Method of steepest descent

Remarks

Let $z_0$ be a simple saddle with $g(z_0) \ne 0$, and let $\theta = \frac12 \big(\pi - \arg f''(z_0)\big)$. The angle $\theta$, defined modulo $\pi$, is the direction in which the path crosses $z_0$, taken with the orientation of the contour. The contribution of $z_0$ is $$g(z_0)\, e^{\lambda f(z_0)}\, e^{i\theta} \sqrt{\frac{2\pi}{\lambda \abs{f''(z_0)}}}\,\big(1 + o(1)\big).$$ Indeed $f(z_0 + r e^{i\theta}) \approx f(z_0) + \frac12 f''(z_0) r^2 e^{2i\theta}$, and $f''(z_0) e^{2i\theta}$ is real and negative.

Example: $I(\lambda) = \int_{\R} e^{i\lambda x^2}\,dx$, with $f(z) = i z^2$ and $g = 1$. The only saddle is $z_0 = 0$ with $f''(0) = 2i$, so $\theta = \pi/4$. On the line $z = r e^{i\pi/4}$ one has $f(z) = -r^2$, which is real and decreasing in $\abs{r}$, so this line is the steepest descent path. Rotating the contour to it gives exactly $I(\lambda) = e^{i\pi/4} \sqrt{\pi/\lambda}$.

Non-example: for $f(z) = z^2$ the real axis passes through the saddle $0$, but $\operatorname{Re} f = x^2$ increases away from $0$ along it. So it is a path of steepest ascent; the steepest descent path is the imaginary axis.