Norm of Eisenstein Integer
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Theorem
Let $\alpha$ be an Eisenstein integer.
That is, $\alpha = a + b \omega$ for some $a, b \in \Z$, where $\omega = e^{2\pi i /3}$.
Then:
- $\cmod \alpha^2 = a^2 - a b + b^2$
where $\cmod {\, \cdot \,}$ denotes the modulus of a complex number.
Proof
We find that:
\(\ds \cmod \alpha^2\) | \(=\) | \(\ds \alpha \overline \alpha\) | Modulus in Terms of Conjugate | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a + b \omega} \paren {\overline {a + b \omega} }\) | Modulus in Terms of Conjugate | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a + b \omega} \paren {\overline a + \overline b \overline \omega}\) | Sum of Complex Conjugates and Product of Complex Conjugates | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a + b \omega} \paren {a + b \overline \omega}\) | Complex Number equals Conjugate iff Wholly Real | |||||||||||
\(\ds \) | \(=\) | \(\ds a^2 + \paren {\omega + \overline \omega} a b + \omega \overline \omega b^2\) |
By the definition of the polar form of a complex number:
- $\omega = \exp \paren {\dfrac {2 \pi i} 3} = \map \cos {\dfrac {2 \pi} 3} + i \, \map \sin {\dfrac {2 \pi} 3} = -\dfrac 1 2 + i \dfrac {\sqrt 3} 2$
Thus by Sum of Complex Number with Conjugate:
- $\omega + \overline \omega = 2 \cdot \paren {-\dfrac 1 2} = -1$
Also:
\(\ds \omega \overline \omega\) | \(=\) | \(\ds \map \exp {\dfrac {2 \pi i} 3} \, \overline {\map \exp {\dfrac {2 \pi i} 3} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \exp {\dfrac {2 \pi i} 3} \, \map \exp {-\dfrac {2 \pi i} 3}\) | Polar Form of Complex Conjugate | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \exp {\dfrac {2 \pi i} 3 - \dfrac {2 \pi i} 3}\) | Exponential of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \exp 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | Exponential of Zero |
Therefore:
- $\cmod \alpha^2 = a^2 + \paren {\omega + \overline \omega} a b + \omega \overline \omega b^2 = a^2 - a b + b^2$
as required.
$\blacksquare$