A line source of strength $k$ has velocity given (in cylindrical polars) by $v_{R}=\frac{k}{R}$, $v_{\theta}=0$. Show that the complex potential for such a source is $w(z)=k\ln(z)$.
My solution so far:
I know that $w=\phi+i\psi$ but i'm unsure where to go from here. (Any hint would be appreciated).
The stream function $\psi(r,\theta)$ in polar coordinates satisfies $$ v_R = \frac{1}{R}\partial_\theta\psi \ \ \textrm{ and } \ \ v_\theta = -\frac{1}{R}\partial_R\psi. $$ Equating this with the given flow field $(v_R,v_\theta) = \left(\dfrac{k}{R},0\right)$ yields $$ \partial_\theta\psi = k \ \ \textrm{ and } \ \ \partial_R\psi = 0. $$ The solution to this two equations is then $\psi(\theta) = k\theta$. Now, we know that the complex potential is given by $w = \phi + i\psi$, where $\phi$ is the velocity potential satisfying $\nabla\phi = (v_R,v_\theta)$. In polar coordinates, $$ (v_R, v_\theta) = \left(\partial_R\phi,\frac{1}{R}\partial_\theta\phi\right) = \left(\frac{k}{R},0\right). $$ Solving for $\phi$ we find that $\phi(R) = k\ln R$. Finally, \begin{align*} w = \phi + i\psi = k\ln R + ik\theta = k\left(\ln R + i\theta\right) = k\ln z. \end{align*} Here, I am choosing the principal branch of the complex logarithm.