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$f$ is analytic, nonzero on a simply connected domain $\Omega \subset \Bbb C$. Show that $\log|f(z)|$ is harmonic on $\Omega$.

I thought of two methods:

  1. $f=u+iv$ so $\log|f(z)|=\log\sqrt{u^2+v^2}$ , then use the definition of harmonic to solve it, i.e. $U_{xx}+U_{yy}=0$.
  2. $f$ can be wrote as something related to $\log|f(z)|+i\arg(f(z))$, but how to continue?
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    $\begingroup$ By simply connected, do you mean the domain of $f$ is simply connected? $\endgroup$ – Danny Pak-Keung Chan Aug 14 '17 at 22:11
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    $\begingroup$ I edited this post to the end of normative usage of the phrase "simply connected". $\endgroup$ – Robert Lewis Aug 14 '17 at 23:06
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    $\begingroup$ $\log |f(z)| = \operatorname{Re}(\log f(z))$; $\log \circ f$ is analytic; and the real part of any analytic function is harmonic. $\endgroup$ – Daniel Schepler Aug 14 '17 at 23:10
  • $\begingroup$ Please edit the title. Functions are not simply connected. $\endgroup$ – zhw. Aug 15 '17 at 3:50
  • $\begingroup$ I also edited the title per the comment of @zhw. $\endgroup$ – Robert Lewis Aug 15 '17 at 5:28
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First, the question of whether a function is harmonic (or not) is purely local, so we do not need to worry about branches, etc. That is, it suffices to take a small disc around a point where $f(z)\not=0$.

Then your first idea would work, using the fact that the real and imaginary parts satisfy the Cauchy-Riemann equations.

A cuter argument (which in reality just amounts to a hidden discussion of the Cauchy-Riemann equations) is to write $2\log |f(z)|=\log f(z)+\log \overline{f(z)}$, and use the idea that the Laplacian is ${\partial\over \partial z}\circ {\partial\over \partial \overline{z}}$. Since $\log f(z)$ is annihilated by $\partial/\partial \overline{z}$, and $\log \overline{f}(z)$ by $\partial /\partial z$, the sum is annihilated by the composition.

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  • $\begingroup$ I like the hint (thanks), but am not completely satisfied because $log(ab)$ is not (in general) $log(a)+log(b)$, you must discuss a bit the branch you take (+1 however). $\endgroup$ – Duchamp Gérard H. E. Aug 15 '17 at 15:09
  • $\begingroup$ @DuchampGérardH.E., yes, but we're only off by constants, which are annihilated by the Laplacian, so the ambiguity is harmless. $\endgroup$ – paul garrett Aug 15 '17 at 15:13
  • $\begingroup$ Maybe the student needs to deliver a tight writing. In this case, your (very nice) argument could begin by "one has just to show it locally i.e. at a neighbourhood of $z_0$. Now with $f(z)=\frac{f(z)}{f(z_0)}f(z_0)$", one can suppose $f(z_0)=1$ and take the principal value of $\log$ for which <your text>". It is one of the numerous options. $\endgroup$ – Duchamp Gérard H. E. Aug 15 '17 at 18:24
  • $\begingroup$ @DuchampGérardH.E., yes, you're certainly right! (I'll not actually edit the answer, though, since that'd bump it to the top of the queue needlessly, I think.) $\endgroup$ – paul garrett Aug 15 '17 at 18:27
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Suppose we write $f(z)$ in polar form, that is

$f(z) = \vert f(z) \vert e^{i \arg(f(z))}; \tag 1$

if we set

$g(z) = \ln (\vert f(z) \vert) + i\arg(f(z)), \tag 2$

then

$e^{g(z)} = e^{ \ln (\vert f(z) \vert) + i\arg(f(z))} = e^{\ln(\vert f(z) \vert} e^{i\arg(f(z))} = \vert f(z) \vert e^{i\arg(f(z))} = f(z). \tag 3$

Now let $z_0 \in \Omega$ and consider the function

$F(z) = \displaystyle \int_{z_0}^z \dfrac{f'(s)}{f(s)}ds; \tag 4$

since $f(z)$ is holomorphic in $\Omega$, so is $f'(z)$, and since $f(z) \ne 0$ in $\Omega$, $f'(z)/ f(z)$ is a well -defined holomorphic function on this domain; furthermore, since $\Omega$ is simply connected, the integral (4) defining $F(z)$ is completely independent of the path along which is taken, just so that path remains in $\Omega$. (Simple connectedness implies that an integral of a holomorphic function over any closed path vanishes, which in turn implies that any two path integrals (of the same holomorphic function) 'twixt the same endpoints are equal. These are standard facts upon which the theory of holomorphic functions is built; they occur in many texts on the subject.) Thus $F(z)$ is a well-defined holomorphic function on $\Omega$. Now

$((f(z))^{-1}e^{F(z)})' = -(f(z))^{-2}f'(z)e^{F(z )} + (f(z))^{-1}(e^{F(z)})'$ $= -(f(z))^{-2}f'(z)e^{F(z )} + (f(z))^{-1}\dfrac{f'(z)}{f(z)} e^{F(z)} = 0; \tag 5$

thus

$(f(z))^{-1}e^{F(z)} = c, \tag 6$

a constant, whence

$e^{F(z)} = cf(z); \tag 7$

since $F(z_0) = 0$,

$cf(z_0) = e^{F(z_0)} = e^0 = 1, \tag 8$

whence

$c = (f(z_0))^{-1}, \tag 9$

leading via (7) to

$f(z) = f(z_0)e^{F(z)}; \tag {10}$

using (3), (10) becomes

$e^{g(z)} = e^{g(z_0)}e^{F(z)} = e^{g(z_0) + F(z)}, \tag{11}$

whence

$e^{g(z) - g(z_0) - F(z)} = 1, \tag{12}$

so

$g(z) - g(z_0) - F(z) = 2n\pi i \tag{13}$

for some $n \in \Bbb Z$; now taking $z = z_0$ in (13) yields, since $F(z_0) = 0$,

$2n\pi i = g(z_0) - g(z_0) = 0, \tag{14}$

whence $n = 0$ and

$g(z) = g(z_0) + F(z). \tag{15}$

In light of (2), this yields

$\ln (\vert f(z) \vert) + i \arg(f(z)) = g(z_0) + F(z), \tag{16}$

which implies that $\ln (\vert f(z) \vert) + i \arg(f(z))$ is holomorphic; thus both $\ln (\vert f(z) \vert)$ and $\arg(f(z))$ are harmonic, and are in fact harmonic conjugates of one another.

Note: This question generalizes

Finding Harmonic conjugate for $\arg(z)$

End of Note.

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    $\begingroup$ explained very well...thanks! $\endgroup$ – освящение Sep 9 '18 at 13:33
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Let $a\in U,$ where $U$ is the domain of $f.$ Since $f(a)\ne 0,$ there is an open disc $D(f(a),r)$ that does not contain $0.$ This implies there exists a ray from the origin that misses this disc. Hence a branch of $\log z,$ which I'll just denote by $ \log z,$ is holomorphic in $D(f(a),r)$. It follows that $\log f$ is holomorphic in $f^{-1}(D(f(a),r)),$ which is an open subset of $U$ containing $a.$ The real part of $\log f,$ which is $\ln |f|,$ is therefore harmonic in $f^{-1}(D(f(a),r)).$ We are done, because harmonicity is a local property. (Note that the simple connectivity of $U$ is not needed.)

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