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For $0<|z|<R$, set $w=R^2/\bar{z}$, which lies outside the closed disk. On $|\zeta |=R$, direct algebra gives the Poisson kernel identityUsing $d\theta =d\zeta /(i\zeta )$, the proposed Poisson formula for a holomorphic function becomesThe first term is $f(z)$ by the Cauchy integral formula; the second vanishes by the Cauchy integral theorem, since its pole lies outside the disk. At $z=0$ the kernel is one, and the same identity is the mean value property for holomorphic functions. Therefore

For $f$ holomorphic on a neighbourhood of the radius-$R$ closed disk and $|z|<R$, the displayed representation holds even though $f$ may be complex-valued. For $z\ne 0$, put $w=R^2/\bar{z}$. On $|\zeta |=R$, the Poisson kernel equals $\zeta /(\zeta -z)-\zeta /(\zeta -w)$. Replacing $d\theta$ by $d\zeta /(i\zeta )$ reduces the formula to the Cauchy integral formula minus a zero contour integral whose pole lies outside. The case $z=0$ is the mean value property for holomorphic functions.