= Solution
Complete the square: $z^2-2iz=(z-i)^2+1$. The pole of $(1-iz)^{-1}$ is at $z=-i$, so the contour may be shifted upward through the <saddle point> $z=i$ without crossing a singularity. Setting $z=t+i$ gives the exact steepest-descent representation
$$
I(x)=e^{-x}\int_{-∞}^{∞}{e^{-xt^2}\over2-it}\,dt.
$$
Near the saddle,
$$
{1\over2-it}=\sum_{n\geq0}{(it)^n\over2^{n+1}}.
$$
Odd terms integrate to zero, while
$$
\int_{-∞}^{∞}t^{2m}e^{-xt^2}\,dt=Γ(m+1/2)x^{-m-1/2}.
$$
The <Watson lemma> therefore yields the full <asymptotic expansion>
$$
\boxed{I(x)\sim e^{-x}\sum_{m=0}^{∞}{(-1)^mΓ(m+1/2)\over2^{2m+1}x^{m+1/2}}.}
$$
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