 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,Thus an image source lies at $b$ and an image sink at zero. With a counterclockwise contour around the cylinder, Blasius' force theorem isIf $q=s/(z-z_0)+h(z)$ locally, the residue of $q^2$ is $2sh(z_0)$. The enclosed poles are zero and $b$, with residuesTheir sum is $2{\kappa }^2\{1/a+1/(b-a)\}=-{\kappa }^2(1-i)$. The residue theorem therefore yieldsThe force points towards the source. For the moment formula the double poles also contribute: ${\mathrm{Res}}_{z_0}(z{q}^2)=s^2+2z_{0}sh(z_0)$. HenceIt follows that $\oint z{q}^{2}dz=-4\pi i{\kappa }^2$ andThe zero torque agrees with pressure acting normally to a circle, hence radially through its centre.