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:

\(\displaystyle \cmod \alpha^2\) \(=\) \(\displaystyle \alpha \overline \alpha\) Modulus in Terms of Conjugate
\(\displaystyle \) \(=\) \(\displaystyle \paren {a + b \omega} \paren {\overline {a + b \omega} }\) Modulus in Terms of Conjugate
\(\displaystyle \) \(=\) \(\displaystyle \paren {a + b \omega} \paren {\overline a + \overline b \overline \omega}\) Sum of Complex Conjugates and Product of Complex Conjugates
\(\displaystyle \) \(=\) \(\displaystyle \paren {a + b \omega} \paren {a + b \overline \omega}\) Complex Number equals Conjugate iff Wholly Real
\(\displaystyle \) \(=\) \(\displaystyle a^2 + \paren {\omega + \overline \omega} a b + \omega \overline \omega b^2\)

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:

\(\displaystyle \omega \overline \omega\) \(=\) \(\displaystyle \map \exp {\dfrac {2 \pi i} 3} \, \overline {\map \exp {\dfrac {2 \pi i} 3} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map \exp {\dfrac {2 \pi i} 3} \, \map \exp {-\dfrac {2 \pi i} 3}\) Polar Form of Complex Conjugate
\(\displaystyle \) \(=\) \(\displaystyle \map \exp {\dfrac {2 \pi i} 3 - \dfrac {2 \pi i} 3}\) Exponential of Sum
\(\displaystyle \) \(=\) \(\displaystyle \map \exp 0\)
\(\displaystyle \) \(=\) \(\displaystyle 1\) 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.

$\blacksquare$