= Conformal transformation rule for the Poisson kernel
For smooth planar domains and a <conformal map> $\Phi:D\to\Omega$ extending as a $C^1$ boundary diffeomorphism with nonzero derivative, the <Poisson kernel> as a density against boundary arclength satisfies
$$
P_D(z,\zeta)=P_\Omega(\Phi(z),\Phi(\zeta))\,|\Phi'(\zeta)|.
$$
The derivative is evaluated at the boundary point, not at the interior observation point.
To derive it, use conformal invariance of <harmonic measure> and change boundary variables: $ds_{\Phi(\zeta)}=|\Phi'(\zeta)|ds_\zeta$. The two Poisson integrals must give the same <harmonic extension> for every continuous boundary datum, identifying the densities. Kernels defined against an angular measure instead require the corresponding measure conversion.
For $\Phi(z)=Rz$ from the unit disk to the radius-$R$ disk, the rule gives $P_{R\mathbb D}(Rz,R\zeta)=P_{\mathbb D}(z,\zeta)/R$. At the center this is $1/(2\pi R)$, whose integral around circumference $2\pi R$ is one. This normalization check detects a misplaced derivative factor.
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