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$ |z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$ show that $z_1^2+z_2^2+z_3^2=0$ $z_1,z_2,z_3 $-complex numbers. I think that since the module is 1, these numbers are on the unit circle and $z_1+z_2+z_3=0$ means that the difference of their angles is $120^{\circ}$ also $(z_1+z_2+z_3)^2=0$ so $z_1^2+z_2^2+z_3^2=-2(z_1*z_2+z_2*z_3+z_3*z_1)=-2\{z_2(z_1+z_3)+z_3*z_1\}=-2(-(z_1+z_3)^2+z_3*z_1)$

could anyone help please?

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From $z_1+z_2+z_3=0$, we get $$ \bar{z_1}+\bar{z_2}+\bar{z_3}=0 \qquad\qquad\qquad\qquad\;\;\;\; $$ But $|z_k|=1$ implies $\bar{z_k}={\large{\frac{1}{z_k}}}$, hence \begin{align*} &\bar{z_1}+\bar{z_2}+\bar{z_3}=0\\[4pt] \implies\;&\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=0\\[4pt] \implies\;&\frac{z_1z_2+z_2z_3+z_3z_1}{z_1z_2z_2}=0\\[4pt] \implies\;&z_1z_2+z_2z_3+z_3z_1=0\\[4pt] \implies\;&(z_1+z_2+z_3)^2-2(z_1z_2+z_2z_3+z_3z_1)=0\\[4pt] \implies\;&z_1^2+z_2^2 +z_3^2=0\\[4pt] \end{align*}

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Suppose $z_1=e^{i\theta}$ for some argument $\theta$. Then $z_2=\zeta e^{i\theta}$ and $z_3=\zeta^2e^{i\theta}$ where $\zeta$ is a complex cube root of 1 - this because the arguments of $z_1, z_2, z_3$ differ by 120 degrees and 240 degrees. So

$z_1^2+z_2^2+z_3^2 = e^{2i\theta}(1 + \zeta^2 + \zeta^4)$

$= e^{2i\theta}(1 + \zeta^2 + \zeta) = e^{i\theta}(z_1+z_3+z_2) = 0$

In other words $z_1^2,z_2^2, z_3^2$ form an equilateral triangle just like $z_1, z_2, z_3$, but rotated through an angle $\theta$.

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