Finding the velocity field from the complex potential 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.
 A: 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.
