 Solution

For a two-dimensional complex orthogonal matrix, let $\epsilon =\det A\in \{+1,-1\}$. Since $A^{-1}=A^T$, comparison with the two-by-two matrix inverse formula forces the second row to be $(-\epsilon b,\epsilon a)$ if the first is $(a,b)$. Thus the complete family isConversely direct multiplication gives $A^{T}A=(a^2+b^2)I=I$ and $\det A=\epsilon$, proving sufficiency. The two parameters obey a bilinear constraint, not $|a{|}^2+|b{|}^2=1$; replacing squares by absolute squares would incorrectly impose unitarity.