Let $\omega$ be a complex number such that $|\omega| = 1,$ and the equation \[z^2 + z + \omega = 0\]has a pure imaginary root $z.$ Find $\omega + \overline{\omega}.$

Guest Jan 26, 2021

#1**+1 **

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?

Guest Jan 26, 2021

edited by
Guest
Jan 26, 2021

edited by Guest Jan 26, 2021

edited by Guest Jan 26, 2021