Why does the asymptotic expansion of the real-valued Kummer function contain complex terms? Working on a problem in spectral theory, I need to study the asymptotics
of a confluent hypergeometric function 
(here $(a)_0=1$ and $(a)_s=a(a+1)\cdots(a+s-1)$ denote the Pochhammer symbol)
$$
\mathbf{M}(a,b,z)=\sum_{s=0}^{+\infty}\frac{(a)_s}{\Gamma(b+s)s!}z^s,\quad\text{as}\ z\to+\infty.
$$
In my case $-1<a<0$ and $b=1$, and I'm only interested in real $z$. 
I had a look in Abramowitz–Stegun (13.5.1, where
$M(a,b,z)=\Gamma(b)\mathbf{M}(a,b,z)$), and found that, as $z\to+\infty$,
we have the expansion
$$
\mathbf{M}(a,1,z)\sim 
\frac{e^{i\pi a}}{\Gamma(1-a)}z^{-a}\sum_{s=0}^{+\infty}\frac{\bigl((a)_s\bigr)^2}{s!(-z)^s}
+
\frac{e^z z^{a-1}}{\Gamma(a)}\sum_{s=0}^{+\infty}\frac{\bigl((1-a)_s\bigr)^2}{s!z^s}.
$$
What is worrying me is the factor $e^{i\pi a}$ in the first term. It is
complex (in fact, non-real), and everything else in the expansion is real for
positive $z$. Also,
from the definition of $\mathbf{M}$ we see that it should be real for
positive $z$. I have also had a look in
13.7.2, where the
same expansion is given. It is also the same in the book Asymptotics and
special functions by Frank Olver, and I get the same from Mathematica. Thus,
I believe that the expansion above is correct.
In the asymptotic expansion above the term with the $e^{i\pi a}$ factor is 
small in comparison with the second one. In fact, some sources 
hint that it can be neglected  (compare 13.7.1). As it 
happens, I want to keep that term, even if it is small. Thus, I think I can state my questions as follows:

  
*
  
*Why is the real-valued function having complex terms in its asymptotic expansion?
  
*I'm only considering real $z$. Will the expansion of $\mathbf{M}(a,1,z)$ above still be valid if I replace $e^{i\pi a}$ with its real part, $\cos(\pi a)$?
  
*Could it be that the imaginary part somehow cancels? (I don't see how it could.)
  

 A: When z is real and  tends to $\infty$ (that is, $z>0$) the second part of this double series is dominant, that is, is exponentially bigger that the first part and then, in asymptotic sense, you can avoid it to obtain
$$\mathbf{M}(a,1,z)\sim 
\frac{e^z z^{a-1}}{\Gamma(a)}\sum_{s=0}^{+\infty}\frac{\bigl((1-a)_s\bigr)^2}{s!z^s}$$
so, the complex terms in its asymptotic expansion dissapear. Note that when $z$ is real and negative, that is, $z\to-\infty$, is the first part that is dominant and taking account that $z^{-a}=e^{i\,\pi\,a}(-z)^{-a}$, the imaginary part also dissapear
$$\mathbf{M}(a,1,z)\sim 
\frac{1}{\Gamma(1-a)}(-z)^{-a}\sum_{s=0}^{+\infty}\frac{\bigl((a)_s\bigr)^2}{s!(-z)^s}$$
where here $-z$ is positive.
A: The statement is that
$$
{\bf M}(a,1,z) \sim \frac{{e^z z^{a - 1} }}{{\Gamma (a)}}\sum\limits_{n = 0}^\infty  {\frac{{((1 - a)_n )^2 }}{{n!z^n }}}  + \frac{{e^{ \pm \pi ia} z^{ - a} }}{{\Gamma (1 - a)}}\sum\limits_{n = 0}^\infty  {\frac{{((a)_n )^2 }}{{n!( - z)^n }}}, 
$$
as $z\to \infty$ in the sectors $ - \frac{\pi }{2} + \delta  \le  \pm \arg z \le \frac{{3\pi }}{2} - \delta$. The region of validity is in the sense of Poincaré, i.e., it ignores the Stokes lines $\arg z =0$, $\arg z =\pm \pi$ and extend the expansion up to the anti Stokes lines $\arg z = \mp \frac{\pi }{2}$ and $\arg z =\pm \frac{3\pi }{2}$. A more precise statement is that these expansions hold whenever $0<\pm \arg z < \pi$. On the Stokes line $\arg z=0$, the correct expansion is the average of the expansions on the two sides. Thus, in summary
$$
{\bf M}(a,1,z) \sim \frac{{e^z z^{a - 1} }}{{\Gamma (a)}}\sum\limits_{n = 0}^\infty  {\frac{{((1 - a)_n )^2 }}{{n!z^n }}}  + S(\theta )\frac{{z^{ - a} }}{{\Gamma (1 - a)}}\sum\limits_{n = 0}^\infty  {\frac{{((a)_n )^2 }}{{n!( - z)^n }}} 
$$
as $z \to \infty$, with $\theta =\arg z$ and the Stokes multiplier
$$
S(\theta)=\begin{cases} e^{-\pi i a} & \text{if }\; -\pi < \theta <0, \\[0.25em] \cos(\pi a) & \text{if }\; \theta = 0, \\[0.25em]  e^{\pi i a} & \text{if }\; 0 < \theta < \pi. \end{cases}
$$
For an exponentially improved, optimally truncated asymptotic expansion on the line $\arg z=0$, see this paper.
