Blasius Theorem why do we always take complex conjugate to represent the net force? Suppose fluid flows steadily past an obstacle $B$ with simple closed boundary $∂B$. If gravity is neglected, the net force $(F_x,F_y)$ exerted on $B$ by the fluid (per unit length out of the plane) is given by
$F_x + iF_y = -\frac{i \rho}{2} \oint_{∂B} \left|\frac{dw}{dz}\right|^2 dz$
Then in the definition of Blasius Theorem, the net force exerted on $B$ is represented by taking the complex conjugate of above equation, I know the fact that the complex conjugate is also a solution of F, but in this case, why do we take the complex conjugate form?
 A: We can see how the conjugate arises by deriving the Blasius formula, where to be perfectly clear, the squared derivative of the potential $w$ should appear -- not the squared modulus.
In inviscid flow, the fluid velocity is tangent to the boundary $\partial B$ which is represented as a closed contour in the complex plane.  Let $\alpha$ be the angle between the tangent and the real axis. The normal vector directed into the body is then, $\mathbb{n} = -\sin \alpha + i\cos \alpha = i e^{i \alpha}.$ 
The components of velocity are related to the complex potential by
$$\frac{dw}{dz} = -u_x + i u_y = \sqrt{u_x^2 + u_y^2}(- \cos \alpha + i \sin \alpha) = -\sqrt{u_x^2 + u_y^2}e^{-i \alpha}.$$  
The pressure is given by Bernoulli's equation
$$p = p_0 - \frac{\rho}{2}\left( u_x^2 + u_y^2\right) = -\frac{\rho}{2}\left( u_x^2 + u_y^2\right),$$
where we take the constant $p_0=0$, since there is no net contribution when integrated around a closed contour.
Hence,
$$p = -\frac{\rho}{2}\left( \frac{dw}{dz}\right)^2e^{2i\alpha}.$$
Note that the above expression involves the squared derivative of the potential, not the squared modulus of the derivative.
The complex representation of the force on the body is obtained as the integral with respect to arclength $s$ around the contour.
$$F = F_x + i F_y = \int_{\partial B} p\, n \, ds = \int_{\partial B} p\, ie^{i \alpha} \, ds. $$
We take the conjugate in order to get a cancellation of the complex exponential when we later substitute for pressure:
$$\overline{F} = -i\int_{\partial B} p\, e^{-i \alpha} \, ds.$$
Note that the pressure is real-valued and is not affected upon taking the conjugate.
The differential of the complex variable is related to arclength differential by $dz = e^{i\alpha} ds.$
Hence,
$$\overline{F} = -i\int_{\partial B} p\, e^{-2i \alpha} \, dz.$$
Substituting for pressure , we get
$$\overline{F} = F_x - iF_y = -i\int_{\partial B} \frac{\rho}{2}\left( \frac{dw}{dz}\right)^2e^{2i\alpha}\, e^{-2i \alpha} \, dz \\ = -\frac{i\rho}{2}\int_{\partial B} \left( \frac{dw}{dz}\right)^2 \, dz .$$
