= Solution
In the common sector $0<\arg z<\pi/2$, rotate the positive real contour clockwise to the negative imaginary axis. The large quarter-circle contributes zero because $\operatorname{Re}(z\zeta)>0$ there. The rotated ray passes through the simple pole $\zeta=-i$, whose <residue> is
$$
R=\operatorname{Res}_{\zeta=-i}\frac{e^{-z\zeta}}{1+\zeta^2}=\frac{e^{iz}}{-2i}.
$$
Take a small semicircular indentation into the right half-plane, excluding the pole from the clockwise contour. The downward ray traverses this semicircle clockwise, contributing $-i\pi R$ in the limit. Thus the deformed downward integral is $\widetilde F-i\pi R$, and it equals the original integral. Consequently
$$
\boxed{F(z)-\widetilde F(z)=-i\pi R=\frac\pi2e^{iz}.}
$$
This half-residue, rather than a full $2\pi i$ residue, results from a pole lying on the rotated integration ray.
Parametrizing $\zeta=-iu$ gives
$$
\widetilde F(z)=-i\,\operatorname{PV}\int_0^\infty\frac{e^{izu}}{1-u^2}\,du.
$$
For $\operatorname{Im}z>0$ its tail decays exponentially. Near $u=1$, subtract the pole coefficient before integrating; the remaining integrand is regular, and the principal-value pole term is explicit. This proves holomorphic dependence on $z$ throughout the upper half-plane, locally uniformly on compact subsets. Agreement in the overlap and the <identity theorem> therefore give the requested <analytic continuation>:
$$
\boxed{F_{\rm continued}(z)=-i\,\operatorname{PV}\int_0^\infty\frac{e^{izu}}{1-u^2}\,du+\frac\pi2e^{iz},\qquad\frac\pi2\le\arg z<\pi.}
$$
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