I am not sure if all the calculations are correct.Could you check for me please ?
$$\frac{1}{r}\frac{d}{dr}\left(r\frac{dg}{dr}\right)+\frac{d^2g}{dz^2}+\left(m^2-\frac{n^2}{r^2}\right)g=\frac{\delta(r-r')\delta(z-z')}{r} \tag{1a}$$
where $0<r,r'<r_0$, and $0<z,z'<z_0$
Taking the fourier transform of (1a) with respect to z
$$\frac{1}{r}\frac{d}{dr}\left(r\frac{dG}{dr}\right)+\left(K^2-\frac{n^2}{r^2}\right)G=\frac{\delta(r-r')e^{-ikz'}}{r} \tag{1b}$$
where $K^2=m^2 - k^2$
Homogeneous solution of $(1b)$ is
$$G(r,k\mid r',z')= \begin{cases} AJ_n(Kr) & \text{for }0 \le r \le r'\\ BH^{(1)}_n(Kr) & \text{for }r' \le r \le r_0 \end{cases}$$
Due to the solution is continuous at $r=r'$
$$G(r'^-,k\mid r',z')=G(r'^+,k\mid r',z')$$
$$\left.\frac{dG(r,k\mid r',z')}{dr}\right|_{r=r'^-}^{r=r'^+}=-\frac{e^{-ikz'}}{r'}$$
The coefficient A and B are determined by
$$\begin{bmatrix}J_n(Kr') & H^{(1)}_{n}(Kr')\\-KJ'_n(Kr') & -KH'^{(1)}_{n}(Kr')\end{bmatrix}\begin{bmatrix}A \\B \end{bmatrix}=\begin{bmatrix}0 \\\frac{e^{-ikz'}}{r'} \end{bmatrix}$$
Wronskian determinant
$$W[J_n(Kr'),H^{(1)}_{n}(Kr')]=\frac{2i}{\pi Kr'}$$
$$A=\frac{\pi i}{2}H^{(1)}_{n}(Kr') \text{ and }B=\frac{\pi i}{2}J_n(Kr') $$
$$G(r,k\mid r',z')=\frac{\pi i}{2}e^{-ikz'}J_n(Kr<) H^{(1)}_{n}(Kr>)$$
$$g(r,z\mid r',z')=\frac{i}{4}\int_{-\infty}^{\infty} J_n(Kr<) H^{(1)}_{n}(Kr>)e^{ik(z-z')}dk$$