# Evanescent field, condition of $z$ component of $k$?

I am studying evanescent field and diffraction limit and I have one question.

Given a field $$U(x,y,0)$$ we can decompose into 2D plane waves.

$$U(x,y,0)= \int \int dk_x dk_y \tilde{U}(k_x,k_y) e^{+i(k_x x + k_yy)}$$

if we want to have $$U(x,y,z_0)$$ we can propagate each wave to the plane $$z_0$$ in this way.

$$U(x,y,z_0)= \int \int dk_x dk_y \tilde{U}(k_x,k_y) e^{+i(k_x x + k_yy)} e^{i k_z z_0}$$

This is clear to me.

It is not clear why $$k_z = \sqrt{ (\frac{2\pi}{\lambda})^2 - k_x^2 -k_y^2 }$$.

Why suddenly the plane waves should be monochromatic with wavelength $$\lambda$$ ?

The wave $$u(x,y,z,t)$$ satisfies the Helmholtz wave equation $$\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}-\frac{1}{c^2}\frac{\partial^2u}{\partial t^2}=0 \tag{1}\label {1}.$$ This equation is solved by plane waves of the form $$U(x,y,z,t) = A e^{\mathfrak j (\omega t - (k_x x+k_yy+k_zz))}\tag{2}\label {2}.$$ The dispresion relation between $$\omega$$ and $$\mathbf k(k_x,k_y,k_z)$$ can be had by substituting $$\eqref{2}$$ in to $$\eqref{1}$$ and noting that $$\frac{d^2 e^{\mathfrak jas}}{ds^2} = -a^2e^{\mathfrak jas}$$ : $$-\frac{\omega^2}{c^2}+ k_x^2+k_y^2+k_z^2=0$$ or
$$\left(\frac{2\pi}{\lambda}\right)^2 = k_x^2+k_y^2+k_z^2 \tag{3}\label {3}.$$ The values of $$\omega$$ or $$\mathbf k$$ are not restricted to be real numbers, they just have to satisfy the dipersion relation $$\eqref{3}$$.