I am trying to find the stagnation point of a fluid flow from a complex potential. The complex potential is given by: $\Omega(z) = Uz + \cfrac{m}{2\pi}\ln z$. From this I found the streamfunction to be $\psi=Ur\sin\theta + \cfrac{m}{2\pi}\theta$ and the velocity potential to be $\phi=Ur\cos\theta + \cfrac{m}{2\pi}\ln r$.
I think the stagnation points occur when $u=v=0$, where $u = \cfrac{\partial \phi}{\partial x}$ and $v = \cfrac{\partial \psi}{\partial y}$. If so, would I have to convert back into Cartesian coords? Any help appreciated, thanks!
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$\begingroup$Yes you could do that, but it's probably easier to use the expression of the gradient in polar coordinates:
$$ u_r =\frac{\partial\phi}{\partial r} =\frac{1}{r}\frac{\partial\psi_z}{\partial \theta} , u_\theta=\frac{1}{r}\frac{\partial\phi}{\partial \theta}=-\frac{\partial \psi_z}{\partial r} $$
Both $u_r$ and $u_\theta$ should be zero.
As you probably recognized, this is the potential flow of a source/sink at the origin and a free stream parallel to $x$.
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