Let $z_1,z_2,z_3\in\mathbb{C}$ such that $|z_1|=|z_2|=|z_3|=1$.
If $z_1+z_2+z_3\ne0$ and $z_1^2+z_2^2+z_3^2=0$ then prove $|z_1+z_2+z_3|=2$.
What I did
$z_1^2+z_2^2+z_3^2=0~~|\cdot(z_1+z_2+z_3)$
$\Rightarrow z_1^3+z_1^2z_2+z_1^2z_3+z_2^2z_1+z_2^3+z_2^2z_3+z_3^2z_1+z_3^2z_2+z_3^3=0$
$z_1^2z_2+z_1^2z_3+z_2^2z_1+z_2^2z_3+z_3^2z_1+z_3^2z_2+(z_1^3+z_2^3+z_3^3)=0~~~(*)$
(HERE IS THE MISTAKE) $z_1^3+z_2^3+z_3^3=(z_1+z_2+z_3)(\underbrace{z_1^2+z_2^2+z_3^2}_0-z_1z_2-z_1z_3-z_2z_3)=$
$=-(z_1+z_2+z_3)(z_1z_2+z_1z_3+z_2z_3)=$
$=-(3z_1z_2z_3+z_1^2z_2+z_1^3z_3+z_2^2z_1+z_2^2z_3+z_3^2z_1+z_3^2z_2)$
Substituting $(z_1^3+z_2^3+z_3^3)$ in $(*)~(z_1^2z_2+z_1^2z_3+z_2^2z_1+z_2^2z_3+z_3^2z_1+z_3^2z_2)$ reduces $\Rightarrow$
$-3z_1z_2z_3=0\Rightarrow z_1z_2z_3=0$. That means one of them must be $0$. Let's prove it.
$z_1=a_1+b_1i,~z_2=a_2+b_2i~(a_1,b_1,a_2,b_2\in\mathbb{R})$. We want to see when $z_1z_2=0$.
$(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i=0$
$ \begin{cases} a_1a_2-b_1b_2=0\Rightarrow a_1a_2=b_1b_2&(1) \\ a_1b_2+a_2b_1=0&(2) \end{cases} $
Let's consider $a_1,b_2\ne0$ ,so we can divide by $a_1b_2$ in $(1)$.
$\frac{a_2}{b_2}=\frac{b_1}{a_1}\Rightarrow a_2=\frac{b_1}{a_1}\cdot b_2$
Substitute in $(2)$ and $a_1b_2+\frac{b_1}{a_1}\cdot b_2b_1=0\Rightarrow b_2\cdot\frac{a_1^2+b_1^2}{a_1}=0$. $b_1\ne0\Rightarrow a_1^2+b_1^2=0$, and since $a_1,b_1\in\mathbb{R}$ the only possibility is that $a_1=b_1=0\Rightarrow z_1=0$
So when $z_2$ is complex ($b_2\ne0$) $z_1=0$ and by symmetry when $z_1$ is complex $z_2=0$.
Conclusion: $z_1z_2=0\Leftrightarrow z_1=0$ or $z_2=0$. The relation applies for multiple variables also.
Thus, one of $z_1,z_2,z_3$ must be $0$. But in the hypothesis $|z_1|=|z_2|=|z_3|=1$.
How is that possible?