= Solution
Let $J=A/a$, so $J^2=-I$. The even and odd powers in the <matrix exponential> give
$$
\begin{aligned}
D
&=I\sum_{k=0}^\infty\frac{(-1)^ka^{2k}}{(2k)!}
+J\sum_{k=0}^\infty\frac{(-1)^ka^{2k+1}}{(2k+1)!}\\
&=I\cos a+J\sin a.
\end{aligned}
$$
Here the two scalar sums are the <Taylor series> for <cosine> and <sine>. Consequently
$$
\boxed{D=\begin{pmatrix}\cos a&\sin a\\-\sin a&\cos a\end{pmatrix},\qquad \theta=a\pmod{2\pi}}.
$$
With the usual convention that positive angles turn anticlockwise, this <rotation matrix> rotates by $-a$; the parameter in the form printed in the paper is $\theta=a$.
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