Example of Saddle-Point method I am trying to solve using the saddle point method (large a>0):
$$I(\alpha)= \int_{-i\pi/2}^{\pi/2}dz\, (1+z^2)e^{-a\cos(z)}$$
So I find that the point I want to expand about is z=0, because $\partial_z\cos(z)=0\implies z=0,n\pi$  So at $z_0=0$, I get
$$I(\alpha)=1\int_{-\epsilon}^\epsilon e^{-a(1-z^2/2+...)}\approx e^{-a}\int_{-\infty}^\infty e^{az^2/2}\, dz\approx i\frac{\sqrt{2\pi}}{\sqrt{a}}e^{-a}$$
My question is if this is a valid approach.  Mostly, did I correctly choose to expand about z=0.  I get confused on which saddle point to select, because I can deform the integral in many ways.
And then if I want $a<0$, would I approach it the same way?
 A: I) Let us define 
$$ f(z)~:=~1+z^2 \tag{1} $$
for later convenience.
OP's integrand $ f(z) \exp\left(-a\cos z\right)$ is an entire function, so the integral is independent on the integration contour (as long as the endpoints are the same). Let us choose an $L$-shaped contour:
$$I~:=~ \int_{-\frac{i\pi}{2}}^{\frac{\pi}{2}} \!\mathrm{d}z~f(z)\exp\left(-a\cos z\right)~=~J+iK, \tag{2}$$
where
$$J~:=~\int_0^{\frac{\pi}{2}}\!\mathrm{d}x~f(x)\exp\left(-a\cos x\right) \tag{3} $$
$$~=~\int_0^{\frac{\pi}{2}}\!\mathrm{d}x~f\left(\frac{\pi}{2}-x\right)\exp\left(-a\sin x\right) \tag{4} $$
$$~=~e^{-a} \int_0^{\frac{\pi}{2}}\!\mathrm{d}x~f(x)\exp\left(2a\sin^2\frac{x}{2}\right), \tag{5}$$
and where
$$K~:=~\int_{-\frac{\pi}{2}}^0\!\mathrm{d}y~f(iy)\exp\left(-a\cosh y\right)
~=~\int_0^{\frac{\pi}{2}}\!\mathrm{d}y~f(-iy)\exp\left(-a\cosh y\right). \tag{6}$$
$$~=~\int_0^{\frac{\pi}{2}}\!\mathrm{d}y~f\left(iy-\frac{i\pi}{2}\right)\exp\left(-a\cosh \left(\frac{\pi}{2}-y\right)\right). \tag{7}$$
This $L$-shaped contour is aligned with the steepest descents of the endpoints.
II) Case $a\gg 0$:
$$ J~\stackrel{(4)}{\approx}~\int_0^{\infty}\!\mathrm{d}x~f\left(\frac{\pi}{2}-x\right)\exp\left(-|a|x \right)
~=~\left(1+\frac{\pi^2}{4}\right)|a|^{-1}+O(|a|^{-2}) , \tag{8} $$
$$K~\stackrel{(6)}{\approx}~\frac{1}{2}f(0)\int_{-\infty}^{\infty} \!\mathrm{d}y~\exp\left(-|a|\left(1+\frac{y^2}{2}\right)\right)
~=~e^{-|a|}\sqrt{\frac{\pi}{2|a|}}. \tag{9}$$
To leading order, OP's integral (2) reads

$$ I~\sim~\left(1+\frac{\pi^2}{4}\right)|a|^{-1} \qquad\text{for}\qquad a~\to ~\infty.\tag{10} $$

III) Case $a \ll 0$:
$$J~\stackrel{(5)}{\approx}~\frac{e^{|a|}}{2}f(0)\int_{-\infty}^{\infty} \!\mathrm{d}x~\exp\left(-2|a|\left(\frac{x}{2}\right)^2\right)~=~e^{|a|}\sqrt{\frac{\pi}{2|a|}},\tag{11}$$
$$K~\stackrel{(7)}{\approx}~\int_0^{\infty}\!\mathrm{d}y~f\left(iy-\frac{i\pi}{2}\right)\exp\left(|a|\cosh \left(\frac{\pi}{2}\right)-|a|y\sinh \left(\frac{\pi}{2}\right) \right)$$
$$~=~\exp\left(|a|\cosh \left(\frac{\pi}{2}\right)\right)\left(\frac{1-\frac{\pi^2}{4}}{|a|\sinh \left(\frac{\pi}{2}\right)}+O(|a|^{-2})\right).\tag{12}$$
To leading order, OP's integral (2) reads

$$ I~\sim~i\exp\left(|a|\cosh \left(\frac{\pi}{2}\right)\right) \frac{1-\frac{\pi^2}{4}}{|a|\sinh \left(\frac{\pi}{2}\right)} \qquad\text{for}\qquad a~\to ~-\infty.\tag{13} $$

