# Norm of Eisenstein Integer

## 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:

$\omega = \exp \paren {\dfrac {2 \pi i} 3} = \map \cos {\dfrac {2 \pi} 3} + i \, \map \sin {\dfrac {2 \pi} 3} = -\dfrac 1 2 + i \dfrac {\sqrt 3} 2$
$\omega + \overline \omega = 2 \cdot \paren {-\dfrac 1 2} = -1$

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.

$\blacksquare$