= Solution
Write $A=Q/(2\pi\epsilon_0a^2)$. The <Taylor expansion> $\sqrt{a^2+z^2}=a+z^2/(2a)+O(z^4/a^3)$ gives the near-axis result
$$
\boxed{\psi(0,z)=A\left[a-|z|+\frac{z^2}{2a}+O(z^4/a^3)\right].}
$$
To extend off the axis, work separately above and below the charged plane. In either source-free half-space the regular axisymmetric <solid harmonics> are $r^nP_n(\cos\theta)$. Matching the positive-axis coefficients, using $P_n(1)=1$, and then reflecting in the disk gives
$$
\boxed{\psi(r,\theta)=A\left[a-r|\cos\theta|+\frac{r^2}{2a}P_2(\cos\theta)+O(r^4/a^3)\right],\qquad r\ll a.}
$$
Here $P_2(t)=(3t^2-1)/2$ is a <Legendre polynomial>. The quadratic term is $A(z^2-\rho^2/2)/(2a)$. The linear term is $-Az$ above and $+Az$ below; it represents the discontinuity of the normal <electric field>. A single regular spherical expansion across the entire charged sheet would incorrectly remove that jump.
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