= 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,
$$
w(z)=c_Q[\log(z-a)+\log(z-b)-\log z],\qquad q(z)=w'(z)=c_Q\left[\frac1{z-a}+\frac1{z-b}-\frac1z\right].
$$
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 is
$$
F_x-iF_y=\frac{i\rho}{2}\oint_C q(z)^2\,dz.
$$
Only 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 $2[1/a+1/(b-a)]=-2/a=-(1-i)$. Therefore
$$
\boxed{F_x-iF_y=\frac{\rho Q^2}{4\pi}(1-i),\qquad F_x=F_y=\frac{\rho Q^2}{4\pi}.}
$$
For the second integral, the <residue> of $zq^2/c_Q^2$ 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 gives
$$
M-iN=-\frac\rho2(2\pi i)(-2c_Q^2)=\frac{i\rho Q^2}{2\pi},\qquad\boxed{M=0,\quad N=-\frac{\rho Q^2}{2\pi}.}
$$
The 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 $2\pi$.
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