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Home » \[ 2u^2 + 3u - 2 = 0 \] - Esdistancia

\[ 2u^2 + 3u - 2 = 0 \] - Esdistancia

📅 March 17, 2026 👤 scraface

Mar 17, 2026

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z = e^{i(\pi + 2k\pi)/4} = e^{i\pi(2k+1)/4}, \quad k = 0, 1, 2, 3
So the roots are:
$ k=0 $: $ e^{i\pi/4} = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} $
$ k=1 $: $ e^{i3\pi/4} = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} $
$ k=2 $: $ e^{i5\pi/4} = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} $
$ k=3 $: $ e^{i7\pi/4} = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} $
The imaginary parts are $ \pm \frac{\sqrt{2}}{2} $. The maximum imagin...
We are told this maximum is $ \sin \theta $ for $ 0 < \theta < \pi $.
\sin \theta = \frac{\sqrt{2}}{2} \quad \Rightarrow \quad \theta = \fra...
Both are in $ (0, \pi) $, but since the maximum imaginary part is posi...

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