Quotient Gaussian Integers Following Quotient ring of Gaussian integers, their extended conclusion is  $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$. However it does not convince me, at least, one example below:
Let $a=2,b=0$, I cannot find explicit isomorphism between   $\mathbb{Z}[i]/2\mathbb{Z}[i]$ and $\mathbb{Z}/4\mathbb{Z}$. 
The coset leaders of $\mathbb{Z}[i]/2\mathbb{Z}[i]=\{0,1,i,1+i\}$, and the coset leaders of $\mathbb{Z}/4\mathbb{Z}=\{0,1,2,3\}$. 
I appreciate if anyone could give a bijection mapping between the two quotient rings.  
I am not sure I am right or not but will be happy to discuss with anyone who is interested in.
From my point of view, $\mathbb{Z}/4\mathbb{Z}=\{0,1,2,3\}$ (I know it is not good to give this expression, just for convinence) has 2 units $1,3$ where the sum is $0$, but $\mathbb{Z}[i]/2\mathbb{Z}[i]$ has two units, say, $i,1$ and their sum is not zero. The structure of the two quotient rings are different, and hence the conclusion $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$ is not correct. 
Anyone here could make things clearer. Thanks a lot.

You'll never find such an isomorphism, because $x+x=0$ for all $x\in\mathbb Z[i]/2\mathbb Z[i]$, but not for all  $x\in\mathbb Z/4\mathbb Z$.
The theorem you state is only true if $\gcd(a,b)=1$.
See this answer from the same question.
