Justification that steepest descent paths are paths with constant imaginary part.

If $f(u,v)$ is a differentiable function, $\nabla f(u,v)$ points towards the direction of most rapid change of $f$. In particular, $\nabla f$ is perpendicular to the lines of $f$ constant(level curves). If $\rho(z)=\phi(z)+i\psi(z)$, the Cauchy-Riemann equation implies:

$$i\frac{\partial\phi(z)}{\partial u}=\frac{\partial\psi(z)}{\partial v} \space ,$$ (2)

which is equivalent to

$$\frac{\partial\phi}{\partial u}=\frac{\partial\psi}{\partial v} \... ...{2mm} \frac{\partial\phi}{\partial v}=-\frac{\partial\psi}{\partial u} \space .$$ (3)

Hence,

$$\frac{\partial\phi}{\partial u}\frac{\partial\psi}{\partial u}+\frac{\partial\phi}{\partial v}\frac{\partial\psi}{\partial v}=0$$ (4)

that in vector notation becomes

$$\nabla\phi \cdot \nabla\psi=0$$ (5)

At any point $(u,v)$, the steepest path is the one of $\nabla e^{K\phi(z)}$ which is parallel to $\nabla \phi$; the constant phase contour are perpendicular to $\nabla \psi$ and because of (5) the conclusion is that constant-phase lines are also steepest lines.

A point at which $\rho'(z)=0$ is called saddle point. In order to understand what happens near a saddle point $z_{s}$, let us write $\rho(z)$ as:

$$\rho(z)=\rho(z_{s})+\frac{(z-z_{s})^2}{2}\rho''(z_{s})$$ (6)

Along a steepest path, the phase of (6) is constant and its real part must be such that

$$\frac{(z-z_{s})^{2}}{2}\rho''(z_{s})$$ (7)

be negative. Let

$$z-z_{s}=se^{j\gamma}$$ (8)

we notice that as the angle of $z-z_{s}$ increases, the angle of $\rho(z)-\rho(z_{s})$ is twice that for $z-z_{s}$. The valley region, where $\phi(z)<\phi(z_{s})$, is represented by two regions in the $z$ plane about the saddle point. The mountain region $\phi(z)>\phi(z_{s})$ is represented by two regions in the $z$ plane, and these four regions meet at the saddle point, each occupying the angle of $\pi/2$, as shown in Fig. 1.

Figure:

The approximation based on the method of steepest descent is derived as follows. From(7) and (8),

$$\frac{(z-z_{s})^{2}}{2}\rho''(z_{s})=-\frac{\left[-e^{+j2\gamma}\rho''(z_{s})\right]s^{2}}{2}$$ (9)

and we choose $\gamma$ so that:

$$-e^{+j2\gamma}\rho''(z_{s})=T>0$$ (10)

Then the integral becomes:

$$I(K)=h(z_{s})e^{K\rho(z_{s})}\int_{SDC}^{}e^{-\dfrac{KTs^{2}}{2}+j\gamma}ds$$ (11)

where the most contribution comes from a neighborhood around $s=0$, hence:

$$I(K)=h(z_{s})e^{K\rho(z_{s})+j\gamma}\int_{-\varepsilon}^{\varepsilon}e^{-\dfrac{KTs^{2}}{2}}ds$$ (12)

The change of variable $\frac{KT}{2}s^{2}=t^{2}$ yields:

$$I(K)=h(z_{s})e^{K\rho(z_{s})+j\gamma}\sqrt{\frac{2}{KT}}\int_{-\sqrt{\frac{KT}{2}}\varepsilon}^{\sqrt{\frac{KT}{2}}\varepsilon}e^{-t^{2}}dt$$ (13)

Extending the limits of integration to $-\infty$ and $\infty$, using

$$\int_{-\infty}^{\infty}e^{-t^{2}}dt=\sqrt{\pi}$$ (14)

the result is:

$$I(K)\sim h(z_{s})e^{K\rho(z_{s})+j\gamma}\sqrt{\frac{2\pi}{K\vert\rho''(z_{s})\vert}}$$ (15)

which is the approximation of the method of the saddle point.

The method of the stationary phase is equivalent to the method of the steepest descent. The former goes through the saddle point along $Re\left[\rho(z)\right]=\phi(z)=const$, whereas the latter crosses it along $Im\left[\rho(z)\right]=\phi(z)=const$.