= Solution
The claim is false because vanishing of the amplitude at one spatial point need not control its derivatives nearby. Take $X=\mathbb R$, $k=1$,
$$
\Phi(x,\theta)=x\theta,
\qquad
a(x,\theta)=x\theta.
$$
Then $0\in Z(a)$, but distributionally
$$
I_\Phi(a)
=x\int_{\mathbb R}e^{ix\theta}\theta\,d\theta
=-2\pi i\,x\delta'(x)
=2\pi i\,\delta(x)
$$
up to the Fourier-transform sign convention. Thus $0$ lies in the <singular support> of $I_\Phi(a)$ despite belonging to $Z(a)$.
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