I'm trying to understand how to get from ($6$) to ($7$) in the excerpt below. Note that this is from a fluid dynamics paper, and $\zeta$ is the vorticity (the $k^{\text{th}}$ component of $\nabla\times\vec{u}$). It's treated as a function here, but for my purposes we can just assume it's constant.
What I'm not able to do is to explicitly show that $w^{\prime} = \psi_{y_0} + \psi_{x_0}$, which should be true, by equation ($3$). Here is my attempt:
\begin{align*} \psi_{x_0} &= \frac{\zeta}{4\pi}\int \int \frac{2(x_0-x)}{r^2} d\mathbf{x} \\ &= \frac{\zeta}{2\pi}\int \int \frac{(x_0-x)}{r^2} d\mathbf{x} \\ \psi_{y_0} &= \frac{\zeta}{2\pi}\int \int \frac{(y_0-y)}{r^2} d\mathbf{x} \end{align*}
where $r^2 = (x_0-x)^2 + (y_0-y)^2$
And then by differentiating and multiplying by the conjugate of the integrand:
\begin{align*} w^{\prime}(z_0) &= \frac{-i}{2\pi} \int \int \frac{1}{z_0-z} d\mathbf{x} \\ &= \frac{-i}{2\pi} \int \int \frac{1}{x_0-x +i(y_0-y)}*\frac{x_0-x + i(y-y_0)}{x_0-x + i(y-y_0)} d\mathbf{x} \\ &= \frac{-i}{2\pi} \int \int \frac{x_0-x +i(y-y_0)}{r^2} \\ &= \frac{\zeta}{2\pi} \int\int \frac{(y-y_0) + i(x-x_0)}{r^2} \\ &= -\psi_{y_0} -\psi_{x_0} \end{align*} So unfortunately I have $w'(z_0) = -\psi_{y_0} -\psi_{x_0}$, which is off by a sign.
I've n-tuple checked this so many times, and I still cannot see where the error is. Can anyone spot it? Or is there a more fundamental misunderstanding going on?
I did the calculations myself; indeed, the formulas $\nabla \psi = (-v, u)$ and $w' = u-iv$ are inconsistent. You did drop $i$ in the computation, the result should be $$ w' = - \psi_y - i \psi_x $$ A quick way to check this is to consider the potential of delta-function at $0$, so that $\psi(x_0) = \frac{1}{4\pi}\log|x_0|^2$ and $w(z_0) = \frac{-i}{2\pi} \log z_0$. Then $$\nabla \psi(x_0) = \frac{x_0 }{2\pi|x_0|^2}$$ and $$w'(z_0) = -\frac{i}{2\pi z_0} = -\frac{i \overline{z_0}}{2\pi |z_0|}$$ In real notation, $-i \overline{z_0} = -i(x-iy) = -y - ix$, hence the result.