The velocity potential formed by adding source and sink is:
$$\phi = \frac{m}{r} - \frac{m}{(r^2 - 2z\delta + \delta^2)^{1/2}}$$
Expanding the potential to first order in $\delta$ gives:
$$\phi = \frac{m}{r} - \frac{m}{r}\left(1 + \frac{z \delta}{r^2} + O(\delta^2)\right)$$
Would someone tell me how this is achieved, please? I have tried several times and I am not near the answer.
The main idea is to use the binomial expansion, at least as far as $$ (1+x)^a = 1 + ax + O(x^2). $$ To do so, we pull the $r^2$ out of the square root to get the $1$: $$ \frac{m}{\sqrt{r^2-2z\delta+\delta^2}} = \frac{m}{r} \left( 1- \delta\frac{2z+\delta}{r^2} \right)^{-1/2} $$ Now, $x=\delta\frac{2z+\delta}{r^2}$ is more complicated that before, but it is $O(\delta)$, so its square will be $O(\delta^2)$ and so on. Hence we can write $$ \frac{m}{r} \left( 1- \delta\frac{2z+\delta}{r^2} \right)^{-1/2} = \frac{m}{r} \left( 1 - \left(-\frac{1}{2}\right)\delta\frac{2z+\delta}{r^2} + O(\delta^2) \right) = \frac{m}{r} \left( 1 + \frac{z\delta}{r^2} + O(\delta^2) \right). $$