Let \(w\) and \(z\) be complex numbers such that \(w = z + \frac{1}{z} = z^2 + \frac{1}{z^2}\). Find all possible values of \(w\).
+6248 \(z+\dfrac 1 z = z^2 + \dfrac{1}{z^2}\\ z^4-z^3-z+1=0\)
\(\text{A quick eyeball solution is }z=1\\ \text{So let's divide through by }(z-1)\\ z^3-1 = 0\\ \text{This also clearly has a solution at }z=1 \text{ so divide again}\\ z^2+z+1 = 0\\ \text{This is just quadratic so we can apply the quadratic formula}\\ z = \dfrac{-1\pm \sqrt{1-4}}{2} = -\dfrac 1 2 \pm i\sqrt{3}\\ \text{So in total we have}\\ z = 1, ~z= -\dfrac 1 2 + \dfrac{i\sqrt{3}}{2}, ~z = -\dfrac 1 2 - \dfrac{i\sqrt{3}}{2}\)
\(w = 1 + \dfrac 1 1 = 2\\ -\dfrac 1 2 \pm i \sqrt{3} = e^{\pm i\frac{2\pi}{3}}\\ e^{\pm i\frac{2\pi}{3}}+e^{\mp i\frac{2\pi}{3}} = 2\cos\left(\dfrac{2\pi}{3}\right) = -1 \)
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