Let $\vec{v} \in \mathbb{C}^4$ be the vector given by $\vec{v} = (1, i, −1, −i)$. Find the matrix (with respect to the canonical basis on $\mathbb{C}^4$) of the orthogonal projection $P \in L(\mathbb{C}^{4})$ such that
$$\operatorname{null}(P) = {v}_{\perp} $$
First determine an orthonormal basis for $v_\perp$, let's say this basis is $B$ (what this comes down to really, is use something like Gram-Schmidt to find an orthonormal basis for $\mathbb{C}^4$ with $v/\|v\|$ as the first vector and all the other vectors you determine will be in $B$). Then every vector in $B$ is an eigenvector of $P$ associated with the eigenvalue $0$, and $v/\|v\|$ itself is the only eigenvector of $P$ associated with the eigenvalue $1$.
So let $Q$ be the orthogonal matrix with columns $v/\|v\|$ and the vectors in $B$. So then to get the matrix representation we have: \begin{equation} \text{Mtx}_{\epsilon,\epsilon}(P)=Q\text{Dg}[1,0,0,0]Q^T, \end{equation} where $\epsilon$ is the standard basis of $\mathbb{C}^4$.