Changes

Then <math>A(r')=\frac{-z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')</math>

Then <math>A(r')=\frac{-z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')</math>

<math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}=\sqrt{r^2+r'^2+2r\rho^2cos\phi}</math><br><br>

<math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}=\sqrt{r^2+r'^2+2r\rho^2cos\phi}</math><br><br>

<math>=r'-\frac{2r\rho'cos\phi}{2r'}, \frac{\rho'}{r'}=sin\theta'</math>

<math>=r'-\frac{2r\rho'cos\phi}{2r'}, \frac{\rho'}{r'}=sin\theta'</math><br><br>

<math>\frac{1}{|\mathbf{r}-\mathbf{r}'|} \approx \frac{1}{r'^2}\left(\frac{1+2rsin\theta'cos\phi}{r'}\right)</math>