Prove that the Gaussian Integer's ring is a Euclidean domain

Let $a=\alpha_1+\alpha_2 i, b=\beta_1+\beta_2i$ where $\alpha_1,\alpha_2,\beta_1,\beta_2\in\Bbb Z$. Then $$ \frac ab=\frac{\alpha_1+\alpha_2i}{\beta_1+\beta_2i}=\frac{(\alpha_1+\alpha_2i)(\beta_1-\beta_2i)}{N(b)}=\frac{(\alpha_1\beta_1+\alpha_2\beta_2)-(\alpha_1\beta_2-\alpha_2\beta_1)i}{N(b)} $$ By a modified form of the division algorithm on the integers, $\exists q_1,q_2,r_1,r_2\in\Bbb Z$ such that $$ \begin{align}\alpha_1\beta_1+\alpha_2\beta_2&=N(b)q_1+r_1\\\alpha_1\beta_2-\alpha_2\beta_1&=N(b)q_2+r_2\end{align} $$ Where $-\frac12N(b)\le r_\ell\le\frac12N(b)$.

Then our quotient is $q=q_1-q_2i$ and our remainder is $r=r_1-r_2i$. Then $\frac ab=\frac{N(b)q+r}{N(b)}$ or $$ a=bq-\frac{r}{\overline b} $$ By closure, $\frac{r}{\overline b}\in\Bbb Z[i]$, so $\frac{r}{\overline b}$ is the remainder. $$ N\left(\frac{r}{\overline b}\right)=N\left(\overline {b^{-1}}\right)N(r)=N(b)^{-1}N(r) $$ While $N(r)=r_1^2+r_2^2\le2\left(\frac12N(b)\right)^2=\frac12N(b)^2$. Thus the remainder satisfies $$N\left(\frac{r}{\overline b}\right)\le \frac12N(b)^{-1}N(b)^2=\frac12N(b)$$

Think about this problem geometrically:

For example : $\alpha=3+2i$ and $\beta=-10+6i$

Consider a circle of radius $\sqrt{3^2+2^2}$ by centered at $\beta$, and then move from $\alpha$ to $\beta$ !

$$\beta=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{DE}+\overrightarrow{Ef}+\overrightarrow{FG}+\overrightarrow{GH}+(-1-i) =$$ $$\alpha+i\alpha-\alpha+i\alpha-\alpha+i\alpha+(-1-i)=$$ $$(-1+3i)(3+2i)+(-1-i)$$

It's easy to generalize this idea to get a complete proof.
And by this way it's easy to see why $q$ and $r$ in $\beta = q\alpha+r$ are not necessarily unique. Because you can move form $\alpha$ into the circle around the $\beta$, by different ways!

Here is a different geometric proof. We have two Gaussian integers $a$ and $b$, and we have to prove that there exists a Gaussian integer $z$ such that

$$|az-b|<|a|$$

Well, let's consider the set $A=\{az\mid z\in\mathbb Z[i]\}$. What does it look like? Writing $z=x+yi$, we see that $az=xa+y(ai)$. But $ai$ is just $a$, rotated clockwise by $90$ degrees, so $a$ and $ai$ form a pair of orthogonal vectors of length $|a|$. We now see that our set $A$ is a square lattice, it is the set of gridpoints of a square grid of mesh size $|a|$.

We have to prove that at least one of these gridpoints is inside the open disc of radius $|a|$ centered at $b$. But in fact, something slightly stronger is true: given a square grid of mesh size $s$, any disc of radius $s$ must contain a grid point, no matter where it's placed. This is visually obvious: such a disc has diameter $2s$ whereas a square has a diameter of only $\sqrt 2 s$. There clearly isn't enough room to place a disc without overlapping a gridpoint.