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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 $\phi \ne \pm 1$, 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 $\phi =\pm 1$.