Quotient ring of Gaussian integers

This diagram shows the Gaussian integers modulo $3-i$.

The red points shown are all considered to be $0$ but their locations in $\mathbb Z[i]$ are $0$, $3-i$, $i(3-i)$ and $3-i + i(3-i)$. Every congruence class must be inside that box once and you can see there are $10$ of them.

The arrows show adding by $1$ each time. Doing that takes you through every equivalence class and then back to the start.

So $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$.

Define $$\phi: \mathbb{Z} \rightarrow \mathbb{Z}[i]/(3-i) \text{ where } \phi(z) = z + (3-i)\mathbb{Z}[i].$$ It follows simply that $\ker \phi = (3-i)\mathbb{Z}[i] \cap \mathbb{Z}$. So for any such $z \in \ker \phi$, we have $z = (3-i)(a+bi)$ for some $a,b \in \mathbb{Z}$. But $(3-i)(a+bi) \in \mathbb{Z}$ happens if and only if $3b-a=0$. So $$\begin{align*} \ker \phi = (3-i)\mathbb{Z}[i] \cap \mathbb{Z} &= \{(3-i)(3b+bi)\mid b \in \mathbb{Z}\}\\ &= \{(9b + b) + i(3b-3b)\mid b \in \mathbb{Z}\}\\ &= \{10b\mid b \in \mathbb{Z}\}\\ &= 10\mathbb{Z}. \end{align*}$$

To see $\phi$ is surjective, let $(a+bi) + (3-i)\mathbb{Z}[i] \in \mathbb{Z}[i]/(3-i)$. Then $a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$, so $\phi(a+3b) = (a+bi) + (3-i)\mathbb{Z}[i]$.

Hence $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$.

Firstly: it is not true in general that $\mathbb Z[i]/(a - ib) \cong \mathbb Z/(a^2 + b^2).$ (Consider the case of $3 - 0\cdot i$.)

The claimed isomorphism does hold if $a$ and $b$ are coprime.

Here is a sketch of how to see this:

To begin with, note that it is much easier to consider maps from $\mathbb Z$ to other rings, rather than maps in the opposite direction (as you suggested in your answer), because $\mathbb Z$ maps to any ring with unity in a canonical way, by sending $1$ to $1$.

So consider the canonical map $\mathbb Z \to \mathbb Z[i]/(a - i b).$

The target is finite of order $a^2 + b^2$, and so this map factors to give an injection $\mathbb Z/(n) \hookrightarrow \mathbb Z[i]/(a - i b)$ for some $n$ dividing $a^2 + b^2$.

Now if $a$ and $b$ are coprime then $b$ is coprime to $a^2 + b^2$, hence coprime to $n$, and so $b$ is invertible in $\mathbb Z/(n)$. Combining this observation with the equation $a - i b = 0$ (which holds in $\mathbb Z[i]/(a - i b)$) one finds (and I leave this as an exercise!) that the map $\mathbb Z/(n) \hookrightarrow \mathbb Z[i]/(a - ib)$ contains $i$ in its image, and hence is surjective as well as injective, and so we are done.