I was wondering if anyone could help me with the following problem, as I'm unsure on how to begin. The problem is the following.
Two equal line sources of strength $k$ are located at $x=3a$ and $x=-3a$, near a circular cylinder of radius $a$ with axis normal to the $x-y$ plane and passing through the origin. The fluid is incompressible and the flow is irrotational and inviscid. I need to use the Milne-Thomson circle theorem to show that the complex potential for this flow is $$ w(z) = k\ln\left(a^{4}-9a^{2}z^{2}-9\frac{a^{6}}{z^{2}}+81a^{4}\right). $$ I know that a line source of strength $k$ has a complex potential $w(z)=k\ln(z)$. I just don't understand how to apply this with the theorem to get the required result.
Here is a link to the theorem https://en.wikipedia.org/wiki/Milne-Thomson_circle_theorem
Any suggestions would be appreciated, thanks for reading this.
Let $f_1(z) = k\ln(z-3a)$ and $f_2(z)$ be the complex potential for the line source at $x=3a$ and $x=-3a$ respectively (without the cylinder). The complex potential $f(z)$ for these double line sources (without the cylinder) is just the sum of the individual complex potential, i.e. $$ f(z) = f_1(z) + f_2(z) = k\ln\left(z^2 - 9a^2\right). $$ By virtue of the Milne-Thomson Circle theorem, we just need to compute $\overline{f\left(\dfrac{a^2}{\bar z}\right)}$: \begin{align*} f\left(\frac{a^2}{\bar z}\right) & = k\ln\left(\frac{a^4}{\bar z^2} - 9a^2\right) \\ \overline{f\left(\frac{a^2}{\bar z}\right)} & = k\ln\left(\frac{a^4}{z^2} - 9a^2\right). \end{align*} The desired expression for $w(z)$ follows by adding $f(z)$ and $\overline{f\left(\dfrac{a^2}{\bar z}\right)}$ together and I will leave this computation for you.