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This Möbius transformation maps the upper half-plane onto the unit disk. For $\mathrm{Im}z>0$, the identity $1-|C(z){|}^2=4\mathrm{Im}z/|z+i{|}^2$ puts the image inside the unit disk. Its inverse is $z=i(1+w)/(1-w)$, so the map is a conformal bijection. On the extended real boundary it maps onto the unit circle. It is the scalar complex-analytic version of a Cayley map, rather than the matrix parametrization used in Cayley transform.

For a real skew-symmetric matrix $A$, $I-A$ is invertible: $(I-A)x=0$ would imply $\Vert x{\Vert }^2=x^{T}Ax=0$. Its Cayley transform is an orthogonal matrix, since transposition gives $Q^T=(I+A{)}^{-1}(I-A)$ and the commuting factors cancel in $Q^{T}Q$. Also $\det (I+A)=\det ((I+A{)}^T)=\det (I-A)$, so $\det Q=1$ in every dimension. If $Av=\lambda v$, then $Qv=(1+\lambda )(1-\lambda {)}^{-1}v$. For $\lambda =it$, the transformed eigenvalue is $e^{2i\arctan t}$. In odd dimension $\det A=\det (-A)=-\det A$, so $A$ has a nonzero kernel and $Q$ fixes that direction.

For $\phi \ne \pm 1$, the Cayley transform isWriting its diagonal and off-diagonal entries as $a,b$, one has $a^2-b^2=1$, so $C^T\eta C=\eta$ and $\det C=1$. For $|\phi |<1$, put $\phi =\tanh (t/2)$ to obtain $C(\phi K)=\Lambda (t)$, so this interval covers the identity component. For $|\phi |>1$ the image lies in the other component and covers it except for $-I$, approached only as $|\phi |\to \infty$. The Exponential map of a Lie group reaches only the identity component, whereas the Cayley transform also reaches nonidentity-component elements but omits $-I$ and is undefined at $\phi =\pm 1$.

Using part (i) and $A^{†}=-A$ in the Cayley transform with argument $-A$ givesThe factors commute because $I-A$ and $I+A$ are polynomials in the same matrix, and an invertible commuting matrix also commutes with the other factor's matrix inverse. Both inverses exist by part (ii). Thus $U$ is unitary.

Taking determinants and using invariance under transpose,This determinant identity actually holds in every dimension. In dimension three, skew symmetry additionally givesso $\det A=0$. Hence there is a real nonzero vector $x$ in the kernel:The eigenvalue relation then gives $Nx=x$. Thus the Cayley transform fixes an axis and is a proper three-dimensional rotation, including the identity as a degenerate case.