4. If $z_{1}, z_{2}, z_{3} \in \mathbb{C}$ such that $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1, z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=0$ and $z_{1}+z_{2}+z_{3} \neq 0$, calculate $\left|z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\right|$. R.M.T. $4 / 2015$
Solution. $\left(z_{1}+z_{2}+z_{3}\right)^{2}=z_{1}^{2}+z_{2}^{2}+z_{3}^{2}+2 z_{1} z_{2}+2 z_{2} z_{3}+2 z_{1} z_{3}=2 z_{1} z_{2}+2 z_{2} z_{3}+2 z_{1} z_{3} \Rightarrow$

$$
\begin{aligned}
& \Rightarrow\left|z_{1}+z_{2}+z_{3}\right|^{2}=2\left|z_{1} z_{2}+z_{2} z_{3}+z_{1} z_{3}\right|
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_abadef4c0df516f9f3bbg-3.jpg?height=65&width=1501&top_left_y=976&top_left_x=274)

$$
\begin{aligned}
& =3+\overline{z_{1}} z_{3}+z_{1} \overline{z_{3}}+\overline{z_{2}} z_{1}+z_{2} \overline{z_{1}}+\overline{z_{2} z_{3}}+z_{2} \overline{z_{3}}(2) \\
& \left|z_{1}+z_{2}+z_{3}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+\left.|| z_{3}\right|^{2}+\overline{z_{1} z_{3}}+z_{1} \overline{z_{3}}+\overline{z_{2}} z_{1}+z_{2} \overline{z_{1}}+\overline{z_{2} z_{3}}+z_{2} \overline{z_{3}}=3+\overline{z_{1} z_{3}}+z_{1} \overline{z_{3}}+\overline{z_{2} z_{1}}+z_{2} \overline{z_{1}}+\overline{z_{2}} z_{3}+z_{2} \overline{z_{3}}
\end{aligned}
$$

From relations (1), (2), (3), and $z_{1}+z_{2}+z_{3} \neq 0$, it follows that $\left|z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\right|=2$.

## Grading.

| Deduce relation (1) | $2 \mathrm{p}$ |
| :--- | :--- |
| Deduce relation (2) | $2 \mathrm{p}$ |
| Deduce relation (3) | $2 \mathrm{p}$ |
| Finalize $\left\|z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\right\|=2$ | $1 \mathrm{p}$ |

Note: Any other correct solution will be graded accordingly.