= Solution
{c}
Put $b=\Delta\omega_b>0$ and $s=t-t_m$. At the peak time, $e^{-ip\Delta\omega t_m}=1$, so the exact comb sum becomes
$$
E(t_m+s)=Ae^{i\phi_0-i\omega_{\rm ce}t_m}
\sum_p e^{-[(\omega_p-\omega_0)/b]^2}e^{-i\omega_ps}.
$$
For many occupied <normal modes>, $b\gg\Delta\omega$, replace the sum by an <integral> with density $1/\Delta\omega$. Setting $x=\omega-\omega_0$ and evaluating the <Fourier transform of a Gaussian> gives
$$
\begin{aligned}
E(t_m+s)&\simeq\frac{A}{\Delta\omega}e^{i\phi_0-i\omega_{\rm ce}t_m}
\int_{-\infty}^{\infty}e^{-[(\omega-\omega_0)/b]^2}e^{-i\omega s}\,d\omega\\
&=\frac{A\sqrt\pi b}{\Delta\omega}
e^{-b^2s^2/4}e^{-i\omega_0s}e^{i\phi_0-i\omega_{\rm ce}t_m}.
\end{aligned}
$$
Thus the isolated-<optical pulse> approximation has
$$
\boxed{F(s)=\frac{A\sqrt\pi\,\Delta\omega_b}{\Delta\omega}
e^{-(\Delta\omega_b s/2)^2},\qquad
\psi_m=\phi_0-\omega_{\rm ce}t_m.}
$$
The <optical pulse> has a <Gaussian function> envelope and no <chirp>. Its carrier <angular frequency> is $\omega_0$, and its <light intensity> <full width at half maximum> is $2\sqrt{2\ln2}/b$. Successive envelope maxima have different <carrier-envelope phases>, consistent with <1/b>.
\b[Approximation in the printed assertion.] A discrete <optical frequency comb> gives a train, not one exact isolated <Gaussian pulse>. Applying the <Poisson summation formula> to the ideal sum over all <integers> instead gives
$$
E(t)=\frac{A\sqrt\pi b}{\Delta\omega}
\sum_{j\in\mathbb Z}e^{-b^2(t-jT)^2/4}
e^{-i\omega_0(t-jT)}e^{i\phi_0-i\omega_{\rm ce}jT}.
$$
Near $t_m$, the $j=m$ term dominates when $bT\gg1$. The other terms are exponentially small but not identically zero. A single <Gaussian function> used globally would decay between $0$ and $T$, whereas the exact <light intensity> has equal maxima there; this supplies a concrete reason the isolated form cannot be an exact global identity. The extension to negative <frequencies> is the usual narrow-positive-band analytic-field approximation when $\omega_0\gg b$.
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