> [!NOTE] Lemma > The [[Complex Modulus|complex modulus]] is a [[Euclidean Domain|Euclidean valuation]] on the [[Gaussian integers|ring of Gaussian integers]] $\mathbb{Z}[i]$. ###### Proof Clearly for all $a+bi\in \mathbb{Z}[i]$, $|a+bi|=a^2+b^2\in \mathbb{N}$. Let $\alpha,\beta\in \mathbb{Z}[i] \setminus \{ 0 \}$. Then $|\alpha\beta |=|\alpha|\cdot|\beta|\geq |\alpha|$since $|\beta|$ is a positive integer. Thus $|\cdot|$ satisfies (EF1). Now let $\alpha,\beta\in \mathbb{Z}[i]$ with $\alpha \neq0$. We may write $\beta=b_{1}+b_{2}i$ and $\alpha=a_{1}+a_{2}i$ with $b_{1},b_{2},a_{1},a_{2}\in \mathbb{Z}$, and $a_{1},a_{2}$ not both zero. Now $\beta/\alpha\in \mathbb{C}$ and we can write $\beta/\alpha=u_{1}+u_{2}i$ where $u_{1},u_{2}\in \mathbb{R}$. Let $u_{j}=v_{j}+w_{j}$ where $v_{j}\in \mathbb{Z}$ and $-\frac{1}{2}\leq w_{j}\leq \frac{1}{2}$. Thus $\beta/\alpha=(v_{1}+v_{2}i)+(w_{1}+w_{2}i)$. We write $q=v_{1}+v_{2}i\in \mathbb{Z}[i]$ and $r=(w_{1}+w_{2})\alpha$. Then $\beta=q\alpha+r.$Observe that $r=\beta-q\alpha\in \mathbb{Z}[i]$ since $\mathbb{Z}[i]$ is a ring. Finally suppose $r\neq 0$. Then $|r|=\lvert \alpha(w_{1}+w_{2}i) \rvert = \lvert \alpha \rvert \cdot (w_{1}^2+w^2) .$However $w_{1}^2+w_{2}^2 \leq \frac{1}{4}+\frac{1}{4}=\frac{1}{2}$Hence $\lvert r \rvert<\lvert \alpha \rvert$ as required.