= Solution
Choose the positive <Green function> convention
$$
\boxed{-\Delta_xG(x,\xi)=\delta_\xi(x),\quad G=0\text{ on }\partial D,\quad
G(x,\xi)\sim-\frac1{2\pi}\log|x-\xi|\text{ near }\xi.}
$$
Away from its pole, $G$ is harmonic, and the <maximum principle for harmonic functions> makes it positive inside $D$. The sign of its <logarithmic singularity> is important: it is the convention for $-\Delta$, whereas a <Green function> for $+\Delta$ has the opposite sign.
<Green's second identity>, justified by removing a small disk about the pole and taking its radius to zero, gives
$$
\int_D(u\Delta G-G\Delta u)\,dx
=\int_{\partial D}(u\partial_nG-G\partial_nu)\,ds.
$$
Since $\Delta u=-f$, $\Delta G=-\delta_\xi$ and $G$ vanishes on the outer boundary, this rearranges to
$$
\boxed{u(\xi)=\int_DG(x,\xi)f(x)\,dx-\int_{\partial D}g(x)\partial_nG(x,\xi)\,ds.}
$$
The local <logarithmic singularity> is integrable, so continuous $f$ gives a well-defined volume <integral>.
In <complex coordinates> $z=x+iy$, $\zeta=\xi+i\eta$ for the <unit disk>, use
$$
G(z,\zeta)=\frac1{2\pi}\log\left|\frac{1-z\bar\zeta}{z-\zeta}\right|.
$$
The numerator has no zeros inside the disk, so its logarithm is harmonic there. On $|z|=1$, the numerator and denominator have equal modulus, giving the required zero boundary value. Differentiating in the outward radial direction at $z=e^{i\theta}$ gives
$$
\boxed{-\partial_nG(e^{i\theta},\zeta)=\frac{1-|\zeta|^2}{2\pi|e^{i\theta}-\zeta|^2}.}
$$
With zero forcing the representation therefore becomes the <Poisson integral>
$$
\boxed{u(\xi,\eta)=\frac{1-\xi^2-\eta^2}{2\pi}\int_0^{2\pi}
\frac{g(\theta)\,d\theta}{(\xi-\cos\theta)^2+(\eta-\sin\theta)^2}.}
$$
Finally replace $\eta$ by $-\eta$ and change variable $\theta\mapsto2\pi-\theta$. Cosine stays the same, sine changes sign and the assumed boundary symmetry leaves $g$ unchanged. The <integral> is therefore invariant, so $\boxed{u(\xi,-\eta)=u(\xi,\eta)}$. Equivalently, <reflection> produces a <harmonic function> with the same Dirichlet data, and uniqueness identifies it with the original solution.
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