= Solution
Write $B=b+ic$ with $b>0$. Normalization of the <Gaussian wave packet> gives $|A|^2\int e^{-2bx^2}\,dx=1$, hence $|A|^2=\sqrt{2b/\pi}$. The supplied <Gaussian integral> ratio then gives
$$
\boxed{\langle x^2\rangle=\frac1{4b}}.
$$
The <momentum operator> obeys $p\Psi=2i\hbar Bx\Psi$, and direct second differentiation gives $p^2\Psi=(2\hbar^2B-4\hbar^2B^2x^2)\Psi$. Taking the <expectation value> and substituting the first result gives
$$
\langle p^2\rangle=2\hbar^2B-\frac{\hbar^2B^2}{b}=\frac{\hbar^2B(2b-B)}b=\frac{\hbar^2|B|^2}b=\boxed{4\hbar^2|B|^2\langle x^2\rangle},
$$
since $2b-B=\bar B$. These are the <second moments of a complex Gaussian wave packet>.
For the solution in part (b), the trigonometric identity
$$
\tan(\phi-it)=\frac{\sin(2\phi)-i\sinh(2t)}{\cos(2\phi)+\cosh(2t)}
$$
gives, writing $D=\cosh(2t)+\cos(2\phi)$,
$$
b=\frac{\sin(2\phi)}{2\hbar D},\qquad |B|^2=\frac{\cosh(2t)-\cos(2\phi)}{4\hbar^2D}.
$$
Thus normalizability requires $\sin(2\phi)>0$, and the exact <expectation values> are
$$
\langle x^2\rangle=\frac{\hbar[\cosh(2t)+\cos(2\phi)]}{2\sin(2\phi)},\qquad \langle p^2\rangle=\frac{\hbar[\cosh(2t)-\cos(2\phi)]}{2\sin(2\phi)}.
$$
Since $\cosh(2t)\sim e^{2t}/2$,
$$
\boxed{\langle x^2\rangle\sim\langle p^2\rangle\sim\frac{\hbar}{4\sin(2\phi)}e^{2t}}.
$$
Their difference is the constant $-\hbar\cot(2\phi)$, so the mean <energy> $\langle H\rangle=-\hbar\cot(2\phi)/2$ stays constant despite exponential spreading in both observables.
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