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Prove that the function $G=\ln|f(z)|$ is harmonic in a region $R$ if $f(z)$ is analytic in $R$ and also $f(z)\cdot f'(z)$ does not equal zero in $R$.

My difficulty here is that the expression for the Laplacian of $G$ is very big and ugly, and I know that I have to apply the Cauchy-Riemann Equations somewhere , but it is not clear to me how and where. Also the condition of the multiplication of the complex function with its derivative not being zero looks rather mysterious . Any help would be appreciated.

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    $\begingroup$ Saying that $f(z)f'(z) \neq 0$ just means that neither $f(z)$ nor $f'(z)$ are zero. $\endgroup$ – user98602 May 12 '15 at 21:13
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    $\begingroup$ However, zeros of the derivative wouldn't hurt at all here. $\endgroup$ – Daniel Fischer May 12 '15 at 21:14
  • $\begingroup$ @MikeMiller : I'd have said neither $f(z)$ nor $f'(z)$ is zero. I wonder if using "are" rather than "is" is on its way toward being considered standard usage? $\endgroup$ – Michael Hardy May 12 '15 at 21:37
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Trigger warning: If the phrase "multiple-valued function" causes you pain, then stop reading here.

Suppose we define $$ \operatorname{Log} z = \log |z| + i\arg z \tag 1 $$ where $\operatorname{Log}$ is "multiple-valued" because $\arg$ is multiple-valued. The different branches of $\operatorname{Log}$ differ by a constant and so all have the same derivative, which is $1/z$. If you like, just restrict attention to some region within which $\operatorname{Log}$ is single-valued. Then it's holomorphic, and thus harmonic. If $f$ is holomorphic then the chain rule tells us $\operatorname{Log}\circ f$ is holomorphic.

So suppose you can prove

  • $(1)$ is holomorphic (maybe by showing that it's locally an inverse of $\exp$?), and
  • The real part of a holomorphic function is harmonic.

Then you've got it. If your question is how to prove these two points, then I would need to add more, but otherwise maybe I can hope this answers your question.

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    $\begingroup$ But what lovely pain! You're first line buys my vote! Endorsed!!! $\endgroup$ – Robert Lewis May 12 '15 at 22:15
  • $\begingroup$ Hi and thanks for the response ,but I still do not get it $\endgroup$ – herashefat May 13 '15 at 0:16

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