## How to differentiate Complex Fluid Potential

Let's get some formulas straight in the first place. You have $$F(z) = \phi + i\psi$$ Also, it is given that $$u = \frac{\partial \phi}{\partial x}, \ \ v = \frac{\partial \phi}{\partial y}$$ And no, what you have instead is this: $$u = \frac{\partial \psi}{\partial y}, \ \ v = -\frac{\partial \psi}{\partial x}$$ Differentiate in the $x$-direction or in the $y$-direction - it doesn't matter (why?): $$\frac{dF}{dz} = \frac{\partial F}{\partial x}\frac{\partial x}{\partial z} \qquad ; \qquad \frac{dF}{dz} = \frac{\partial F}{\partial y}\frac{\partial y}{\partial z}$$ With $$\frac{\partial z}{\partial x} = 1 \quad \Longrightarrow \quad \frac{\partial x}{\partial z} = 1/1 = 1 \qquad ; \qquad \frac{\partial z}{\partial y} = i \quad \Longrightarrow \quad \frac{\partial y}{\partial z} = 1/i = -i$$ Hence in the $x$- or $y$-direction, take your favorite: $$\frac{dF}{dz} = 1 \left(\frac{\partial \phi}{\partial x} + i \frac{\partial \psi}{\partial x} \right) = -i \left( \frac{\partial \phi}{\partial y} + i \frac{\partial \psi}{\partial y}\right) = u - i v$$ More about this subject in Cauchy-Riemann Equations (PDF).