Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / K / 4 / Solution

Take source strength $Q$ to mean total two-dimensional flux, so an isolated source has potential $Q\log (z-a)/(2\pi )$. Let $a=1+i$, $b=1/\bar{a}=(1+i)/2$, and put $c_Q=Q/(2\pi )$. The circle theorem gives, up to an additive constant,The derivative is single valued, which is what the force integrals require. With a positively oriented contour around the cylinder, the force form of Blasius' theorem isOnly the poles at zero and $b$ are enclosed; the physical source at $a$ lies outside. The residues of $q^2/c_Q^2$ are $2(1/a+1/b)$ at zero and $2[1/(b-a)-1/b]$ at $b$. Their sum is $2[1/a+1/(b-a)]=-2/a=-(1-i)$. ThereforeFor the second integral, the residue of $z{q}^2/c_Q^2$ at zero is one. At $b$ it is $1+2b[1/(b-a)-1/b]=-3$. Their sum is $-2$. Using precisely the supplied complex-moment convention givesThe vanishing physical moment $M$ is also consistent with pressure acting normally to a circle, hence radially about its centre. If strength is instead defined as the coefficient of $\log (z-a)$, replace $c_Q$ by that strength throughout; stating the flux convention fixes all factors of $2\pi$.