= Solution
Let $P(z)=1/(2\sqrt\pi z^{1/4})$, choose a consistent branch, and truncate the corrected series after $r=n-1$, with $n=|\sigma|+O(1)$. The leading factorial tail in part (iii) is represented by the same indented integral as in part (ii):
$$
R_n\sim\frac{P(z)e^{\sigma/2}}{2\pi}\sum_{r=n}^{\infty}\frac{\Gamma(r)}{\sigma^r}
\quad\longleftrightarrow\quad
\frac{P(z)e^{-\sigma/2}}{2\pi}I(\sigma,n).
$$
Here the divergent sum means its lateral <Borel summation>, not ordinary summation. To see why the <pole> approximation controls this remainder, write $a_r=\sigma^rY_r$ and insert $\Gamma(r)=\int_0^\infty e^{-v}v^{r-1}dv$ in the leading late terms. Summing the geometric tail creates $1/(1-v/\sigma)$; $v=\sigma t$ produces $I(\sigma,n)$. The relative $O(r^{-1})$ corrections to the late coefficients give lower-order contributions on its saddle scale. More concretely, the <Borel transform> of the coefficient series is
$$
\sum_{r=1}^\infty\frac{a_r t^{r-1}}{\Gamma(r)}
=\frac d{dt}\,{}_2F_1\left(\frac16,\frac56;1;t\right)
=\frac1{2\pi(1-t)}+O(\log(1-t))\quad(t\to1).
$$
The <simple pole> supplies the leading <Stokes phenomenon>; its weaker <logarithmic singularity> supplies the smaller corrections. This is the integral justification for retaining the factorial tail.
For $\arg\sigma=\phi/\sqrt{|\sigma|}$, expand $\sigma=n+i\phi\sqrt n+O(1)$. Part (i), with $\mu=\phi$, now gives
$$
\boxed{R_n\sim\frac{i e^{-\sigma/2}}{4\sqrt\pi z^{1/4}}
\left[1+\operatorname{erf}\left(\frac\phi{\sqrt2}\right)\right]}.
$$
This formula uses the upper-pole prescription and the upper <Airy function> switching ray: near $\arg z=2\pi/3$, take $\arg\sigma=3\arg z/2-\pi$ near zero. The coefficient of the subdominant exponential changes smoothly from zero to $i$ across this ray and has value $i/2$ on it. The angular transition width is $O(|\sigma|^{-1/2})=O(|z|^{-3/4})$. The conjugate lower ray has the conjugate contour sign, rather than the same $+i$ on both rays. This is <error-function smoothing of a Stokes multiplier>, not a discontinuity of the analytic <Airy function>. With the literal printed $r=1$ series, the missing $P(z)e^{\sigma/2}$ would invalidate this remainder formula.
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