Using the Extended Euclidean Algorithm as implemented in this answer, modified for Gaussian integers, we get
$$
\begin{array}{r}
&&2&-1+2i&1-i\\\hline
1&0&1&1-2i&2+3i\\
0&1&-2&-1+4i&-5-5i\\
5+5i&2+3i&1-i&1&0
\end{array}\tag{1}
$$
This says that
$$
(1-2i+(2+3i)k)\color{#C00000}{(5+5i)}+(-1+4i-(5+5i)k)\color{#C00000}{(2+3i)}=1\tag{2}
$$
where $k$ is a Gaussian integer. For example, $k=i$ gives the solution
$$
\underbrace{(-2)\color{#C00000}{(5+5i)}}_{-10-10i}+\underbrace{(4-i)\color{#C00000}{(2+3i)}}_{11+10i}=1\tag{3}
$$
Since $-10-10i=(-2)(5+5i)$, we have
$$
\begin{align}
-10-10i&\equiv1\pmod{2+3i}\\
-10-10i&\equiv0\pmod{5+5i}
\end{align}\tag{4}
$$
and since $11+10i=(4-i)(2+3i)$
$$
\begin{align}
11+10i&\equiv0\pmod{2+3i}\\
11+10i&\equiv1\pmod{5+5i}
\end{align}\tag{5}
$$
Adding $a$ times $(4)$ and $b$ times $(5)$ gives a solution to
$$
\begin{align}
x&\equiv a\pmod{2+3i}\\
x&\equiv b\pmod{5+5i}
\end{align}\tag{6}
$$
Applying $(6)$ to the particular question above
$$
\begin{align}
-21-20i&\equiv \phantom{-}1\pmod{2+3i}\\
-21-20i&\equiv -1\pmod{5+5i}
\end{align}\tag{7}
$$
Since the solution in $(7)$ is given mod $(2+3i)(5+5i)=-5+25i$, adding $(1-i)(-5+25i)=20+30i$, we also get the solution
$$
\begin{align}
-1+10i&\equiv \phantom{-}1\pmod{2+3i}\\
-1+10i&\equiv -1\pmod{5+5i}
\end{align}\tag{8}
$$
