= Solution
Orient the body <boundary> counterclockwise and write its <tangent> displacement as $(dx,dy)$. The outward normal from the body has displacement-weighted components $(dy,-dx)$. <Pressure> therefore exerts the <force> components $F_x=-\oint p\,dy$ and $F_y=\oint p\,dx$, or
$$
F_x-iF_y=-i\oint_{\partial B}p\,d\bar z.
$$
For steady <irrotational> incompressible flow at constant <density>, Bernoulli gives $p=p_0-\rho|w'(z)|^2/2$, with any constant far-field kinetic term absorbed into $p_0$. Its constant contribution integrates to zero. The <boundary> is a <streamline>, so $\operatorname{Im}w$ is constant there and $w'(z)\,dz$ is real. Thus $w'\,dz=\overline{w'}\,d\bar z$, giving $|w'|^2d\bar z=(w')^2dz$. This <identity element> remains valid at a <stagnation point> as both sides vanish. <Substitution> yields <Blasius force theorem>:
$$
\boxed{F_x-iF_y=\frac{i\rho}2\oint_{\partial B}(w'(z))^2\,dz.}
$$
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