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 have used 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 also know that the speed of the flow at the surface of the cylinder is given by $\left|\frac{dw}{dz}\right|$. This was given in the previous part of the question that I am doing: $$ \left|\frac{dw}{dz}\right| = v = \frac{18k\sin(2\theta)}{a(41-9\cos(2\theta))} $$ However I do not know how to show this. Computing the derivative gives $$ \frac{dw}{dz}=\frac{2kz}{z^{2}-9a^{2}} - \frac{2kz^{2}a^{4}}{a^{4}z^{3}-9a^{2}z^{3}} $$ If anyone could show me how to simplify this expression to the required one i would be extremely grateful.
My main problem is how to determine the positions of the points on the surface of the cylinder at which the pressure is minimum?
Edit: The problem about computing the flow speed on the cylinder has been answered here: How can I show the following trigonometric function?
Wth the complex potential $w(z)$ found using the Milne-Thompson Circle Theorem, we also know that the cylinder $|z| = a$ is a streamline. It follows from Bernoulli streamline theorem for incompressible, irrotational, inviscid fluid in a steady flow that $$ \frac{p}{\rho} + \frac{1}{2}|\mathbf{u}|^2 = \textrm{constant} \qquad\textrm{on $|z|=a$.} $$ From the previous part of the question, $$ |\mathbf{u}|^2\Big|_{r=a} = \left|\frac{dw}{dz}\right|^2\bigg|_{r=a} = \frac{324k^2\sin^2(2\theta)}{a^2\left(41 - 9\cos(2\theta)\right)^2}. $$ Denoting $\lambda = 324k^2/a^2$, we have that $$ \frac{p}{\lambda\rho} = \textrm{constant} - \frac{\sin^2(2\theta)}{\left(41 - 9\cos(2\theta)\right)^2} \qquad \textrm{ on $r=a$}. $$ The pressure $p$ is now a function of $\theta$, i.e. $p = p(\theta)$ and finding the location on the cylinder in which the pressure is minimum, is now simply a calculus question.