= Solution
For points on the unit circle, their convex hull contains the origin exactly when they are not all contained in an open semicircle. Equivalently, the shortest closed arc containing them has length at least $\pi$. Applying this to $V=U_2^\dagger U_1$ and using
$$
N(V)=\operatorname{conv}(\operatorname{spec}V)
$$
gives
$$
0\in N(U_2^\dagger U_1)
\quad\Longleftrightarrow\quad
\theta(U_2^\dagger U_1)\geq\pi.
$$
Part (b) now proves the claimed perfect-discrimination criterion.
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