Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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,


share|cite|improve this question
up vote 8 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.