= Solution
For the <Joukowski map> $g(z)=z+z^{-1}$,
$$
\operatorname{Im}g(z)=\operatorname{Im}z\left(1-\frac1{|z|^2}\right).
$$
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. Therefore
$$
\boxed{K_0=\{z\in\mathbb H:|z|\leq1\},\qquad\operatorname{hcap}(K_0)=1.}
$$
The <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>.
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