= Solution
Part (c) gives
$$
\nabla\times(\mathbf A_+-\mathbf A_-)=0.
$$
The half-space $x_1>0$ is a <simply connected domain>, so the <Poincaré lemma> makes this curl-free field a <conservative vector field>: it is the gradient of a single-valued scalar potential.
Using $\tan u+\cot u=2/\sin(2u)$,
$$
\mathbf A_+-\mathbf A_-
=\frac2{r\sin\theta}\mathbf e_\phi.
$$
The spherical-coordinate formula from part (b) shows that a potential must satisfy $f_\phi=2$ and may be independent of $r,\theta$. On $x_1>0$ the azimuthal angle has the single-valued branch
$$
\phi=\arctan\frac{x_2}{x_1}.
$$
Hence one solution, up to an <additive constant>, is
$$
\boxed{f(\mathbf x)=2\phi
=2\arctan\frac{x_2}{x_1}}.
$$
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