= Solution
The identity theorem says <analytic functions> agreeing on a set with an interior accumulation point agree throughout a connected domain. Analytic continuation is an <analytic function> on a larger connected domain agreeing on a nonempty overlap. Factor the denominator as $(t+i)(t-z)$. For $\Im z<0$, contour closure below gives
$$
F(z)=-2\pi i\frac{e^{-5iz}-e^{-5}}{z+i},
$$
with a removable value at $-i$; this entire expression is the continuation. For $\Im z>0$, the displayed real-axis <integral> encloses only $-i$ and equals $2\pi i e^{-5}/(z+i)$, so it is not that continuation. Finally $1/n\to0$ only at a boundary point of $D$, so the zeros do not determine $G$; for example $G(z)=\sin(\pi/z)$ is nonzero and has them all.
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