= Solution
Orient the body boundary counterclockwise. The outward normal from the body satisfies $n\,ds=(dy,-dx)$, so pressure gives
$$
F_x-iF_y=-i\oint_Cp\,d\bar z.
$$
For steady <irrotational flow>, Bernoulli's equation makes $p+\rho|u|^2/2$ constant in the <connected> fluid region. Its constant contribution integrates to zero. Write $q=w'=u-iv$. Impermeability says the velocity is tangent to the boundary, so $q\,dz=\bar q\,d\bar z$ there. Consequently
$$
F_x-iF_y=\frac{i\rho}2\oint_C|q|^2d\bar z=\boxed{\frac{i\rho}2\oint_Cq^2\,dz},
$$
which proves Blasius' theorem with this contour convention.
The background <stream function> is the imaginary part of $w_0(z)=\alpha(z-c)^3$. Since its coefficients are real, the circle theorem gives
$$
\boxed{w(z)=\alpha\{(z-c)^3+(a^2/z-c)^3\}}.
$$
Its derivative is
$$
q=\alpha(3z^2-6cz+3c^2-3a^6z^{-4}+6ca^4z^{-3}-3c^2a^2z^{-2}).
$$
In $q^2$ the coefficient of $z^{-1}$ comes from the exponent pairs $(2,-3)$ and $(1,-2)$, including both orders. It is $36\alpha^2ca^4+36\alpha^2c^3a^2$. The <residue theorem> then gives
$$
\boxed{F_x=-36\pi\rho\alpha^2ca^2(a^2+c^2),\qquad F_y=0}.
$$
The vanishing transverse force also agrees with <reflection> symmetry about the real axis.
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