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

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)?

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 $$z = x + iy,$$ $$$$F(z) = \frac{Q}{2\pi}(\mathrm{ln}(r) + i \theta),$$$$ where $$r = |z| = \sqrt{x^2 + y^2}$$ and $$\theta = \mathrm{Arg}(z)$$. Comparing to $$$$F(z) = \phi + i \psi$$$$ and equating the real and imaginary parts gives $$(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.

• many thanks @Zac. For the sake of interest, what is the atan2 function. Could you provide a insighful link? – ecjb Jan 25 at 12:33
• The Wikipedia page here on the $\mathrm{atan2}$ function gives a reasonable explanation. A problem with using the $\mathrm{arctan}$ formula is that the range of $\mathrm{arctan}$ is $(- \pi/2, \pi/2)$, so it will never give a correct argument for a complex number whose real component is less than zero (for example $-1 + i$.) The $\mathrm{atan2}$ function fixes this by adding or subtracting $\pi$ as appropriate. Often things will still work out if we stick with the $\mathrm{arctan}$ formula, but it can be good to know about its shortcomings! – Zac Jan 25 at 13:20