Roots of Complex Number/Examples/z^5 + 1 = 0

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 $9$