# Cauchy's Integral formula real part confusion

Suppose that $$f$$ is holomorphic in $$D_{R_0}(0) = \{|z| < R_0\}$$. Show that whenever $$0 < R < R_0$$ and $$|z| < R$$, then $$f(z) = \frac{1}{2\pi}\int_0^{2\pi} f(Re^{i\theta})\text{Re}(\frac{Re^{i\theta}+z}{Re^{i\theta}-z})d\theta$$.

It seems as if they want you to use Cauchy's Integral formula and I set $$\zeta = Re^{i\theta}$$, and so $$f(z) = \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta - z}d\zeta =\frac{1}{2\pi}\int_0^{2\pi} \frac{f(Re^{i\theta})}{Re^{i\theta} - z}(Re^{i\theta})d\theta$$. I'm not quite sure if this is where I should go with it. I've also tried using analytic-ness for $$f$$ and replacing $$\zeta = Re^{i\theta}-z$$ but I don't know how to get that real part it says in the answer. Any hints appreciated, thanks.

• It is much easier if you use Poisson kernels and the fact that harmonic function in 2 variables are locally real parts of holomorphic functions. Sep 24, 2018 at 21:25

Consider the linear fractal transform $$T(\xi) = \frac{R(R\xi+z)}{\overline{z}\xi + R}$$ . Because $$T$$ has its singularity at $$\xi = -\frac{R}{\overline{z}}$$ the function $$f \circ T$$ is holomorphic in a $$B_r(0)$$ where $$r$$ is choosen slightly larger then $$1$$. Further we have
$$T^ {-1}(\omega) = \frac{R\omega - Rz}{-\overline{z}\omega+R^2}$$ and $$(T^{-1})'(\omega) = R \frac{R^2-|z|^2}{(R^2-\overline{z}\omega)^2}$$ and the substitution formula $$\int_{T^{-1} \circ (T \circ \partial B_1(0))} {\frac{(f \circ T)(\xi)}{\xi} d\xi = \int_{T \circ \partial B_1(0)} {\frac{f(x)}{T^{-1}(x)}\cdot {(T^{-1})'(x)}dx} } = \int_{T \circ \partial B_1(0)} {\frac{f(x)\cdot(R^ 2-|z|^2)}{(x-z)\cdot(R^2 - \overline{z}x)}dx}$$ which gives in Combiation with Cauchy's Integral formula $$f(z) = f(T(0)) = \frac{1}{2 \pi i} \int_{\partial B_1(0)} { \frac{f(T(\xi))}{\xi} d\xi } = \frac{1}{2 \pi i} \int_{T \circ \partial B_1(0)} {\frac{f(x)\cdot(R^ 2-|z|^2)}{(x-z)\cdot(R^2 - \overline{z}x)}dx}$$
Now we calculate $$T \circ \partial B_1(0)$$. A direct computation shows that $$|T(e^{i\theta})|=R$$. We conclude this linear fractional transformation carries $$\partial B_1(0)$$ into $$\partial B_R(0)$$. Further we have
$$\text{Ind}_{T \circ \partial B_1(0)}(0) = \frac{1}{2 \pi i} \int_{\partial B_1(0)} {\frac{T'(x)}{T(x)}}dx = 1$$
because the integral is equal to the number of zeros of $$T$$ in $$B_1(0)$$ which can be calculated directly. This show that $$T$$ carries $$B_1(0)$$ onto $$B_R(0)$$. Otherwise there would be a "gab" on $$\partial B_R(0)$$ which connects $$0$$ to the unbounded component. Now it is easy to conclude that $$T \circ \partial B_1(0)$$ and $$\partial B_R(0)$$ are homotopic. So we can calculate
$$f(z) = \frac{1}{2\pi i} \int_{0}^{2 \pi} {f(Re^{i\theta}) \frac{R^2-|z|^2}{(Re^{i\theta}-z)\cdot(R^2 - \overline{z} Re^{i \theta})} Rie^{i\theta} d \theta} = \frac{1}{2\pi} \int_{0}^{2 \pi} {f(Re^{i\theta}) \frac{R^2-|z|^2}{(Re^{i\theta}-z)\cdot(Re^{-i\theta} - \overline{z} e^{i \theta - i \theta})} d \theta} = \frac{1}{2\pi} \int_{0}^{2 \pi} {f(Re^{i\theta}) \frac{R^2-|z|^2}{(Re^{i\theta}-z)\cdot(Re^{-i\theta} - \overline{z})} d \theta} = \frac{1}{2\pi} \int_{0}^{2 \pi} {f(Re^{i\theta}) \frac{R^2-|z|^2}{|Re^ {i\theta}-z|^ 2} d \theta} = \frac{1}{2\pi} \int_{0}^{2 \pi} {f(Re^{i\theta}) \Re{\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right)} d \theta}$$ because of the following equalities $$\Re{\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right)} = \frac{1}{2 i}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z} + \frac{\overline{Re^{i\theta} + z}}{\overline{Re^{i\theta} - z}}\right) = \frac{1}{2 i}\left( \frac{ \left(Re^{i\theta}+z \right) \cdot \overline{\left( Re^{i\theta} - z\right)} + \overline{\left(Re^{i\theta} + z\right)} \cdot \left( Re^{i\theta} - z\right) }{|Re^{i\theta}-z|^2} \right) = \frac{1}{2 i}\left( \frac{2 |Re^{i\theta}| + 2 |z|^ 2}{|Re^ {i\theta} - z|^2} \right) = \frac{R-|z|^2}{|Re^{i\theta}-z|^2}$$