Past exam of the mathematical methods I course of the University of Oxford / 2017 resit / 4

An orthogonal matrix satisfies $A^{T}A=I$. Taking determinants and using $\det A^T=\det A$ gives $(\det A{)}^2=1$, soThis calculation remains valid over the complex numbers: orthogonality uses the transpose, whereas a unitary matrix uses the conjugate transpose.

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.

The hyperbolic functions satisfy ${\cosh }^2\theta -{\sinh }^2\theta =1$, so a useful family isFor real $\theta$ this gives a one-real-parameter family of complex orthogonal matrices, which is enough to exhibit the requested hyperbolic parametrization. It does not parametrize every matrix in part (ii). For example $a=b=1/\sqrt{2}$ obeys the constraint but cannot have $a=\cosh \theta$ with real $\theta$, since then $a\ge 1$.

Allowing complex $\theta$ gives the full parametrization: $(a-ib)(a+ib)=1$, so choose a complex logarithm $\theta$ with $e^{\theta }=a-ib$. Then $a=\cosh \theta$ and $b=i\sinh \theta$. A general complex pair therefore needs a complex parameter; the real-parameter restriction in the question should not be mistaken for a completeness claim.