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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
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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