= Solution
The generator acts only in the $ct,z$ plane, with its square equal to the identity on that two-dimensional subspace. Its exponential is
$$
\boxed{\Lambda=
\begin{pmatrix}
\cosh\kappa&0&0&\sinh\kappa\\
0&1&0&0\\
0&0&1&0\\
\sinh\kappa&0&0&\cosh\kappa
\end{pmatrix}.}
$$
For the <vector> in (c), this gives
$$
ct'=\tfrac12(\cosh\kappa+\sinh\kappa\cos\theta),\qquad
z'=\tfrac12(\sinh\kappa+\cosh\kappa\cos\theta),
$$
while $x'=x$ and $y'=y$. Since the <vector> is future-null, its spatial magnitude is $ct'$. Therefore
$$
\frac{z'}{ct'}=\frac{\cos\theta+\tanh\kappa}
{1+\tanh\kappa\cos\theta}=\cos\theta',
\qquad
\frac{\sqrt{x'^2+y'^2}}{ct'}=\frac{\sin\theta}{\mathcal R}.
$$
The transverse <azimuth> does not change. These are exactly the direction and scale obtained from the <spinor> congruence. In particular $ct'= \mathcal R\,ct$ and $\tan(\theta'/2)=e^{-\kappa}\tan(\theta/2)$. The azimuth-sign qualification in (c) is compatible with this check because a boost along $z$ leaves both transverse coordinates unchanged.
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