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Home » Similarly, test residues modulo 5: - Coaching Toolbox

Similarly, test residues modulo 5: - Coaching Toolbox

📅 May 4, 2026 👤 scraface

May 04, 2026

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$ z = \pm 1 $, both real, so imaginary part is 0.
Roots from $ z^4 = -1 $:
Write $ -1 = e^{i\pi} $, so:
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...

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