I was going through this link for the proof "Gaussian Integers being an Euclidean Domain" with the function $f(z)=|z|^2$ https://www.cmi.ac.in/~shreejit/Gaussian.pdf
We have $$Z[i]=\left\{a+bi: a,b \in \mathbb{Z}\right\}$$
Condition $(a):$ If $\alpha$ and $\beta$ are two non zero complex numbers we need to prove $$f(\alpha \beta)\ge f(\alpha)$$ is Straight forward
Condition $(b):$
Case $1.$ Let there exists $q, r \in Z[i]$ such that:
$\beta=q\alpha+r$ such that $\alpha=n$ where $n \in \mathbb{Z^+}$
Now if $\beta=a+ib$
By Division algorithm we get:
$a=un+u_1,b=vn+v_1$ where $|u_1|,|v_1|\le \frac{n}{2}$
Hence $\beta=a+ib=(u+iv)n+u_1+iv_1$
Hence $r=u_1+iv_1$ so we get:
$$f(r)=u_1^2+v_1^2\le \frac{2n^2}{4}<n^2=f(n)=f(\alpha)$$
Hence $f(r)<f(\alpha)$
Up to here i am fine since it is same as author's proof. But for case $2.$ i modified the proof
Case $2.$ Now let $\alpha$ be an arbitrary complex number from the Ring of Gaussian Integers
We have $\beta=q\alpha+r$
Multiplying both sides with $\bar{\alpha}$
we get:
$\beta \bar{\alpha}=q|\alpha|^2+r\bar{\alpha}$
Now since again $|\alpha|^2=m$ where $m \in \mathbb{Z^+}$ we can apply the analysis in Case $1.$
That is:
$$f(r\bar{\alpha})<f(m)=m^2=|\alpha|^4$$
Hence $$f(r)f(\bar{\alpha)}<|\alpha|^4$$
So $$f(r)<f(\alpha)$$
So from both the cases condition $(b)$ is also TRUE.
So the conclusion is $Z[i]$ is an Euclidean Domain.
Is my modified case $2.$ Valid?