= Solution
{c}
The <number operator> is $\hat N=a^\dagger a$. The <number state> expansion gives $a|\alpha\rangle=\alpha|\alpha\rangle$ by shifting the summation index; taking the adjoint gives $\langle\alpha|a^\dagger=\alpha^*\langle\alpha|$. Thus the <expectation value> is
$$
\overline N=\langle a^\dagger a\rangle=|\alpha|^2=\mu.
$$
For the second moment, use $aa^\dagger=a^\dagger a+I$:
$$
\hat N^2=a^\dagger aa^\dagger a=(a^\dagger)^2a^2+a^\dagger a.
$$
Its <expectation value> in the <coherent state> is $\langle\hat N^2\rangle=|\alpha|^4+|\alpha|^2$. Therefore
$$
\boxed{\overline N=|\alpha|^2,\qquad
\operatorname{Var}(N)=\langle\hat N^2\rangle-\overline N^2=|\alpha|^2,\qquad
\sigma_N=|\alpha|.}
$$
The <variance> equals the mean, exactly as for the <Poisson distribution> found in <3/b>. For $\mu>0$ the fractional <standard deviation> is $1/\sqrt\mu$, the usual <photon shot noise> scaling. For the <vacuum state>, both the mean <photon number> and its <variance> vanish; this does not imply zero <optical quadrature> fluctuations. At $\mu=0$ the fractional fluctuation is undefined.
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