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I'd like to show that $\ln |f| $ is harmonic, where $f$ is holomorphic defined on a domain of the complex plane and never takes the value 0. My idea was to use the fact that $\ln |f(z)| = \operatorname{Log} f(z) - i*\operatorname{Arg}(f(z)) $, but $Log$ is only holomorphic on some part of the complex plane and $\operatorname{Arg}$ is not holomorphic at all. Any help is welcome!

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This is a local result; you need to show that given a $z_0$ with $f(z_0) \neq 0$ there is a neighborhood of $z_0$ on which $\ln|f(z)|$ is harmonic.

Fix $z_0$ with $f(z_0) \neq 0$. Let $\log(z)$ denote an analytic branch of the logarithm defined on a neighborhood of $f(z_0)$. Then the real part of $\log(z)$ is $\ln|z|$; any two branches of the logarithm differ by an integer multiple of $2\pi i$. The function $\log(f(z))$, being the composition of analytic functions, is analytic on a neighborhood of $z_0$. The real part of this function is $\ln|f(z)|$, which is therefore harmonic on a neighborhood of $z_0$, which is what you need to show.

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  • $\begingroup$ and the harmonicity of real/imaginary parts comes from the cauchy-schwartz equations $\endgroup$
    – Kris
    Aug 17 '12 at 13:39
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    $\begingroup$ Are you sure that Log is everywhere (apart from 0) differentiable? In the course I have, Log is only differentiable in the complex plane with a "cut" corresponding to a half line starting at 0. $\endgroup$
    – Kim
    Aug 17 '12 at 14:00
  • $\begingroup$ Any continuous branch of the complex logarithm is differentiable. You can put the branch cut wherever you wish (within reason; it must extend from the origin to infinity). Moreover, the ambiguity of the complex logarithm is only in the imaginary part (the argument of $z$), so it doesn't really enter into the discussion in any serious way. $\endgroup$ Aug 17 '12 at 14:09
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    $\begingroup$ $log(f(z))$ is not necessarily well-defined on $\{z \mid f(z) \ne 0\}$. You can make this argument rigorous by making it local, but there are still some steps to complete. E.g. at any $z$ such that $f(z) \ne 0$ show there is a neighborhood $U$ of $z$ such that $f(U)$ is contained in a simply-connected domain not containing the origin. Then use the fact that in any simply-connected domain not containing the origin there is a branch of the logarithm. Now complete as above.... $\endgroup$ Aug 17 '12 at 14:10
  • $\begingroup$ Given $a \neq 0$, there is a way of defining $\log(z)$ on a neighborhood of $z = a$ such that this $\log(z)$ is analytic. Each way of defining this $\log(z)$ has the same real part, namely $\ln|z|$ (they differ by multiples of $2\pi i$). So what you do here is use such a $\log$ that is defined on a neighborhood of a given $a = f(z_0)$. Then the above applies. $\endgroup$
    – Zarrax
    Aug 17 '12 at 14:12
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Hint: Write $f = u + iv$ where $u,v$ are real functions on the plane. Now your function is $\frac12 \ln (u^2 + v^2)$. Now compute directly the laplacian of this and make sure to use at some point that $u$ and $v$ satisfy the Cauchy-Riemann equations....

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$\ln |f|$ is real valued, hence holomorphic iff it is constant. (Nonconstant holomorphic maps are locally open, at least when their derivative in not zero, which is true for all but countably many points)

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  • $\begingroup$ I made a mistake in my post: I meant show that $ln |f|$ is harmonic...I corrected it now. $\endgroup$
    – Kim
    Aug 17 '12 at 13:25

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