Sorry I didn't give what I think about this question, here was my go at it,
First of all we make the unknowns $a, b, z$
in which $|a+bi|=\omega$,
solving for magnitude we get $a^2+b^2=1$
That is our first equation,
Our second equation is
$z^2+z+(a+bi)=0$
this is because $\omega=a+bi$, we can plug $a+bi$ in to make our third equation,
finally we find use the pure imaginary root $z$ as the last equation
\(z = {-1 \pm \sqrt{1^2-4(1)4(a+bi)} \over 2}\)
This is because the root is $z$, and the
$a=1$
$b=1$
and
$c=(a+bi)$
from the second equation: $z^2+z+(a+bi)=0$
and those are three equations with 3 unkowns, did I make a mistake?
