Let $Q$ be the square with corners $0$,$1$,$i$,$1+i$ and $R$ the rectangle with corners $0$,$2$,$i$,$2+i$. Prove there's no conformal map $Q$ to $R$ (on the interiors) that extends to a surjective homeomorphism on the closure, and takes corners to corners.
This is from an old qualifying exam. I'd appreciate a hint about what kind of theorem to use.
On
Here's another argument that might be interesting. Apply the Schwarz reflection principle to extend along an edge of $Q$ to a map from a rectangle twice as big as $Q$ to a rectangle twice as big as $P.$ Continue extending in this way to extend to an automorphism of the whole complex plane. This must be a Möbius transformation. This is contradiction because, for example, it doesn't preserve the cross-ratio of the corners of $Q.$
Suppose that there is a conformal map $w=f(z):Q \to R$ satisfying the described conditions. We may assume that $f(0)=0, f(1)=2,f(i)=i$ and $f(1+i)=2+i$ for corners correspondence. Let $C_y=\{x+iy: 0\le x\le 1\}$ $(0\le y\le 1)$ be a horizontal segment joining two points $iy$ and $1+iy$ in $Q$. Its image $f(C_y)$ is a curve joining $f(iy)$ and $f(1+iy)$ in $R$. Obviously the length of $f(C_y)$ is not less than $2$: $$ \text{the length of }f(C_y)=\int_0^1 |f^\prime(x+iy)|dx\ge 2.$$ Therefore we have $$ \int_0^1 \left(\int_0^1 |f^\prime(x+iy)|dx\right)dy \ge 2.\tag{1}$$ On the other hand we have \begin{align} \int_0^1 \left(\int_0^1 |f^\prime(x+iy)|dx\right)dy&=\iint_Q |f^\prime(x+iy)| dxdy\\ &\le \left(\iint_Q |f^\prime(x+iy)|^2 dxdy\right)^\frac{1}{2}\left(\iint_Q dxdy\right)^\frac{1}{2}\\ &=\left(\text{Area of }R\right)^\frac{1}{2}\left(\text{Area of }Q\right)^\frac{1}{2}=\sqrt{2}, \end{align} which contradicts the result of $(1)$.