= Solution
{c}
Set $A=\alpha a^\dagger$ and $B=-\alpha^*a$. Their <commutator> is
$$
[A,B]=- |\alpha|^2[a^\dagger,a]=|\alpha|^2I.
$$
It commutes with both operators, so the stated <Baker--Campbell--Hausdorff formula> terminates:
$$
\boxed{D(\alpha)=e^{-|\alpha|^2/2}e^{\alpha a^\dagger}e^{-\alpha^*a}.}
$$
In particular $e^{-\alpha^*a}|0\rangle=|0\rangle$. Expanding the remaining exponential and using $(a^\dagger)^n|0\rangle=\sqrt{n!}|n\rangle$ gives the <number state> expansion
$$
\boxed{|\alpha\rangle=e^{-|\alpha|^2/2}\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}|n\rangle.}
$$
The orthonormality of the <number states> gives
$$
\boxed{P_n=e^{-|\alpha|^2}\frac{|\alpha|^{2n}}{n!}.}
$$
This is a <Poisson distribution> with parameter $\mu=|\alpha|^2$. Its probabilities sum to $e^{-\mu}e^\mu=1$. The phase of $\alpha$ affects the <coherent state>'s relative <probability amplitudes> but not its <photon number> probabilities. At $\alpha=0$ the distribution is concentrated at $n=0$.
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