= Solution
{c}
Differentiate the argument through the nonvanishing curve, using $g'=\operatorname{Im}(\beta'/\beta)$. Since $\beta'=-ie^{i\theta}$,
$$
g'(\theta)=\operatorname{Im}\frac{-ie^{i\theta}(\overline w-e^{-i\theta})}{|w-e^{i\theta}|^2}
=\frac{1-w_1\cos\theta-w_2\sin\theta}{|w-e^{i\theta}|^2}.
$$
Also $|w-e^{i\theta}|^2=1+|w|^2-2(w_1\cos\theta+w_2\sin\theta)$. Subtracting one therefore gives
$$
\boxed{-1+2g'(\theta)=\frac{1-|w|^2}{|w-e^{i\theta}|^2}>0.}
$$
Thus the lifted boundary angle $\phi(\theta)=-\theta+2g(\theta)$ 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>.
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