= Solution
Writing $\theta=\omega t$ and
$$
D=\cosh(2a)+\cos(2\theta),
$$
the supplied identities give
$$
\operatorname{Re}\tanh(a+i\theta)
=\frac{\sinh(2a)}D,
\qquad
\operatorname{Re}\operatorname{sech}(a+i\theta)
=\frac{2\cosh a\cos\theta}{D}.
$$
Using $A=m\omega/(2\hbar)$ in part (d),
$$
\boxed{
h(t)=
\sqrt{\frac{\hbar}{m\omega}}\,
\frac{\cos(\omega t)}{\sinh a}}.
$$
The peak executes <simple harmonic motion> about the origin with angular frequency $\omega$ and amplitude $\sqrt{\hbar/(m\omega)}/\sinh a$.
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