In Complex Analysis by Elias M. Stein, I'm having trouble proving $\text{Theorem (1)}$. Stein gives a key clue to a proof attempted on my part which can be found in $\text{Lemma (1)}$
$\text{Theorem 1} \, \, \text{(Poisson's Formula)}$
Assume that $u$ is defined and continuous on the closed disk $z$ such that $|z| \leq r$ and is harmonic on the open disk $D(0;r)=z$ such that $|z| < r$ Then for $p < r$, we have the real form of Poisson's Formula
$$u(pe^{i \psi}) = \frac{r^{2}-p^{2}}{2 \pi }\int_{0}^{2 \pi} \frac{u(re^{i \psi})}{r^{2}-2rp \cos(\psi - \theta)} + p^{2}$$
which is equivalent to the complex form of Poisson's Formula
$$u(z) = \frac{1}{2 \pi}\int_{0}^{2 \pi} u(re^{i \psi}) \frac{r^{2}-|z|}{|re^{i \theta}-z|^{2}}d \theta.$$
Stein indicates the name of the game is to aim for an integral representation in $(1.1)$, $\text{Lemma (1)}$ is an attempt to achieve this goal
$(1.1)$
$$f(z) = \frac{1}{2\pi i} \oint_{D_{R_0}} f(\zeta)/(\zeta-z) - f(\zeta)/(\zeta-R^2/\bar{z}) \; d\zeta .$$
$\text{Lemma (1)}$
For the sake of discussion we introduce the Poisson Kernel in a new form.Let $f$ be holomorphic on the closed disc $\bar{D}_R$. Let $z\in D_R$. Then in $(1.3)$
$(1.3)$
$$f(z) = \frac{1}{2 \pi}\int_{0}^{2 \pi}f(Re^{i \phi})Re\left(\frac{re^{i \phi}+z}{re^{i \phi} - z}\right)d \phi.$$
One can observe in $(1.3)$ can be broken into real and complex parts as follows in $(1.4)$
$(1.4)$
$$u(v, \theta) + iv(r, \theta)=\frac{1}{2 \pi} \int_{0}^{2 \pi}f(re^{i \theta}) Re\left(\frac{re^{i \theta} - (R^{2}/ z) + re^{i \theta}}{(re^{i \theta}-re^{i \theta})(re^{i \theta} - (R^{2}/ r)re^{i \theta}}\right)\\= \frac{1}{2 \pi}\int_{0}^{2 \pi}f(re^{i \phi})Re\left(\frac{re^{i \phi}(v(v, \theta))+z}{re^{i \phi}(v(v, \theta)) - z}\right)d \phi \\+ \frac{i}{2 \pi}\int_{0}^{2 \pi}f(re^{i \phi})Re\left(\frac{re^{i \phi}(u(u, \theta))+z}{re^{i \phi}(u(u, \theta)) - z}\right)d \phi.$$