I'm trying to show that $u(x, \, y) = \ln(x^2 + y^2)$ is a harmonic function, without explicitly computing the partial derivatives and showing that $u_{xx} + u_{yy} = 0$.

I believe that it would suffice to show that $u$ is an analytic function. Should I just just use the Cauchy-Riemann equations? Could someone point me in the right direction please.

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    $\begingroup$ $u$ is a real part of an analytic function $\endgroup$ – Hamou Jul 22 '14 at 21:29
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    $\begingroup$ $\ln z = \ln |z| + i \arg z$. $\endgroup$ – Zarrax Jul 22 '14 at 21:30

The general idea is to express $u$ in terms of $z$ and $\bar z$, guess which analytic function $f$ satisfies $2u=f+\bar f$ (which is $u=\operatorname{Re}f$), and verify the guess. The preliminary computations can be a bit sloppy (not paying attention to branching), since at the end the result is checked.


  • $u(x,y)=x^2-y^2 = \frac14 ((z+\bar z)^2+(z-\bar z)^2) = \frac12 z^2+\frac12 \bar z^2$. So we try $f(z) = z^2$.
  • $u(x,y)=\ln(x^2+y^2) = \ln (z\bar z) = \ln z+\ln \bar z$. So we try $f(z) = 2\ln z$.

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