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Note that $\operatorname{Log}|z|$ is the real part of any branch of $\operatorname{log}|z|$, so we pick any of the branches. This branch is analytic in $\mathbb{C}\setminus\{0\}$, which means that it's real part is harmonic, and thus the result follows.

That seems too trivial to be correct, is there something wrong with my proof?

mrf
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user270468
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3 Answers3

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$\log|z|$ is not complex analytic in $\mathbb{C}-\{0\}$, since $\log|z|$ is real-valued but not constant. The proof does work if you replace $\log|z|$ with $\log z$.

Gyu Eun Lee
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    $\log z$ is not holomorphic on the punctured plane. – mrf Sep 16 '15 at 20:11
  • @mrf Yes, but the OP's argument is to choose a branch of $\log z$ to obtain the result in the slit plane. We can then choose two different branches of $\log$, which lets us conclude that the real part is harmonic in the punctured plane, since the real part of a branch of the logarithm doesn't depend on the branch. – Gyu Eun Lee Sep 16 '15 at 20:33
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You can't choose a branch of $\log z$ that is holomorphic on $\mathbb{C}\setminus\{0\}$, but that's ok. The problem is local and for each point in the punctured plane, you can find a neighbourhood of the point and a corresponding branch of $\log z$ which shows that your function (locally the real part of a holomorphic function) is harmonic.

mrf
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  • Ah, I think I understand it now. But would it not be enough to find a branch of $\operatorname{log}z$ which is complex-differentiable in that point, rather than finding a nbd of that point where it is analytic? – user270468 Sep 16 '15 at 20:32
  • The condition of being harmonic (or holomorphic) is really something that happens on open sets. It doesn't make sense to talk about functions being "harmonic at a single point" (but there are functions that are complex differentiable only at say $z=0$). – mrf Sep 16 '15 at 20:35
  • Sure, the definition says that a function can be harmonic only on open sets, but could I not deduce that the function satisfies Laplace's equation at every point if I reason like this? – user270468 Sep 16 '15 at 20:38
  • If we are picking nits: there are functions that are complex-differentiable at a single point but where its real and imaginary parts are not twice differentiable. Even more surprising, there are functions satisfying Laplace's equation everywhere without being harmonic. See for example this. Work on open sets, and you don't have to worry. – mrf Sep 16 '15 at 20:45
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It's almost OK. If you want to be more careful, start by picking a branch. With that you prove that $\log \lvert z\rvert$ is harmonic in $\mathbb{R}^2\setminus \text{a half-line}$. Now pick another branch. With that you prove that $\log\lvert z\rvert$ is harmonic in $\mathbb{R}^2\setminus\text{another half-line}$. From that conclude that $\log\lvert z\rvert$ is harmonic in the union of the two domains.

Giuseppe Negro
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