 Solution

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.