I'm supposed to prove the Poisson integral formula $$f(z)=\frac{1}{2\pi i}\int_{\partial\mathbb{D}}\text{Re}\left(\frac{w+z}{w-z}\right)\frac{f(w)}{w}dw,$$ where $U$ is an open set such that $\overline{\mathbb{D}}\subseteq U \subseteq \mathbb{C}$, and $f\in H(U)$.
I want to prove this using a Möbius transformation $\mathbb{D} \to \mathbb{D}$, where $\mathbb{D}$ is the unit disk.
My Möbius transformation is the following.$$T_{z_0}(z) = \frac{z_0 - z}{1 - \overline{z_0}z}$$
So far I have been able to compute $$(f\circ T_{z_0})(0)=\frac{1}{2\pi i}\int_{\partial\mathbb{D}}f\left(\frac{z_0-w}{1-\overline{z_0}w}\right)\frac{dw}{w},$$ but now I'm stuck. I don't see how I can prove the formula from here on.
The next part of my question is to prove from this the Schwarz formula: $$f(z)=\frac{1}{2\pi i}\int_{\partial\mathbb{D}}\frac{w+z}{w-z}\frac{\text{Re}f(w)}{w}dw + i\ \text{Im}f(0), \qquad z\in\mathbb{D}.$$They put a hint on the worksheet to use $$g(z)=\frac{1}{2\pi i}\int_{\partial\mathbb{D}}\frac{w+z}{w-z}\frac{\text{Re}f(w)}{w}dw $$ and the Poisson formula to show that $f-g$ is constant on $\mathbb{D}$.
I'm completely stuck with this exercise. Any hints on how I can continue?
Concerning the first part of your question: notice that $T_{z_0}$ interchanges points $0$ and $z_0$ hence $$ f(z_0) = f(T_{z_0}(0)) = \frac{1}{2\pi i}\int_{\partial \mathbb{D}}f(T_{z_0}(w))\frac{dw}{w} $$ by the simple Cauchy formula. Next make change of variables $u = T_{z_0}(w)$, then $$ u = \frac{z_0 - w}{1 - \bar{z_0}w}, \quad w = \frac{z_0 - u}{1 - \bar{z_0}u}, \quad dw = \frac{-1+ |z_0|^2}{(1 - \bar{z_0}u)^2}du \\ f(z_0) = \frac{1}{2\pi i}\int_{\partial \mathbb{D}}f(u)\frac{\frac{-1+ |z_0|^2}{(1 - \bar{z_0}u)^2}du}{\frac{z_0 - u}{1 - \bar{z_0}u}} = \frac{1}{2\pi i}\int_{\partial \mathbb{D}}f(u)\frac{(-1+ |z_0|^2)du}{(1 - \bar{z_0}u)(z_0 - u)} \\ = \frac{1}{2\pi i}\int_{\partial \mathbb{D}}f(u)\frac{(-1+ |z_0|^2)du}{(\bar u - \bar{z_0})(z_0 - u)u} = \frac{1}{2\pi i}\int_{\partial \mathbb{D}}f(u)\frac{(1 - |z_0|^2)du}{|z_0- u|^2u}, $$ which is the required form since $$ \Re\left(\frac{u + z_0}{u - z_0}\right) = \frac{1 - |z_0|^2}{|u - z_0|^2}. $$
UPD: to deal with the second problem, rewrite the complex intagrals in the real form, i.e. write $$ f(z) = \frac{1}{2\pi i}\int_{\partial \mathbb{D}}\Re\left(\frac{w + z}{w - z}\right)\frac{f(u)du}{u} = \frac{1}{2\pi}\int_{0}^{2\pi}f(w)\Re\left(\frac{w + z}{w - z}\right)d\phi,\quad w = e^{i\phi}; \\ g(z) = \frac{1}{2\pi i}\int_{\partial \mathbb{D}}\frac{w + z}{w - z}\Re\left(f(u)\right)\frac{du}{u} = \frac{1}{2\pi}\int_{0}^{2\pi}\Re(f(w))\frac{w + z}{w - z}d\phi,\quad w = e^{i\phi}. $$ From this we see that $$ \Re (f(z)) = \Re(g(z)) = \frac{1}{2\pi}\int_{0}^{2\pi}\Re(f(w))\Re\left(\frac{w + z}{w - z}\right)d\phi. $$ Hence $f - g$ is purely imaginary analytic function. The only such function is constant. To calculate this constant put $z = 0$. It gives $g(0) =\Re(f(0))$, therefore $f(z) - g(z) = i\Im(f(0))$.