= Solution
For a <two-dimensional complex orthogonal matrix>, let $\varepsilon=\det A\in\{+1,-1\}$. Since $A^{-1}=A^T$, comparison with the two-by-two <matrix inverse> formula forces the second row to be $(-\varepsilon b,\varepsilon a)$ if the first is $(a,b)$. Thus the complete family is
$$
\boxed{A=\begin{pmatrix}a&b\\-\varepsilon b&\varepsilon a\end{pmatrix},\qquad a,b\in\mathbb C,\qquad a^2+b^2=1,\qquad\varepsilon=\pm1.}
$$
Conversely direct multiplication gives $A^TA=(a^2+b^2)I=I$ and $\det A=\varepsilon$, proving sufficiency. The two parameters obey a bilinear constraint, not $|a|^2+|b|^2=1$; replacing squares by absolute squares would incorrectly impose <unitarity>.
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