= Solution
Put $\kappa=Q/(2\pi)$, $a=1+i$ and $b=1/\bar a=a/2$. The circle theorem gives, up to an additive constant,
$$
W(z)=\kappa\{\log(z-a)+\log(z-b)-\log z\},\qquad q(z)=W'(z)=\kappa\left\{\frac1{z-a}+\frac1{z-b}-\frac1z\right\}.
$$
Thus an image source lies at $b$ and an image sink at zero. With a counterclockwise contour around the cylinder, Blasius' force theorem is
$$
F_x-iF_y=\frac{i\rho}{2}\oint q(z)^2\,dz.
$$
If $q=s/(z-z_0)+h(z)$ locally, the <residue> of $q^2$ is $2s h(z_0)$. The enclosed poles are zero and $b$, with residues
$$
\operatorname{Res}_0q^2=2\kappa^2(1/a+1/b),\qquad
\operatorname{Res}_bq^2=2\kappa^2\{1/(b-a)-1/b\}.
$$
Their sum is $2\kappa^2\{1/a+1/(b-a)\}=-\kappa^2(1-i)$. The <residue theorem> therefore yields
$$
\boxed{F_x-iF_y=\frac{\rho Q^2}{4\pi}(1-i),\qquad F_x=F_y=\frac{\rho Q^2}{4\pi}}.
$$
The force points towards the source. For the moment formula the double poles also contribute: $\operatorname{Res}_{z_0}(zq^2)=s^2+2z_0s h(z_0)$. Hence
$$
\sum\operatorname{Res}(zq^2)=\kappa^2+\kappa^2+2\kappa^2b\{1/(b-a)-1/b\}=
2\kappa^2\frac{b}{b-a}=-2\kappa^2.
$$
It follows that $\oint zq^2\,dz=-4\pi i\kappa^2$ and
$$
\boxed{M=0,\qquad N=-2\pi\rho\kappa^2=-\frac{\rho Q^2}{2\pi}}.
$$
The zero <torque> agrees with pressure acting normally to a circle, hence radially through its centre.
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