 Solution

For the Joukowski map $g(z)=z+z^{-1}$,It maps the exterior upper half-disc $D_0=\{z\in \mathbb{H}:|z|>1\}$ into $\mathbb{H}$. It is injective there: $g(z)=g(w)$ gives $(z-w)(1-1/(zw))=0$, and the second factor cannot vanish when both moduli exceed one. It is onto as well. For $v\in \mathbb{H}$, the two roots of $z^2-vz+1=0$ have product one. Neither lies on the unit circle, since $z+z^{-1}$ is real there. Exactly one root has modulus greater than one, and the displayed imaginary-part formula makes its imaginary part positive. That root belongs to $D_0$ and maps to $v$.

Since $g_{K_0}$ takes values in $\mathbb{H}$, its domain must be contained in $D_0$. The established injectivity and surjectivity of $g:D_0\to \mathbb{H}$ mean that no proper subdomain of $D_0$ can map onto all of $\mathbb{H}$ using this same function. ThereforeThe half-plane capacity is the coefficient of $1/z$ in the normalized expansion. If hulls are regarded as closed subsets of $\overline{\mathbb{H}}$, take the closure of this half-disc, including its diameter; its relative half-plane hull and capacity are unchanged. This is the half-plane capacity of a half-disc.