Roots of Complex Number/Examples/z^5 + 1 = 0
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Theorem
The roots of the polynomial:
- $z^5 + 1 = 0$
are:
- $\set {\cos \dfrac \pi 5 \pm i \sin \dfrac \pi 5, \cos \dfrac {3 \pi} 5 \pm i \sin \dfrac {3 \pi} 5, -1}$
Proof
From Factorisation of $z^n + 1$:
- $z^5 + 1 = \ds \prod_{k \mathop = 0}^4 \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} 5}$
Thus:
- $z = \set {\exp \dfrac {\paren {2 k + 1} i \pi} 5}$
\(\text {(k = 0)}: \quad\) | \(\ds z\) | \(=\) | \(\ds \cos \dfrac \pi 5 + i \sin \dfrac \pi 5\) | |||||||||||
\(\text {(k = 1)}: \quad\) | \(\ds z\) | \(=\) | \(\ds \cos \dfrac {3 \pi} 5 + i \sin \dfrac {3 \pi} 5\) | |||||||||||
\(\text {(k = 2)}: \quad\) | \(\ds z\) | \(=\) | \(\ds \cos \dfrac {5 \pi} 5 + i \sin \dfrac {5 \pi} 5\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \cos \pi + i \sin \pi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1\) | Euler's Identity | |||||||||||
\(\text {(k = 3)}: \quad\) | \(\ds z\) | \(=\) | \(\ds \cos \dfrac {7 \pi} 5 + i \sin \dfrac {7 \pi} 5\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \cos \dfrac {-3 \pi} 5 + i \sin \dfrac {-3 \pi} 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cos \dfrac {3 \pi} 5 - i \sin \dfrac {3 \pi} 5\) | ||||||||||||
\(\text {(k = 4)}: \quad\) | \(\ds z\) | \(=\) | \(\ds \cos \dfrac {9 \pi} 5 + i \sin \dfrac {9 \pi} 5\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \cos \dfrac {-\pi} 5 + i \sin \dfrac {-\pi} 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cos \dfrac \pi 5 - i \sin \dfrac \pi 5\) |
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 3$. Roots of Unity: Exercise $9$