show that $(x,y)\to \ln(\sqrt{x^2+y^2})$ is harmonique over $\mathbb{R}^2\backslash \{(0,0)\} $

Question

by use of following result:

If $f$ is a holomorphic function over the open $O$ of $\mathbb{C}$ then the real part of $f$ is harmonic over the open $O$

show that $(x,y)\to \ln(\sqrt{x^2+y^2})$ is harmonique over $\mathbb{R}^2\backslash \{(0,0)\} $?

we notice that $\ln(| z | ) $ is the real part of any determination of $\ln (z)$, of which it suffices to choose the cut so that $\ln(z)$ is holomorphic on an open neighborhood of the point $(x, y) \in \mathbb{R}^2\backslash \{(0,0)\}$. But how can I choose the cut?

Answer 1

We may also show $u(x, y)$ harmonic by direct differentiation as follows:

$u(x, y) = \ln(x^2 + y^2)^{1/2} = \dfrac{1}{2} \ln(x^2 + y^2); \tag 1$

$u_x = \dfrac{1}{2} \dfrac{1}{x^2 + y^2} (2x) = \dfrac{x}{x^2 + y^2}; \tag 2$

$u_{xx} = \dfrac{x^2 + y^2 - 2x^2}{(x^2 + y^2)^2} = \dfrac{y^2 - x^2}{(x^2 + y^2)^2}; \tag 3$

$u_y = \dfrac{y}{x^2 + y^2}; \tag 4$

$u_{yy} = \dfrac{x^2 + y^2 - 2y^2}{(x^2 + y^2)^2} = \dfrac{x^2 - y^2}{(x^2 + y^2)^2}; \tag 5$

comparing (3) and (5) we see that

$u_{yy} = -u_{xx}, \tag 6$

whence

$u_{xx} + u_{yy} = 0, \tag 7$

and thus we see that $u(x, y)$ is harmonic.

Notice we haven't introduced any cuts in obtaining this result.

Answer 2

You can show that is harmonique on $\mathbb{R}^2 \backslash [0, \infty)$, and on $\mathbb{R}^2 \backslash (-\infty, 0]$.

Maybe the catch is that there does not exist a holomorphic function on $\mathbb{C}\backslash\{0\}$ with real part $\log|z|$. But "locally" ( on complements of some cuts) it's OK.