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Represent the Möbius transformation by $A=\left(\begin{array}{cc}3& 2\\ 7& 5\end{array}\right)$, which has determinant one and trace eight. Its distinct eigenvalues are ${\lambda }_{\pm }=4\pm \sqrt{15}$ and their product is one. A change of basis to eigenvectors conjugates its projective action to a dilation by ${\lambda }_{+}/{\lambda }_{-}=(4+\sqrt{15}{)}^2$ or its reciprocal. Translation length is invariant under conjugacy, so part (i)'s dilation calculation givesThe fixed boundary points solve $7z^2+2z-2=0$, namely $(-1\pm \sqrt{15})/7$. The axis of a hyperbolic-space isometry is the semicircular geodesic with those endpoints in the vertical real plane.

A model of three-dimensional hyperbolic space is ${\mathbb{H}}^3=\{(z,t):z\in \mathbb{C},t>0\}$ with metric $d{s}^2=(|dz{|}^2+d{t}^2)/t^2$, the complete simply connected metric of curvature $-1$. The translation length of an isometry $g$ is $\ell (g)={\inf }_{P}d(P,gP)$. A hyperbolic isometry has positive translation length attained on its invariant complete geodesic, the axis of a hyperbolic-space isometry. It joins the two fixed ideal points and is translated by distance $\ell (g)$; a three-dimensional axial isometry can also rotate around it.

For the boundary map $z\mapsto kz$, the hyperbolic extension of a Möbius transformation is $(z,t)\mapsto (kz,|k|t)$. Every path from $P$ to $gP$ has length at least the absolute change of $\log t$, namely $|\log |k||$, because the metric length dominates $|dt|/t$. On the vertical line $z=0$ a vertical segment attains that bound. HenceIf $|k|\ne 1$ the vertical line is the axis. If $|k|=1$ the map is a rotation with zero translation length, including the identity when $k=1$; it is not a positive-length hyperbolic isometry.