Is $ \ln |f| $ harmonic? 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!
 A: 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....
A: $\ln |f|$ 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)
A: 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.
