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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!

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You are mostly correct (except that $v$ is actually $\frac{\partial \phi}{\partial y}$). However, it is easiest to deal with $\Omega(z)$ directly.

Since the velocity components are $u=\cfrac{\partial \phi}{\partial x}=\cfrac{\partial \psi}{\partial y}$ and $v=\cfrac{\partial \phi}{\partial y}=-\cfrac{\partial \psi}{\partial x}$, a stagnation point with zero velocity needs both to vanish. You can translate this back to the complex derivative of $\Omega$ as $$\frac{d}{dz}\Omega=\frac{\partial \phi}{\partial x}+i\frac{\partial \psi}{\partial x}=\frac{\partial \psi}{\partial y}-i\frac{\partial \phi}{\partial y}=0.$$ This means that you can work directly in (complex) cartesian coordinates to find the stagnation point easily: $$ 0=\frac{d\Omega}{dz}=U+\frac m{2\pi}\frac{1}{z},\quad\text{so}\quad z=x+iy=-\frac{2\pi}m U+0i. $$ Easy!

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  • $\begingroup$ Super, much easier than I was thinking :-) $\endgroup$ – Brit Miller Sep 6 '13 at 13:32
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The working for that last solution was correct except for the final answer. It should be:

$ z = \frac{-m}{2 \pi U}+0i $

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