Complex Velocity on a solid body I wish to find the complex velocity on a solid body in the flow (to find stagnation points). I have started with angular flow past a cylinder $$\Omega(z)=U\left(ze^{-i\alpha}+\frac{e^{i\alpha}}{z}\right).$$ 
I then applied the Joukowski mapping to an offset cylinder $w=az+b$, where $a=|1-b|$, so the Joukowski mapping is given by $$\zeta(z)=az+b+\frac{1}{az+b}$$ with inverse mapping $$z=\frac{\zeta-b+\sqrt{\zeta^2-1}}{a}$$ which results in a complex potential $$\Omega(\zeta)=U\left(\frac{\zeta-b+\sqrt{\zeta^2-1}}{a}e^{-i\alpha}+\frac{\left(\zeta-b-\sqrt{\zeta^2-1}\right)ae^{i\alpha}}{b^2-2b\zeta+1}\right).$$ From here it becomes a bit cumbersome. I have tried directly differentiating the potential to find the complex velocity, but the result is pretty huge. My other thought was that $\frac{d\Omega}{d\zeta}=\frac{d\Omega}{dz}\left(\frac{d\zeta}{dz}\right)^{-1}$. I expect that on the solid body $z=e^{i\theta}$ the limit as $\theta\rightarrow 0$ is undetermined as that relates to the trailing edge of an aerofoil, but using that approach, I had a finite limit. Any thoughts welcome, thanks!
 A: Let $$\Omega(z)=U\left((z-z_0)e^{-i\alpha}+\frac{\mu^2e^{i\alpha}}{(z-z_0)}\right)$$ where $z_0 = \epsilon +\delta i$,  $\mu = \sqrt{(b-\epsilon)^2 + \delta^2}$ and b is the critical point of the Joukowski mapping. This represents a flow past an offset cylinder centred at $z_0$ and passes through the critical point $b$. 
Then consider Joukowski mapping $$\zeta(z) = \frac{1}{2}\left(z+\frac{b^2}{z}\right)$$ with inverse $$z = \zeta + \sqrt{\zeta^2-b^2}.$$
Then by the chain rule $$\frac{d\Omega}{d\zeta} = \frac{d\Omega}{dz}\left(\frac{d\zeta}{dz}\right)^{-1}$$
$$= 2U \left[ e^{-i\alpha}-\frac{\mu^2e^{i\alpha}}{(z-z_0)^2}\right]\left(1-\frac{b^2}{z^2}\right)^{-1}$$
$$= 2Uz \left[\frac{ (z-z_0)^2e^{-i\alpha}-\mu^2e^{i\alpha}}{(z^2-b^2)(z-z_0)^2}\right].$$
On the solid body $z=\mu e^{i\theta}+z_0$ and noting that the stagnation points occur when $\frac{d\Omega}{d\zeta}=0$ substituting in $z$ and with some simplification we have $$\frac{d\Omega}{d\zeta}=2U(\mu e^{i\theta}+z_0)e^{-i\alpha}\left[\frac{\mu^2e^{i\theta}e^{-i\alpha}-\mu^2e^{-i\theta}e^{i\alpha}}{((\mu e^{i\theta}+z_0)^2-b^2)\mu^2e^{2i\theta}}\right]=0$$ $$\implies \mu^2(e^{i(\theta-\alpha)}-e^{-i(\theta-\alpha)} = 0$$ $$\implies 2\mu^2sin(\theta-\alpha) = 0$$ $$\implies \theta = \alpha + k\pi,$$ where $k=0,1$ and we are done. 
