Problem :
Let a,b,c be distinct non zero complex numbers with $|a|=|b|=|c|$ If each of the equations $az^2+bz+c=0$ and $bz^2+cz+a=0$ has a root having modulus 1, then prove that : $|a-b|=|b-c|=|c-a|$
My approach :
Let $z_1,z_2$ be the roots of the equation with $|z_1|=1$ From $z_2=\frac{c}{a}.\frac{1}{z_1}$
we g et $|z_2|=|\frac{c}{a}|\frac{1}{|z_1|} =1$ As $z_1+z_2=-\frac{b}{a}, |a|=|b|$
we get $|z_1+z_2|^2=1$
$\Rightarrow (z_1+z_2)^2=z_1.z_2$ = $(-\frac{b}{a})^2=\frac{c}{a}$
$\Rightarrow b^2=ac$
Similarly $a^2=bc$
Now please suggest how to proceed further, will be of great help thanks.
Best Answer
Let $a=\rho e^{i\theta_a}, b=\rho e^{i\theta_b}, c=\rho e^{i\theta_c} $. If $e^{i\eta}$ is a root of $ax^2+bx+c$, by Viète's theorem the other root is $\frac{c}{a}e^{-i\eta}$ and $$ e^{i\eta}+\frac{c}{a}e^{-i\eta} = -\frac{b}{a}.$$ All the terms in this equation are complex numbers with unit modulus, but the sum of two complex numbers with unit modulus is still a complex number with unit modulus in very few cases, namely just the ones in which the difference of the arguments is $\frac{2\pi}{3}$. That gives that the angle between $a$ and $c$ is $\frac{2\pi}{3}$, and by a similar argument with the second polynomial we also have that the angle between $a$ and $b$ is $\frac{2\pi}{3}$, so the triangle with vertices at $a,b,c$ is equilateral and the claim follows.