If $F(x + iy) = \frac{Q}{2\pi}\ln (x + iy)$, how to prove that $\phi(x,y) = \frac{Q}{2\pi}\ln (\sqrt{x^2 + y^2})$?

Question

The potential flow theory states that $$ F(z) = \phi + i\psi $$ where $z = x + iy$ In case of a source/sink, we have $$ F(z) = \frac{Q}{2\pi}\ln z\tag{1} $$ $$ \phi(x,y) = \frac{Q}{2\pi}\ln (\sqrt{x^2 + y^2}) \quad \text{or}\quad \phi(r,\theta) = \frac{Q}{2\pi}\ln(r)\tag{2} $$ $$ \psi(x,y) = \frac{Q}{2\pi}\arctan{\frac{y}{x}} \quad \text{or}\quad \phi(r,\theta) = \frac{Q}{2\pi}\theta\tag{3} $$ how do we go from (1) to (2) and from (1) to (3)?

Answer 1

This is a matter of the definition of the complex logarithm. If $z = r e^{i \theta}$ for $\theta \in (-\pi, \pi],$ the principal value of the complex logarithm of $z$ is defined as \begin{equation*} \mathrm{Log}(z)= \mathrm{ln}(r) + i\theta. \end{equation*} So given $z = x + iy,$ \begin{equation} F(z) = \frac{Q}{2\pi}(\mathrm{ln}(r) + i \theta), \end{equation} where $r = |z| = \sqrt{x^2 + y^2}$ and $\theta = \mathrm{Arg}(z)$. Comparing to \begin{equation} F(z) = \phi + i \psi \end{equation} and equating the real and imaginary parts gives $(2)$ and $(3)$.

Note also that the formula $\mathrm{Arg}{(x + iy)} = \mathrm{arctan}{\frac{y}{x}}$ is only valid for $x > 0$. A more general formula makes use of the $\mathrm{atan2}$ function.