 Solution

Differentiate the argument through the nonvanishing curve, using $g^{\prime }=\mathrm{Im}({\beta }^{\prime }/\beta )$. Since ${\beta }^{\prime }=-i{e}^{i\theta }$,Also $|w-e^{i\theta }{|}^2=1+|w{|}^2-2(w_1\cos \theta +w_2\sin \theta )$. Subtracting one therefore givesThus the lifted boundary angle $\varphi (\theta )=-\theta +2g(\theta )$ is strictly increasing. Its total change around the circle is $2\pi$, because the disk automorphism has boundary degree one. The same total change follows directly from the winding number one of $w-e^{i\theta }$. This is the boundary-angle derivative of a disk automorphism.