# Harmonic Function which cannot be described as real part of a holomorphic function

We define $f:\mathbb{C}\rightarrow\mathbb{C},\ f(z)=\log|z|$. $f$ is harmonic. Why can't we describe $f$ as a real part of a holomorphic (analytic) function?

Thank you very much for your time,

Chris

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Any harmonic function on a simply connected region in $\mathbb{R}^2$ (seen as $\mathbb{C}$) is indeed the real part of a holomorphic function on the same region. The function $f(z) = \log |z|$ is not defined at zero, and so the domain you're referring to is actually $\mathbb{C}\backslash\{0\}$. This is the reason why it doesn't globally represent the real part of a holomorphic function.

The complex logarithm is the typical example of many related notions in complex analysis, notably "multi-valued" functions, functions defined on Riemann surfaces (multi-sheated surfaces), and functions that fail to be analytically continued properly. The harmonic function $\log |z|$ is indeed the real part of the complex logarithm, but its imaginary part is not well defined because of its reliance on the complex argument (which is the same mod $2\pi$). Any introductory text on complex analysis will discuss all of this in depth.

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Maybe one should add that the function admits, locally, an extension to a holomorphic function. A problem only occurs if the domain of definition will contain a closed path surrounding the origin. –  user20266 Jun 17 '12 at 5:16
The fact that $\log|z|$ is not defined at zero is the reason why it is not the real part of a holomorphic function on the same region? –  Chris Jun 17 '12 at 13:00
A harmonic function defined on a simply connected, open region is the real (or imaginary) part of a holomorphic complex function on the same region. log |z| is not defined at the origin, which presents a big problem. If you take a simply connected region, like the complex plane, and then take out the origin, it is no longer simply connected, and consequently log |z| isn't the real part of a holomorphic complex function on the complex plane minus 0. –  Greg Zitelli Jun 18 '12 at 14:29