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 $z^n - a$:

$z^4 - a = \displaystyle \prod_{k \mathop = 0}^3 \paren {z - \alpha^k b}$

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:

$z = \set {3 \exp \dfrac {k i \pi} 2}$


\((k = 0):\quad\) \(\displaystyle z\) \(=\) \(\displaystyle 3 \paren {\cos \dfrac 0 \pi 2 + i \sin \dfrac 0 \pi 2}\)
\(\displaystyle \) \(=\) \(\displaystyle 3\)
\((k = 1):\quad\) \(\displaystyle z\) \(=\) \(\displaystyle \cos \dfrac {\pi} 2 + i \sin \dfrac {\pi} 2\)
\(\displaystyle \) \(=\) \(\displaystyle 3 i\)
\((k = 2):\quad\) \(\displaystyle z\) \(=\) \(\displaystyle \cos \dfrac {2 \pi} 2 + i \sin \dfrac {2 \pi} 2\)
\(\displaystyle \) \(=\) \(\displaystyle -3\)
\((k = 3):\quad\) \(\displaystyle z\) \(=\) \(\displaystyle \cos \dfrac {3 \pi} 2 + i \sin \dfrac {3 \pi} 2\)
\(\displaystyle \) \(=\) \(\displaystyle -3 i\)

$\blacksquare$


Sources