Roots of Complex Number/Examples/z^4 - 81 = 0
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Theorem
The roots of the polynomial:
- $z^4 - 81$
are:
- $\set {3, 3 i, -3, -3 i}$
Proof
From Factorisation of $z^n - a$:
- $z^4 - a = \ds \prod_{k \mathop = 0}^3 \paren {z - \alpha^k b}$
where:
- $\alpha$ is a primitive complex $4$th root of unity
- $b$ is any complex number such that $b^4 = a$.
Here we can take $b = 3$, as $81 = 3^4$.
Thus:
- $z = \set {3 \exp \dfrac {k i \pi} 2}$
\(\text {(k = 0)}: \quad\) | \(\ds z\) | \(=\) | \(\ds 3 \paren {\cos \dfrac 0 \pi 2 + i \sin \dfrac 0 \pi 2}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 3\) | ||||||||||||
\(\text {(k = 1)}: \quad\) | \(\ds z\) | \(=\) | \(\ds \cos \dfrac {\pi} 2 + i \sin \dfrac {\pi} 2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 3 i\) | ||||||||||||
\(\text {(k = 2)}: \quad\) | \(\ds z\) | \(=\) | \(\ds \cos \dfrac {2 \pi} 2 + i \sin \dfrac {2 \pi} 2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds -3\) | ||||||||||||
\(\text {(k = 3)}: \quad\) | \(\ds z\) | \(=\) | \(\ds \cos \dfrac {3 \pi} 2 + i \sin \dfrac {3 \pi} 2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds -3 i\) |
$\blacksquare$
Sources
- 2008: David Joyner: Adventures in Group Theory (2nd ed.) ... (previous) ... (next): Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Ponderable $2.1.2$