I obtained the complex potential for flow past a flat plate at angle of attack by applying the Joukowski transform $z = \zeta + 1/\zeta$ (centered at the origin, so $\zeta = e^{i\theta}$) to flow past a cylinder.
For reference, the complex potential for flow past a cylinder (at angle of attack $\alpha$, of unit radius, with freestream velocity $U_\infty$, and with circulation $\Gamma$) is $ F(\zeta) = U_\infty \left( \zeta e^{-i \alpha} + \frac{e^{i \alpha}} {\zeta} \right) -\frac{i \Gamma}{2 \pi} \log \zeta $ and after taking the Joukowski transform, I get $ F(z) = U_\infty \left( z \cos \alpha - i (z^2 - 4)^{1/2} \sin \alpha \right) - \frac{i \Gamma}{2 \pi} \log \left[\frac{1}{2}\left( z + (z^2-4)^{1/2} \right)\right] $
Then we should be able to calculate lift around the flat plate with the Blasius formula, $ F_x - i F_y = \frac{i \rho}{2} \oint \left( \frac{dF(z)}{dz} \right)^2 dz $ where the integral is around the surface (flat plate from $z = -2$ to $z = 2$, or the cylinder $\zeta = e^{i \theta}$ if we integrate in the $\zeta$-plane).
What's the best way to evaluate this contour integral? I tried changing the integral to the $\zeta$-plane with something like $ F_x - i F_y = \frac{i \rho}{2} \oint \left( \frac{dF(\zeta)}{d\zeta} \frac{d\zeta}{dz} \right)^2 \frac{dz}{d\zeta} d\zeta = \frac{i \rho}{2} \oint \left( \frac{dF(\zeta)}{d\zeta} \right)^2 \frac{1}{1-1/\zeta^2} d\zeta $ but the $\frac{1}{1-1/\zeta^2}$ factor seems to mess things up by adding poles at $z=\pm1$ and making the residue of the pole at $z=0$ go to $0$. (I know what the answer should come out to by the Kutta-Joukowski theorem, but I want to be able to verify it.) Do I need to evaluate this integral in the $z$-plane?
Thanks in advance, any help is appreciated! My complex analysis and conformal mapping is definitely a bit rusty.
EDIT 1: I decided to try and generate interest by taking the calculations further to show where the problem is, and found out that continuing as I was doing does indeed give the right answer! Valuable lesson for me there that I shouldn't claim to be stuck just because it doesn't "seem" like I will get the right answer.
In the end, you do indeed have poles at $\zeta=\pm 1$ which have nonzero residues, and the pole at $\zeta = 0$ still has a nonzero residue as well. They add up perfectly to give the same sum of residues as in the non-transformed case, so that the Kutta-Joukowski theorem holds as we expect. (I guess so it's no longer unanswered, I'll answer my own question.)