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I'm trying to find all entire functions $f$ such that $|f|$ is harmonic. My attempt is as follows.

Because $f$ is entire, we may write $f(z) = f(x + iy) = u(x,y) + iv(x,y)$ where $u$ and $v$ have continuous first-order partial derivatives and satisfy the Cauchy-Riemann equations. Furthermore, $u$ and $v$ are harmonic. Now $|f| = \sqrt{u^2 + v^2}$ so I thought maybe to look at $|f|^2$ first. Then, $$(|f|^2)_{xx} = 2(u_x)^2 + 2(v_x)^2 + 2(u\cdot u_{xx} + v \cdot v_{xx})$$ and $$(|f|^2)_{yy} = 2(u_y)^2 + 2(v_y)^2 + 2(u\cdot u_{yy} + v \cdot v_{yy}).$$ I need $(|f|^2)_{xx} + (|f|^2)_{yy} = 0$ in order for $|f|^2$ to be harmonic. So adding these equations and using that $u$ and $v$ are harmonic, I obtain $$(u_x)^2 + (u_y)^2 + (v_x)^2 + (v_y)^2 = 0.$$ By the Cauchy-Riemann equations, I can simplify this to $$(u_x)^2 + (v_x)^2 = 0.$$ I'm a little stuck on how to proceed from here. I want to somehow conclude $|f|^2$ is constant but I'm not sure how.

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    $\begingroup$ $u_x$ and $v_x$ are real numbers, so $u_x=v_x=0$ $\endgroup$
    – saulspatz
    Dec 20, 2020 at 17:30
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    $\begingroup$ That makes sense, thank you! To check my logic, I can conclude that $u_x = v_y = v_x = u_y = 0$ and this must imply that $|f|^2$ is constant. Therefore, $|f|$ must be constant and so by Liouville, $f$ is necessarily constant. Does this sound right? $\endgroup$
    – Nick
    Dec 20, 2020 at 17:50
  • $\begingroup$ Yes, that's right. $\endgroup$
    – saulspatz
    Dec 20, 2020 at 19:35
  • $\begingroup$ It is no wrong by evaluating $\Delta(\sqrt{u^2+v^2})$ directly. saulspatz is right. $\endgroup$ Dec 20, 2020 at 23:29
  • $\begingroup$ @NikosBagis So do the computations above for $|f|$ rather than $|f|^2$? $\endgroup$
    – Nick
    Dec 21, 2020 at 15:57

2 Answers 2

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$f(z)=f(x+iy)=u+iv$. Then from analyticity of $f$: $$ u_x=-v_y\textrm{ and }u_y=v_x.\tag 1 $$ Hence if $|f|=\sqrt{u^2+v^2}$ (away from the zeros of $f$), then $$ |f|_x=\frac{u_x+v_x}{\sqrt{u^2+v^2}}\Rightarrow |f|_{xx}=\frac{(u_{xx}+v_{xx})|f|-\frac{(u_x+v_x)^2}{|f|}}{|f|^2}\Rightarrow $$ $$ |f|_{xx}=\frac{u_{xx}+v_{xx}}{|f|}-\frac{(u_x+v_x)^2}{|f|^3}.\tag 2 $$ In the same way $$ |f|_{yy}=\frac{u_{yy}+v_{yy}}{|f|}-\frac{(u_y+v_y)^2}{|f|^3}.\tag 3 $$ Hence using $u_{xx}+v_{yy}=0$, $v_{xx}+v_{yy}=0$ and equations (1), we get $$ \Delta|f|=|f|_{xx}+|f|_{yy}=-\frac{1}{|f|^3}\left((u_x+u_y)^2+(u_y-u_x)^2\right)=-2|f|^{-3}\left((u_x)^2+(u_y)^2\right).\tag 4 $$ Hence $$ \Delta|f|=0\textrm{ iff }(u_x=u_y=v_x=v_y=0)\textrm{ iff }f=const.\tag 5 $$ In case that $f(z)$ is zero in a set $A$, then $A$ will be discrete and from (5) constant in $\textbf{C}-A$. Hence $f(z)$ zero everywhere.

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    $\begingroup$ You must also show first that f has no zeros in order to justify the differentiation. $\endgroup$
    – Martin R
    Dec 22, 2020 at 8:35
  • $\begingroup$ I add the corrections in my answer. Thank you very much. $\endgroup$ Dec 22, 2020 at 10:09
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You have demonstrated that constant functions are the only entire functions for which $|f|^2$ is harmonic.

Now let us investigate the case where $f$ is entire and $ |f|$ is harmonic. We can assume that $f$ is not identically zero.

It is surely possible to calculate $\Delta |f|$ directly, but one can simplify the work by showing first that $f$ has no zeros, so that $f=e^g$ for some entire function $g$: Assume that $f(z_0) = 0$, then $$ 0 = |f(z_0)| = \frac{1}{2\pi} \int_0^{2\pi} |f(z_0+re^{it}| \, dt $$ for all $r > 0$, which implies that $f$ is identically zero, in contrast to the assumption.

So we have $|f| =|e^g| = e^{\operatorname{Re} g} = e^v$ where $v = \operatorname{Re} g$ is harmonic. Then $$ 0 = \Delta |f| = \Delta(e^v) = e^v \left(v_{xx} + (v_x)^2 + v_{yy} + (v_y)^2\right) = e^v \left( (v_x)^2 + (v_y)^2\right) $$ so that $v_x$ and $v_y$ are identically zero. It follows that $v$ is constant. Consequently, $|f|$ is constant, which implies that $f$ is constant.

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  • $\begingroup$ How do we get the implication that $f$ is identically zero? $\endgroup$
    – Nick
    Dec 21, 2020 at 15:56
  • $\begingroup$ $ \int_0^{2\pi} |f(z_0+re^{it}| \, dt = 0$ implies that $f$ is zero on the circle with center $z_0$ and radius $r$, for all $r > 0$. $\endgroup$
    – Martin R
    Dec 21, 2020 at 16:04
  • $\begingroup$ Thank you for clarifying! I appreciate the alternate approach to this problem. $\endgroup$
    – Nick
    Dec 22, 2020 at 14:34

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