Fix the observation point and apply Green's second identity in the source variable . Since , and on the boundary, it gives
The identity can be justified by deleting a small circle about and letting its radius tend to zero; the logarithmic flux contributes .
Set and , and write . The symmetric expression for the Green's function is
At the boundary, the outward normal derivative is at , and hence
Since on the unit circle, the Poisson integral on the unit disk follows, with Poisson kernel

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