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Let \(\omega\) be a complex number such that \(|\omega| = 1\), and the equation has a pure imaginary root \(z\). Find \(\omega + \overline{\omega}.\)

 Oct 5, 2022
 #1
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a + bi = w

a + bi + a - bi = 2a

a + bi = 1, and a - bi is -1

Add together you get 2a = 0 so w + conjugatew = 0

 Oct 5, 2022
 #2
avatar+95 
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Let the pure imaginary root be \(ki,\) where \(k\) is real, so

\(-k^2 + ki + \omega = 0.\)
Thus, \(\omega = k^2 - ki.\) Then \(\overline{\omega} = k^2 + ki,\) so

\(1 = |\omega|^2 = \omega \overline{\omega} = (k^2 - ki)(k^2 + ki) = k^4 + k^2.\)
Then \(k^4 + k^2 - 1 = 0.\) By the quadratic formula,

\(k^2 = \frac{-1 \pm \sqrt{5}}{2}.\)
Since \(k\) is real,

\(k^2 = \frac{-1 + \sqrt{5}}{2}.\)
Therefore,
\(\omega + \overline{\omega} = k^2 - ki + k^2 + ki = 2k^2 = \boxed{\sqrt{5} - 1}.\)

 Oct 5, 2022
edited by WorldEndSymphony  Oct 5, 2022

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