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I would like to show that $ \log(x^2 + y^2) $ is not the real part of any analytic function in $ \mathbb{C} - \{0\} $ A similar question can be found here, but I don't think this argument is satisfactory.

Here are my thoughts on the problem. On the domain $ \mathbb{C} - \{x \geq 0\} $, $ \log(x^2 + y^2) = \mathfrak{Re}(\log(z^2)) $. Therefore, if I can show the only analytic function with real part $ \log(x^2 + y^2) $ is $ \log(z^2) $ then the problem is solved, since $ \log(z^2) $ is not analytic on the positive real-axis. The problem I have is I don't know how to show that the only analytic function with real part $ \log(x^2 + y^2) $ is $ \log(z^2) $

Any tips?

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Max
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    If there was another such function, then their difference would give an analytic function whose real part identically vanishes on an open subset of $\mathbb{C}$. It then remains to use Cauchy-Riemann equations. – Start wearing purple Jun 12 '13 at 14:15
  • Genius! Thanks a lot! – Max Jun 12 '13 at 14:16

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Be careful when saying that $\log(z^2)$ is "the only analytic function" with real part $\log(x^2 + y^2)$, because (a) however you define it you will have some complex angle (branch cut) where it is undefined, and (b) $\log(z^2) + c$ also works for any $c$ imaginary. So really the candidates are some branch of $\log(z^2)$, plus some imaginary constant. (In this case, the imaginary constant can also be included in the branch cut.) Then when you conclude that this doesn't work because "$\log(z^2)$ is not analytic on the positive real axis" you should modify this to say it is not analytic on whatever branch cut you took.

In general, if $a(x,y) + b(x,y)i$ and $a(x,y) + c(x,y)i$ are two analytic functions on some domain, as O.L. mentions in the comments, taking the difference and dividing by $i$ gives that $b(x,y) - c(x,y)$ is analytic. Cauchy-Riemann equations imply that $b(x,y)$ and $c(x,y)$ differ by a constant.

Caleb Stanford
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