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For the compact H-hull $(x,x+ih]$, the square-root branch asymptotic to $z-x$ gives the hydrodynamic normalization at infinity of its mapping-out function of a compact H-hull. Its expansion has coefficient $h^2/2$, the half-plane capacity of a vertical slit.

For the vertical slit, choose the branch of $\sqrt{z^2+1}$ asymptotic to $z$ at infinity. It is the mapping-out function of a compact H-hull for $(0,i]$, andThus half-plane capacity of a vertical slit givesThe endpoint $0$ is merely part of the boundary closure convention.

For the filled upper half-disc $K=\{z\in \mathbb{H}:|z|\le 1\}$, the conformal map $g_K(z)=z+1/z$ maps the exterior half-disc onto $\mathbb{H}$. Indeed, it sends the semicircle to $[-2,2]$, the remaining real boundary to the complementary intervals, and its inverse is the branch of $(w+\sqrt{w^2-4})/2$ asymptotic to $w$. Hence half-plane capacity of a half-disc givesIf $\mathbb{D}$ denotes the open unit disc, the printed $\mathbb{H}\cap \mathbb{D}$ is not relatively closed and is literally not a compact H-hull. The intended value is the one for its filled relative closure $K$. With a closed-disc convention the displayed notation already represents that hull.

For a nonempty compact H-hull $A$, its closure meets $\mathbb{R}$: otherwise a nonempty compact set strictly inside $\mathbb{H}$ would be separated from the lower half-plane in the complement, contradicting the simply connected domain condition for $\mathbb{H}\setminus A$. Choose $x\in \overline{A}\cap \mathbb{R}$ and put $d=\mathrm{diam}(A)$. Taking limits in the definition of diameter shows $|z-x|\le d$ for every $z\in A$. Thus $A$ lies in the filled half-disc of radius $d$ centered at $x$. Its half-plane capacity is $d^2$ by scaling and translation of half-plane capacity. Using monotonicity of half-plane capacity,For the empty compact H-hull, use diameter zero. This proves the unit-constant version of half-plane capacity is bounded by squared diameter; no claim of optimality of the constant is needed.

There is a small closure issue in the printed set: the unit disc is open, so its half-disc is not relatively closed in $\mathbb{H}$. As written, the set is not a compact H-hull. We compute the intended half-plane capacity after taking its relative closure in $\mathbb{H}$.

First remove the closed unit half-disc $K$. Its mapping-out function of a compact H-hull is the Joukowski mapFor $z=iy$ on the remaining portion of the vertical slit, $1<y\le 2$,Thus the image of the remaining slit is $(0,3i/2]$. By the half-plane capacity of a vertical slit, its capacity is $(3/2{)}^2/2=9/8$. The half-plane-capacity composition rule givesAs a direct check, composing with the slit map gives $g_A(z)=\sqrt{(z+z^{-1}{)}^2+9/4}$, with the branch asymptotic to $z$. Its $z^{-1}$ coefficient is $17/8$.