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 / S1 / 1

Complex differentiability gives $u_x=v_y$, $u_y=-v_x$. Differentiating once more, for continuously differentiable first derivatives with the needed second derivatives, yields $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$. Thus both components satisfy Laplace's equation. The result is local on the domain of holomorphy; it makes no assertion across a branch cut.

Choose a continuous argument $\theta$ on a simply connected domain avoiding zero, and put $r=\sqrt{x^2+y^2}$. With $u=x\log r-y\theta$, direct differentiation givesHere ${\theta }_x=-y/r^2$, ${\theta }_y=x/r^2$; these identities specify the branch needed in the written arctangent. Integrating the Cauchy-Riemann equations givesThe literal principal $\arctan (y/x)$ works only separately on suitable half-planes. It cannot stand for a single continuous argument across the imaginary axis.

For the upper half-plane use $0<\theta <\pi$, and interpret the source's arctangent as this argument. Then $V=(V_0/\pi )\mathrm{Im}(z\mathrm{Log}z)$ is harmonic, so $\mathbf{E}=-\nabla V$ has zero curl and divergence where there is no charge. On the positive boundary $\theta \to 0$ and $y\log r\to 0$; on the negative boundary $\theta \to \pi$. Hence the two boundary values are respectively zero and $V_{0}x$.

Branch qualification: the principal real arctangent alone gives zero on both sides of $y=0$, contradicting the stated negative-axis boundary condition, for example at $x=-1$. The continuous upper-half-plane argument repairs this notation; it is not an optional additive constant.

Along a level curve, $V_{x}dx+V_{y}dy=0$, with $V_x=V_0\theta /\pi$ and $V_y=V_0(1+\log r)/\pi$. ThereforeIn the intended 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 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 $V\pi /V_0=r(\sin \theta \log r+\theta \cos \theta )$.