= Solution
Regard $A$ as a <matrix> over the <rational numbers>. Part (a)(ii), with $n=2$, gives its <characteristic polynomial> $X^2+1$. Therefore, writing $i$ for the <imaginary unit>,
$$
\det(iI-A)=i^2+1=0.
$$
The <matrix> $M=iI-A$ is not zero: otherwise $A=iI$ would have noninteger diagonal entries. It consequently has <rank of a matrix> one over the <fraction field> $\mathbb Q(i)$ of the <Gaussian integers>.
By part (b)(i), $\mathbb Z[i]$ is a <euclidean domain>, hence a <principal ideal domain>, so apply <Smith normal form>. There exist $P,Q\in\operatorname{GL}_2(\mathbb Z[i])$ such that
$$
P(iI-A)Q=\begin{pmatrix}d&0\\0&0\end{pmatrix},\qquad d\ne0.
$$
Indeed, invertible multiplication preserves <rank of a matrix>, so exactly one diagonal entry is nonzero. For $e_2=(0,1)^{\mathsf T}$, multiplication on the right gives $P(iI-A)Qe_2=0$. Since $P$ is an <invertible matrix>,
$$
\boxed{(iI-A)Qe_2=0}.
$$
The <Smith normal form supplies a kernel basis> lemma ensures that the null vector $Qe_2$ is a column of an <invertible matrix> over $\mathbb Z[i]$; an arbitrary nonzero solution over $\mathbb Q(i)$ would not itself establish that requirement.
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