# Complex Potential Flow - Two Vortices offset from the Origin

1. Feb 9, 2013

### squire636

1. The problem statement, all variables and given/known data

a. Determine the complex potential for two equal counter-rotating vortices with strength $\Gamma$, the positive one located at z=-a and the negative one at z=a.

b/ Show the shape of the streamlines for this case.

2. Relevant equations

z = x + iy = r*e^(i$\theta$)

W(z) = $\Phi$ + i$\Psi$
where $\Phi$ is the potential function and $\Psi$ is the stream function

3. The attempt at a solution

a. This part is relatively easy. I know that the complex potential for a vortex at the origin is

i$\Gamma$/(2*$\pi$) * ln(z)

Therefore, for the two vortices, we will have:

W(z) = i$\Gamma$/(2*$\pi$) * ln((z+a)/(z-a))

b. This is where I start to have trouble. I need to separate W(z) into the real and imaginary parts, and then I know that the imaginary part is the stream function. However, I don't know how to do this. It is easy for a vortex at the origin, because then I would have

ln(z) = ln(r*e^(i$\theta$)) = ln(r) + i$\theta$

However, the addition and subtraction of 'a' inside of the log is giving me a lot of trouble. I've tried to separate it every way that I can think of but haven't had any success. Any help would be much appreciated. Thanks!

2. Feb 9, 2013

### Staff: Mentor

To separate real and imaginary parts, it is easier to write the log as ln(z+a)-ln(z-a). In both cases, this is just a shift by +-a.

3. Feb 9, 2013

### squire636

I tried that, but still did not make any progress.

ln(z+a) - ln(z-a)
ln(r*e^iθ + a) - ln(r*e^iθ - a)

Now what? I can't split it up into ln(r)+iθ anymore.

4. Feb 9, 2013

### Staff: Mentor

What about $r^+$,$\theta^+$ corresponding to the magnitude and phase of z+a?

5. Feb 9, 2013

### squire636

I'm sorry I don't quite follow you, could you explain in more detail? I'm unfamiliar with that notation. Thanks.

6. Feb 9, 2013

### Staff: Mentor

That is a notation I invented for your specific problem.

z+a can be written as $z+a=r^+ e^{i \theta^+}$ with some real values $\theta^+$ and $r^+$ - there are formulas how to convert an arbitrary complex number to that shape.
In the same way, z-a can be written as $z-a=r^- e^{i \theta^-}$ with some real values $\theta^-$ and $r^-$.

7. Feb 9, 2013

### squire636

Would I do something along these lines?

ln(z+a)
ln(r*e^iθ + a)
ln(r*cos(θ)+i*r*sin(θ) + a)
ln((r*cos(θ)+a) + i*r*sin(θ))

And then try to find a new r and θ in order to put this back into the form of r*e^iθ ? That doesn't seem to me like it will be possible, and I can't find the formulas that you mentioned. Are they available online somewhere?

8. Feb 9, 2013

### Staff: Mentor

That is possible.
The standard formulas how you get r and θ if you have a complex number a+ib? They are everywhere.

9. Feb 9, 2013

### squire636

I think I figured it out, thanks so much!