Finding the velocity field from the complex potential

Question

The question i have is

Consider two vortices with rotation Γ at positions $(a,0)$ and $(-a,0)$. Find the Complex potential and hence find the velocity field.

Now my working is as follows

$w = -\frac{iΓ}{2{\pi}}log(z-a)-\frac{iΓ}{2{\pi}}log(z+a)$

$\frac{dw}{dz}= -\frac{iΓ}{2{\pi}}(\frac{1}{z-a} + \frac{1}{z+a}) = -\frac{iΓ}{2{\pi}}(\frac{2z}{z^2-a^2})$

This is the answer i got for the complex potential, now I know that the complex potential is $w(z) = \phi(z) + i\varphi(z)$ which is the velocity potential as the real part of the complex potential and the streamfunction for the imaginary part. So to find the velocity field $\textbf{u}=\nabla\phi$ however, I'm unsure how to pick out the real part of the complex potential and use this to find the velocity field. I feel like I'm super close but just missing something obvious.

Answer 1

With $z = x + iy$ we have

$$\frac{dw}{dz} = -\frac{iΓ}{2{\pi}}\left(\frac{1}{z-a} + \frac{1}{z+a}\right) = -\frac{iΓ}{2{\pi}}\left(\frac{1}{(x-a) + iy} + \frac{1}{(x+a) + iy}\right) \\ = -\frac{iΓ}{2{\pi}}\left(\frac{(x-a)-iy}{(x-a)^2 + y^2} + \frac{(x+a) - iy}{(x+a)^2 + y^2}\right) $$

Since $w = \phi(x,y) + i\psi(x,y)$, the complex derivative is

$$\frac{dw}{dz} = \frac{\partial \phi}{\partial x} + i \frac{\partial \psi}{\partial x} = \frac{\partial \phi}{\partial x} - i \frac{\partial \phi}{\partial y},$$

where the last equality follows from the Cauchy-Riemannn equations. More details are given here.

Given that the velocity field $\mathbf{u} = (u,v)$ is the gradient of the potential, it follows that

$$u-iv = -\frac{iΓ}{2{\pi}}\left(\frac{(x-a)-iy}{(x-a)^2 + y^2} + \frac{(x+a) - iy}{(x+a)^2 + y^2}\right)\\ = \frac{Γ}{2{\pi}}\left(\frac{-y - i(x-a)}{(x-a)^2 + y^2} + \frac{-y - i(x+a)}{(x+a)^2 + y^2}\right)$$

From here, you should be able to determine $u$ and $v$ by equating real and imaginary parts.