

$\bar{w}=Z_1+Z_2+Z_3$ thus $w=-1 \ \ \ (3) \ $ (by using (1))

Taking into account (1), (2), (3), by Vieta's formulas, $\{Z_1,Z_2,Z_3\}$ is the set of roots of

$$Z^3+Z^2-Z-1=0$$

i.e., $1$ (multiplicity 1) and $-1$ (multiplicity 2).

One can assume, using definitions (*), that $Z_1=1,Z_2=Z_3=-1$ ; thus

$$\begin{cases}z_1^2=z_2z_3\\ z_2^2=-z_3z_1\\ z_3^2=-z_1z_2\end{cases} \ \ \ (4)$$

or, equivalently:

$$\dfrac{z_1}{z_2}=\dfrac{z_3}{z_1}=-\dfrac{z_2}{z_3}=:\zeta \ \ \ (5)$$

(i.e., we have given the name $\zeta$ to this common ratio).

Thus, by elementary operations on (5), one finds

$$\zeta^3=-1 \ \ \ \Longrightarrow \ \ \zeta=-1, e^{i \pi/3}, e^{-i \pi/3} \ \ \ (6)$$

Thus, using (5), the set of solutions of the initial equation is

$$\begin{cases}z_1=z_1\\ z_2=\zeta^2 z_1\\ z_3=\zeta z_1\end{cases} \ \ \ (7)$$

where $z_1$ is any element of $\mathbb{U}$.

Thus $|z_1+z_2+z_3|=|z_1||1+\zeta+\zeta^2|=2|z_1|=2$ whatever the value of $\zeta$.

Answer 3

Given $$\frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}}+1=0$$ and $|z_{1}|=|z_{2}|=|z_{3}| = 1$

So $z^3_{1}+z^3_{2}+z^3_{3}+z_{1}z_{2}z_{3} = 0.$ So $|z^3_{1}+z^3_{2}+z^3_{3}|=1$

Using $\bullet\; (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$

Put $a=z^3_{1}$ and $b=z^3_{2}$ and $c=z^3_{3},$ Then we get

$\left(z^3_{1}+z^3_{2}\right)\cdot \left(z^3_{2}+z^3_{3}\right)\cdot \left(z^3_{3}+z^3_{1}\right)=\left(z^3_{1}+z^3_{2}+z^3_{3}\right)\left(z^3_{1}z^3_{2}+z^3_{2}z^3_{3}+z^3_{3}z^3_{1}\right)-z^3_{1}z^3_{2}z^3_{3}$

$$ = z^3_{1}z^3_{2}z^3_{3}\left(z^3_{1}+z^3_{2}+z^3_{3}\right)\cdot \left(\frac{1}{z^3_{1}}+\frac{1}{z^3_{2}}+\frac{1}{z^3_{3}}\right)-z^3_{1}z^3_{2}z^3_{3}$$ $$ = z^3_{1}z^3_{2}z^3_{3}\left(z^3_{1}+z^3_{2}+z^3_{3}\right)\cdot \left(\bar{z^3_{1}}+\bar{z^3_{2}}+\bar{z^3_{3}}\right)-z^3_{1}z^3_{2}z^3_{3}=0$$

Now Let $z^3_{1}+z^3_{2} = 0\Rightarrow \left(z_{1}+z_{2}\right)=0$ or $z^2_{1}+z^2_{2}-z_{1}z_{2} =0$ So $z^2_{1}+z^2_{2}=-2z_{1}z_{2}$ or $z^2_{1}+z^2_{2}=z_{1}z_{2}$ So from Initial equation , We get $z^3_{3}=-z_{1}z_{2}z_{3}\Rightarrow z^2_{3}=-z_{1}z_{2}$ Now Using $$|z_{1}+z_{2}+z_{3}|^2=\left(z_{1}+z_{2}+z_{3}\right)\cdot \left(\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}\right)$$ $$ = 3+\left(\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}\right)+\left(\frac{z_{1}}{z_{3}}+\frac{z_{3}}{z_{2}}\right)+\left(\frac{z_{2}}{z_{3}}+\frac{z_{3}}{z_{1}}\right)$$ $$ = 3+\frac{z^2_{1}+z^2_{2}}{z_{1}z_{2}}+\frac{z^2_{2}+z_{1}z_{2}}{z_{2}z_{3}}+\frac{z^2_{3}+z_{1}z_{2}}{z_{3}z_{1}} = 3+\frac{z^2_{1}+z^2_{2}}{z_{1}z_{2}}$$ Bcz $z_{1}+z_{2}=0$ and $z^2_{3}=-z_{1}z_{2}$ from above Condition. So We get $|z_{1}+z_{2}+z_{3}|^2=1\Rightarrow \bullet\; |z_{1}+z_{2}+z_{3}|=1,$ When $z^2_{1}+z^2_{2}=-2z_{1}z_{2}$

And $|z_{1}+z_{2}+z_{3}|^2=4\Rightarrow \bullet\; |z_{1}+z_{2}+z_{3}| = 2,$ When $z^2_{1}+z^2_{2}=z_{1}z_{2}$

Answer 4

After the following step in your approach

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

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

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 $|z_1+z_2+z_3|=2$