Finding inverses in quotient rings In $ A=\mathbb{Z}[i]=\{a+bi \ : \ a,b \in \mathbb{Z}\} $ we consider $a=7+56i; \ b=3+3i; \ c=1+8i$. We will write $(a)$ to refer to the ideal generated by $a$

Find out whether the elements $\overline{b}:= b + (a), \ \overline{c} \in \frac{A}{(a)}$ are invertible elements in $\frac{A}{(a)}$ or not. If they are, calculate their inverses. 

I’ve managed to do something:
To see if $\overline{b}$ is invertible we have to find $\overline{t}$ such that $\overline{b} \overline{t}= \overline{1}$ which translates into $\overline{1-bt}=0 \Rightarrow 1-bt \in (a) \Rightarrow \lambda a =1-bt \Rightarrow 1=\lambda a +bt $ and this last expression I know it’s a Bézout Identity but I don’t know how to work with it in $\mathbb{Z}[i]$. 
 A: Perform Euclidean algorithm:
\begin{align}
7+56i&=(3+3i)(10+8i)+1+2i\\
3+3i&=(1+2i)(2-i)-1
\end{align}
to get
$$(7+56i)(2-i)-(3+3i)(29+6i)=1$$
from which
$$(3+3i)^{-1}\equiv -29-6i\pmod{7+56i}$$
On the other hand $a=7c$, hence $c$ is not invertible modulo $a$.
A: $a,b$ have coprime $\rm\color{#0a0}{norms}$ so we can  use the Euclidean algorithm in $\Bbb Z\,$ (vs. $\,\Bbb Z[i]).\,$ Having nontrivial integer factors, they've  even smaller $\rm\color{#0a0}{multiples \in\Bbb Z},\,$ so we get, with a minute of mental arithmetic
$\qquad\begin{align} \overbrace{(1\!-\!8i)\,a}^{\large\ 7(1\,+\,8i)\,=\,a}\!\!\!\! &= 7\cdot 65,\\[.2em]  
\underbrace{(\color{#c00}{1 -\, i)\,b}}_{\large\ \ 3(1\ +\ i)\,=\,b}\!\!\!  &=\color{#c00} 6,\end{align}\ $ $
\begin{align}{\rm thus}\ \ \ \ 7\cdot 65\ - 76\,\cdot\,\color{#c00}{6} \ \ \ \ \  &= -1\ \ \ {\rm by}\ \ \ (7\cdot 65)\div 6\\[.2em]
\Rightarrow\ \, (1\!-\!8i)a - 76\color{#c00}{(1\!-\!i)b} &= -1\ \Rightarrow\  b^{-1}\equiv 76(1\!-\!i)\!\!\pmod{\!a}\end{align}$
Remark $ $ Generally $\,(\bar aa,\bar bb)=1\,\Rightarrow\, j\bar aa + k\bar bb = 1\,$ is a Bezout identity for $\,(a,b)=1\,$ too.
