This is a fragment of my complex analysis book, when he explains how to solve the Dirichlet problem
Theorem. Let $U(\phi)$ continuous for $0\leq \phi \leq 2\pi$. Then the function $$u(z)=\frac{1}{2\pi}\int_{0}^{2\pi} U(\phi) \frac{R^2 - |z|^2}{|Re^{i\phi}-z|^2}d\phi $$ is harmonic in $|z|<R$ and $$\lim_{z \rightarrow Re^{i\phi}}u(z)=U(\phi).$$ Proog. [...] given $\epsilon >0$, there is a $\delta >0$ such that $|U(\phi)-U(\alpha)|<\epsilon$ when $|\phi - \alpha| < \delta$. Then by the Poisson kernel property and the fact that $U$ is $2\pi$-periodic $$|u(z)-U(\alpha)|= \left| \frac{1}{2\pi} \int_{0}^{2\pi} (U(\phi) - U(\alpha)) \frac{R^2 - |z|^2}{|Re^{i\phi}-z|^2}d\phi \right| \leq \frac{1}{2\pi} \int_{0}^{2\pi} \left| (U(\phi) - U(\alpha)) \right| \frac{R^2 - |z|^2}{|Re^{i\phi}-z|^2}d\phi \leq \epsilon + \frac{1}{2\pi} \int_{\delta \leq|\phi - \alpha|\leq\pi} \left| U(\phi) - U(\alpha) \right| \frac{R^2 - |z|^2}{|Re^{i\phi}-z|^2}d\phi.$$ Now, if $|\arg(z)-\alpha|<\delta /2$ and $|\phi - \alpha|\geq \delta$, then $$|Re^{i\phi}-z|\geq R\sin(\delta/2)$$
I can't understand where it comes the bound $$|Re^{i\phi}-z|\geq R\sin(\delta/2)$$
Thank you
We consider the case $\alpha+\delta \le \phi \le \alpha +\pi/2$ for simplicity (we can do similarly even in other cases). See the diagram below. In the diagram $$ \left|Re^{i\phi }-z\right|=\text{the length of the black-colored segment},$$ which is greater than the length of the red-colored segment since $|\arg(z)-\alpha|<\delta /2.$ The length of the red-colored segment is equal to $$ R\sin \left(\phi -\alpha -\frac{\delta}{2}\right).$$ Since $\phi -\alpha -\frac{\delta}{2}\ge (\alpha +\delta) -\alpha -\frac{\delta}{2}=\frac{\delta}{2} $, we see $$ \sin \left(\phi -\alpha -\frac{\delta}{2}\right)\ge \sin \left(\frac{\delta}{2}\right) .$$ Therefore we have $$ \left|Re^{i\phi }-z\right|\ge R\sin \left(\phi -\alpha -\frac{\delta}{2}\right)\ge R\sin \left(\frac{\delta}{2}\right) .$$ Of course $R\sin \left(\frac{\delta}{2}\right)$ is the length of the blue-colored segment.