= Solution
Along a <level curve>, $V_x dx+V_y dy=0$, with $V_x=V_0\theta/\pi$ and $V_y=V_0(1+\log r)/\pi$. Therefore
$$
\boxed{\frac{dy}{dx}=-\frac{\theta}{1+\log r}.}
$$
In the intended <upper half-plane (complex analysis)>[upper half-plane] $0<\theta<\pi$, the slope is positive for $0<r<e^{-1}$, vertical at $r=e^{-1}$, and negative for $r>e^{-1}$. At the positive real boundary $\theta=0$ the slope is zero except at the <stationary point> $(e^{-1},0)$, where both <Derivative>[derivatives] vanish and the quotient is indeterminate. More generally, outside the chosen <half-plane> its sign is the sign of $-\theta(1+\log r)$, with the <argument> branch held fixed. A convenient parametric description of the curves to sketch is $V\pi/V_0=r(\sin\theta\log r+\theta\cos\theta)$.
\Image[/past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2017-complex-potential-and-transmission.png]
{title=Equipotentials for the upper-half-plane logarithm branch and illustrative square-well transmission}
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