= Solution
Orient $\partial B$ counterclockwise, and define positive <circulation> by that orientation. The outward normal from the body satisfies $n\,ds=(dy,-dx)$. Pressure acts inward, so
$$
F_x-iF_y=-\oint p(dy+i\,dx)=-i\oint p\,d\overline z.
$$
The steady <Bernoulli equation> for this constant-density irrotational flow is $p+\rho|v|^2/2=C$. The constant has zero <integral> around a closed curve. With <complex potential of a two-dimensional potential flow> $w=\phi+i\psi$, let $W=w'=u-iv$. The no-penetration condition makes the boundary a streamline, so $dw=Wdz$ is real along it. Thus $Wdz=\overline Wd\overline z$, and $|W|^2d\overline z=W^2dz$. Substitution proves \b[<Blasius force theorem>]:
$$
\boxed{F_x-iF_y=\frac{i\rho}{2}\oint_{\partial B}(w'(z))^2\,dz.}
$$
At infinity the <Laurent expansion> has the form
$$
W(z)=U-iV+\frac{\Gamma}{2\pi iz}+O(z^{-2}).
$$
There is no source term because the impermeable body has zero net flux. The circulation term follows from $\oint Wdz=\Gamma$. The coefficient of $z^{-1}$ in $W^2$ is $\Gamma(U-iV)/(\pi i)$. Analyticity in the fluid lets the body contour be enlarged to a large counterclockwise circle, so the <residue theorem> gives $\oint W^2dz=2\Gamma(U-iV)$. Hence
$$
\boxed{(F_x,F_y)=\rho\Gamma(V,-U).}
$$
Its component along $(U,V)$ is zero: \b[the drag is zero]. The lift magnitude is $\boxed{\rho|\Gamma|\sqrt{U^2+V^2}}$. With the stated positive counterclockwise circulation, the signed lift along the left normal $(-V,U)/\sqrt{U^2+V^2}$ is $-\rho\Gamma\sqrt{U^2+V^2}$. Reversing the circulation convention reverses this signed formula. This is the <Kutta–Joukowski theorem> with its orientation made explicit.
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