 Solution

Orient the body boundary counterclockwise. Its outward body normal satisfies $nds=(dy,-dx)$, so the pressure force has complex formBy steady Bernoulli's equation, $p=p_{\infty }+\rho (U_{\infty }^2-|q{|}^2)/2$, where $q=w^{\prime }=u-iv$. Constant terms integrate to zero around a closed curve. On an impermeable boundary, velocity is tangent, so $qdz=\overline{q}d\overline{z}$. Therefore $|q{|}^{2}d\overline{z}=q^{2}dz$, yielding the Blasius force theoremFor a slit body, the contour encloses both faces and the tips, interpreted by a limiting exterior contour.

Choose the analytic branch $s(z)=\sqrt{z^2-a^2}$ outside the slit with $s(z)\sim z$ at infinity. It may be written $\sqrt{z-a}\sqrt{z+a}$ using the principal individual square roots. For $-a<x<a$, approaching from above gives $s_{+}(x)=i\sqrt{a^2-x^2}$, and from below gives $s_{-}(x)=-i\sqrt{a^2-x^2}$.

NowThere is no $z^{-1}$ term in $q^2$. Since $q^2$ is analytic in the exterior, enlarging the contour to a large circle shows its integral is zero. ThusThis total-force argument retains the slit-tip contribution automatically.

On the upper face, $q_{+}(x)=\cos \alpha -x\sin \alpha /\sqrt{a^2-x^2}$; on the lower face, $q_{-}(x)=\cos \alpha +x\sin \alpha /\sqrt{a^2-x^2}$. For $0<\alpha \le \pi /2$, their zeros areThey are distinct face points even at $\alpha =\pi /2$, when both have coordinate zero. At $\alpha =0$, $w=z$ and $q=1$ everywhere in the exterior: there are no velocity-zero stagnation points. The limiting positions at the two tips must not be substituted into a singular expression and treated as zeros.

For the sketch at $\alpha =\pi /4$, the far-field velocity is $(1,1)/\sqrt{2}$, and streamlines are levels of $\mathrm{Im}w$. The plate is the zero-stream-function boundary, with the two stagnation points at $\pm a/\sqrt{2}$ on the indicated faces.