= Solution
Set $s=|z|^2$. Contraction of the <symplectic form> with the rotation <vector field> gives
$$
\iota_K\omega=\frac{i}{(1+s)^2}\left(iz\,d\bar z+i\bar z\,dz\right)=-\frac{ds}{(1+s)^2}.
$$
With the convention fixed in the unheaded solution, $\iota_K\omega=-dH$, so \b[the <Hamiltonian function> is]
$$
\boxed{H(z)=\frac{|z|^2}{1+|z|^2}+C.}
$$
Its derivative is $dH=ds/(1+s)^2$. In the chart $w=1/z$, $H=1/(1+|w|^2)+C$, so it extends smoothly across infinity; the additive constant is the kernel freedom already identified. Thus $K=X_H$ globally, proving that the action is a <Hamiltonian group action>. With $C=0$, the <moment map for rotation of the complex projective line> takes values in $[0,1]$, reaching its endpoints at the two fixed poles.
One can check the normalization in polar coordinates: $\omega=2r(1+r^2)^{-2}dr\wedge d\phi$ has total <symplectic area> $2\pi$, not $4\pi$, and $H=r^2/(1+r^2)$. Equivalently, in a polar angle with $r=\tan(\vartheta/2)$, $H=(1-\cos\vartheta)/2$. This explains the half-height factor that would be lost by replacing the printed form with the unscaled round-sphere area form. Under the opposite convention $\iota_{X_H}\omega=dH$, the Hamiltonian is the negative of this function, up to a constant.
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