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In the complex plane, consider $z_1,z_2,z_3$ as distinct complex numbers lying on the curve $|z| = 3$. Suppose that a root of $f(z) = z_1z^2+ z_2z+ z_3$ satisfies $|z| = 1$.

I want to prove that $z_1,z_2,z_3$ are in geometric progression.

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Write $z_1 = 3e^{i\alpha}$, $z_2 = 3e^{i\beta}$, $z_3 = 3e^{i\gamma}$

Since for one $z$, $|z| = 1$, let $z = e^{i\theta}$

Thus, $f({e^{i\theta}}) = 3[e^{i\alpha}({e^{i\theta}})^2 + e^{i\beta}({e^{i\theta}}) + e^{i\gamma}] = 0$

Simplify,

$e^{i(\alpha+2\theta)} + e^{i(\beta+\theta)} + e^{i\gamma} = 0$

Treat above three terms as (force) vectors of equal magnitude and for sum to be zero (equilibrium), we get the condition (by Lami's theorem), that they are situated on vertices of an equilateral triangle and angle between any two is equal.

For ease of writing $\omega = e^{i\frac{2\pi}{3}}$

Applying rotation, $e^{i\gamma} = \omega e^{i(\beta+\theta)} = \omega^2e^{i(\alpha+2\theta)}$

Multiplying by 3 and simplifying,

$3e^{i\gamma} = 3(\omega e^{i\theta}) e^{i\beta} = 3(\omega e^{i\theta})^2e^{i\alpha}$

Take $(\omega e^{i\theta}) = r$

and put back $z_1, z_2, z_3$, getting:

$z_3 = r . z_2 = r^2 . z_3$

Hence, they are in geometric progression.

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