I have a question in method of steepest descent
Method as written from where I am studying -> If w is a complex function such that w'($z_0$) =0 and w"($z_0$) =|w"($z_0$)| $ e^{i\alpha_0}$$\neq$ 0.
Denote $\theta$ by the direction of line $\Delta$ passing through $z_0$ that is arg(z-$z_0$) =$\theta$ where z belongs to $\Delta$ .
I am looking for a proof of next line -> Then, there are exactly two paths of greatest descent from Re(w) in $z_0$ of which the tangents of the directions $z_0$ are $\theta_+$= π/2 - $\alpha_0$/2 and $\theta_-$= -π/2 - $\alpha_0$ /2.
How the two paths of greatest descent have exactly these two $\theta$'s .
Can someone please explain!!
Ref. 1 is considering
$$\int_L g(z)e^{nw(z)} \mathrm{d}z~=~\ldots, \qquad n~>~0, \tag{4} $$
which is a standard situation for the method of steepest descent. Let $z_0$ be a stationary point $w^{\prime}(z_0)=0$. Knowing the behaviour of the exponential function, the direction of steepest descent in the complex $w$-plane has argument that belongs to $\pi +2\pi\mathbb{Z}$. The Taylor expansion reads $$ w(z)~=~w(z_0)+ \frac{1}{2}w^{\prime\prime}(z_0)(z-z_0)^2 + O((z-z_0)^3). $$ The quadratic Taylor term $\frac{1}{2}w^{\prime\prime}(z_0)(z-z_0)^2$ in the exponent has by definition argument $\alpha_0+2\theta.$ If this argument belongs to $\pi +2\pi\mathbb{Z}$, then the corresponding angle $$\theta~\in~ \frac{\pi}{2}-\frac{\alpha_0}{2}+\pi\mathbb{Z}$$ is the direction of steepest descent in the complex $z$-plane, cf. the quote from Ref. 1.
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