= Solution
Use the <complex velocity> $q(z)=w'(z)=u-iv$. For this steady, constant-<mass density>, <irrotational flow> with no <body force>, the applicable <Bernoulli equation> is
$$
p+\frac\rho2|q|^2=P_*,
$$
with a single constant $P_*$ throughout the connected regular fluid region. The body's impermeability makes the <velocity> tangent to $C$. Thus $q\,dz$ is real there, giving
$$
q\,dz=\bar q\,d\bar z,\qquad |q|^2d\bar z=q^2dz.
$$
The <complex force> from part (a) is
$$
\begin{aligned}
F_x-iF_y
&=-i\oint_Cp\,d\bar z\\
&=\frac{i\rho}{2}\oint_C|q|^2d\bar z
=\frac{i\rho}{2}\oint_Cq^2dz,
\end{aligned}
$$
where the constant-$P_*$ term vanishes. This proves the <Blasius force theorem>:
$$
\boxed{F_x-iF_y=\frac{i\rho}{2}\oint_C\left(w'(z)\right)^2dz.}
$$
For the moment, observe that
$$
z\,d\bar z=(x\,dx+y\,dy)+i(y\,dx-x\,dy).
$$
Therefore $M=\operatorname{Re}\oint_Cp\,z\,d\bar z$. The constant <pressure> contribution has zero real part, because $x\,dx+y\,dy=\tfrac12d(x^2+y^2)$. Substituting <Bernoulli's principle> and the same tangency identity gives the <Blasius moment theorem>:
$$
\boxed{M=-\frac\rho2\operatorname{Re}\oint_Cz\left(w'(z)\right)^2dz.}
$$
The real-part operation is essential: $\oint z\,d\bar z$ itself need not vanish, but its real part does.
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