= Solution
The <method of images> places an <image charge> $-q$ at the reflected position below the plane. The candidate potential in the physical upper half-space is
$$
\Phi=\frac q{4\pi\varepsilon_0}\left[\frac1{\sqrt{x^2+y^2+(z-d)^2}}-\frac1{\sqrt{x^2+y^2+(z+d)^2}}\right].
$$
At $z=0$ the two terms cancel, so $\Phi=0$ and $E_x=E_y=0$. Direct differentiation gives
$$
E_z(0^+)=-\frac{qd}{2\pi\varepsilon_0(x^2+y^2+d^2)^{3/2}},\qquad
\sigma=-\frac{qd}{2\pi(x^2+y^2+d^2)^{3/2}}.
$$
Thus the upper-side field satisfies exactly the conductor boundary conditions from part (i). The fictitious pair does have a field below the plane, but that part of its solution is not used for the physical conductor.
In $z>0$, the potential has the correct real point source, the correct zero <Dirichlet boundary condition>, and the required decay at infinity. By <Uniqueness of the Dirichlet problem>, it is the physical potential there. This is why the artificial charge can represent the induced conductor response without being a second physical charge.
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