I am working on an exercise in complex analysis and do not know how to proceed. So far, I have proven that if $f$ is holomorphic in $D_{R_0} =\big\{z\in \mathbb{C}\;|\; |z|<R_0\big\}$, then whenever $0<R<R_0$ and $|z|<R$, $$f(z) = \frac{1}{2\pi }\int_0^{2\pi}f(Re^{i\theta})Re\Big(\frac{Re^{i\theta}+z}{Re^{i\theta}-z}\Big)d\theta$$ Note here that $Re\Big(\frac{Re^{i\theta}+z}{Re^{i\theta}-z}\Big)$ is the real part of $\Big(\frac{Re^{i\theta}+z}{Re^{i\theta}-z}\Big)$
Here is where I am stuck: Suppose that $f = u+iv$ and that $f(0)$ is real. Prove that $$f(z) = \frac{1}{2\pi }\int_0^{2\pi}u(Re^{i\theta})\Big(\frac{Re^{i\theta}+z}{Re^{i\theta}-z}\Big)d\theta$$ I have tried to use the fact that $$\int_0^{2\pi}v(Re^{i\theta})d\theta = 0$$ as a consequence of $f(0)$ being real and my above finding, but I do not know how to proceed.
I'm not sure how helpful this is, but this is very similar to Exercise 11 in Chapter 2 from Stein's Complex Analysis.
Let $f = u+iv$ be a holomorphic function in $D_{R_0}(0)$ such that $f(0)$ is real. This implies $v(0) = 0$. From our previous finding, we know that $$u(z) = \frac{1}{2\pi}\int_{0}^{2\pi}u(Re^{i\theta})Re\Big(\frac{Re^{i\theta}+z}{Re^{i\theta }-z}\Big)d\theta$$ Notice that $$\tilde{f}(z):=\frac{1}{2\pi}\int_0^{2\pi}u(Re^{i\theta })\Big(\frac{Re^{i\theta}+z}{Re^{i\theta}-z}\Big)d\theta$$ defines a holomorphic function on $D_R(0)$ since the integrand is holomorphic in $z$. Thus, $Re(\tilde{f}) = Re(f) = u$. This implies that $f-\tilde{f}$ is a purely imaginary holomorphic function. It follows from the Cauchy Riemann equations that $f-\tilde{f}$ is constant. Now, by the mean value property, we know that $$\tilde{f}(0) = \frac{1}{2\pi}\int_0^{2\pi}u(e^{i\theta})\Big(\frac{Re^{i\theta}+0}{Re^{i\theta}-0}\Big) d\theta = \frac{1}{2\pi}\int_0^{2\pi}u(Re^{i\theta})d\theta =u(0)$$ $$\Rightarrow f(0) - \tilde{f}(0) =iv(0) =0$$ Hence, we conclude that $$f(z) = \tilde{f}(z) = \frac{1}{2\pi}\int_0^{2\pi}u(Re^{i\theta})\Big(\frac{Re^{i\theta}+z}{Re^{i\theta}-z}\Big)d\theta$$