Consider the following complex integral
$$ \int_{\mathcal P} e^{v\cdot w(z)} \xi(z) dz, $$
where $v$ is large and positive and the integration path $\mathcal P$ satisfies the following two conditions:
- $\mathcal P$ passes through a zero $z_0=x_0+iy_0$ of the derivative $w'(z_0)$ of $w$;
- the imaginary part of $w$, $\Im w(z)$ is constant.
If we write
$$ z = x+iy, \; w(z)=u(x, y)+iv(x,y); z_0=x_0+iy_0; w'(z_0)=0. $$
Then it is claimed that by the Cauchy-Riemann equations:
$$ u_x = v_y, \; u_y=-v_x. $$
It follows that $(x_0, y_0)$ cannot be a maximum or a minimum but must be a saddle point on the surface $S: (x,y)\mapsto u(x,y)$. Why is this so, please? Thank you!
In order to find whether something is a saddle point, you need to look at the second derivative of the function. There is a test you can use that has 4 possible cases.
$$ 1. \; u_{xx} < 0 \quad u_{xx} u_{yy}- u^2_{xy} > 0 $$ $$ 2. \; u_{xx} > 0 \quad u_{xx} u_{yy} - u^2_{xy} > 0 $$ $$ 3. \; u_{xx} u_{yy} - u^2_{xy} < 0 $$ $$ 4. \; u_{xx} u_{yy} - u^2_{xy} = 0 $$
In case 1 $z_0$ is a max, in case 2 it's a min, in case 3 it's a saddle point, and in case 4 we can't determine what type of maximum it is. Because $z_0$ is analytic we can then use the Cauchy-Riemann conditions to relate $u_{xx}$ and $u_{yy}$ and find $$ u_{xx}+u_{yy} = 0.$$ This proves that $u_{xx} u_{yy}- u^2_{xy} $ is always negative and therefore cases 1 and 2 are impossible if the Cauchy-Riemann conditions are satisfied. The reasoning goes like this. Since $u_{xx}$ and $u_{yy}$ have different signs, their product will always be less than or equal to zero. It's also obvious that $-u^2_{xy}$ is less than or equal to zero. Therefore, the whole term must be less than or equal to zero.