= Solution
\b[Conformality and analyticity.] Use the orientation-preserving convention for a <conformal map>, with a continuously differentiable, nonzero differential. Its <Jacobian matrix> must be a positive scale times a rotation,
$$
D(u,v)=\lambda\begin{pmatrix}\cos\phi&-\sin\phi\\\sin\phi&\cos\phi\end{pmatrix},\qquad\lambda>0.
$$
Thus $u_x=v_y$, $u_y=-v_x$, the <Cauchy-Riemann equations>; continuous partial derivatives then imply that $u+iv$ is a <holomorphic function>. If angle preservation is allowed to reverse orientation, the statement needs this qualification: $z\mapsto\bar z$ preserves unsigned angles but is an <anticonformal map> and is not a <holomorphic function>.
\b[The mapping.] Put $a=\pi/L$. The displayed components combine to give
$$
\boxed{w=\cosh(az),\qquad w'=a\sinh(az).}
$$
Its <derivative> does not vanish in the open half-strip $x>0$, $0<y<L$, so it is a <conformal map> there. Also $\operatorname{Im}w=\sinh(ax)\sin(ay)>0$. The lower edge maps to $[1,\infty)$, the upper edge to $(-\infty,-1]$, and the vertical edge to $[-1,1]$. The open half-strip maps bijectively onto the <upper half-plane>. Indeed, $\cosh\zeta=\cosh\eta$ implies $\zeta=\pm\eta+2\pi i m$; the half-strip allows only equality. Surjectivity follows by solving $t+t^{-1}=2w$ and choosing the root $t=e^{az}$ with $|t|>1$ and $0<\arg t<\pi$. The two corners are boundary critical points, not interior failures of <conformality>.
\b[Transformed boundary values.] A <conformal map> preserves <Laplace's equation> up to a positive scale. Thus the transformed <Dirichlet problem> is to find a real <harmonic function> $U(u,v)$ on $v>0$ with
$$
U(u,0)=\begin{cases}-V_0,&-1<u<1,\\V_0,&|u|>1.\end{cases}
$$
The conflicting values at the original corners are interpreted as one-sided edge limits.
\b[Logarithmic solution.] Choose the <complex logarithms> of $w\pm1$ with arguments between zero and $\pi$ in the <upper half-plane>. A linear combination $F(w)=\alpha\log(w+1)+\beta\log(w-1)+\gamma$ is a complex-valued <harmonic function>, since it is a <holomorphic function>; the real physical potential must be a real or imaginary part. In particular, set
$$
\mathcal F(w)=V_0+\frac{2iV_0}{\pi}\bigl[\log(w-1)-\log(w+1)\bigr].
$$
Then
$$
U=\operatorname{Re}\mathcal F=V_0-\frac{2V_0}{\pi}\bigl[\arg(w-1)-\arg(w+1)\bigr].
$$
On the three real boundary intervals, the argument difference is respectively $0,\pi,0$, which gives the required values. This is the bounded <harmonic> solution, with $U\to V_0$ at infinity; uniqueness holds in this bounded class by the <maximum principle>. Without a boundedness or far-field condition, the edge data alone do not give uniqueness: one may add $C\sinh(\pi x/L)\sin(\pi y/L)$.
\b[Back substitution.] Since
$$
\frac{w-1}{w+1}=\tanh^2(az/2),\qquad
\tanh(az/2)=\frac{\sinh(ax)+i\sin(ay)}{\cosh(ax)+\cos(ay)},
$$
the required continuous argument difference is $2\arctan[\sin(ay)/\sinh(ax)]$ in the open half-strip. Consequently
$$
\boxed{V(x,y)=V_0-\frac{4V_0}{\pi}\arctan\!\left(\frac{\sin(\pi y/L)}{\sinh(\pi x/L)}\right).}
$$
As $y\to0,L$ with $x>0$, the arctangent tends to zero; as $x\to0^+$ with $0<y<L$, it tends to $\pi/2$. These limits verify every open edge of the <Dirichlet problem>.
\Image[/past-exam-of-the-short-options-paper-of-the-physics-prelims-of-the-university-of-oxford/2022-half-strip-potential.png]
{title=The bounded <harmonic function> across the half-strip, with contour labels giving $V/V_0$}
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