We have $a, b \in \mathbb{Z}[i]$
$a=4$
$b=3+3i$
Then, are there any $u, v \in \mathbb{Z}[i]$ such that $ua+vb=−1+5i$?
We have $a, b \in \mathbb{Z}[i]$
$a=4$
$b=3+3i$
Then, are there any $u, v \in \mathbb{Z}[i]$ such that $ua+vb=−1+5i$?
Hint:
$a=(1+i)(2-2i)$
$b=(1+i)3$
Hint: with the same proof as in $\Bbb Z$ (or any PID), $\,\exists\, x,y\!:\ ax+by = c\iff \gcd(a,b)\mid c.\ $ Recall that $\,\Bbb Z[i]\,$ has a constructive (norm) Euclidean algorithm so we can always quickly compute this gcd, which determines solvabilty (and also the general solution - by scaling - follow the link).
Or: $\,(4,3(1\!+\!i)) = (4,i\!+\!1) = 1\!+\!i,\ $ by $\ 1\!+\!i\mid j\!+\!ki\iff 2\mid j\!-\!k\ $ (so $\ 1\!+\!i\mid -1\!+\!5i),\,$ thus
$$\begin{align} &4x + (3\!+\!3i) y \,=\, j\! +\! ki\ \ {\rm is\ solvable}\\[.2em] \iff\ &1\!+\!i\mid j\!+\!ki\ \ \ {\rm by}\ \ \ 1\!+\!i = \gcd(4,\,3\!+\!3i)\\[.2em] \iff\ &\ \ \ \ \ 2\mid j\!-\!k\\[.2em] \iff\ &\ \ \ \ \ \ j,\,k\ \,\text{have equal parity}\end{align}\qquad$$
This is equivalent to saying that $\,j\!+\!ki\,$ is "even" in $\,\Bbb Z[i],\,$ as is explained in this post on notions of parity for algebraic numbers.