# Find all entire functions $f$ such that $|f|$ is harmonic.

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.

• $u_x$ and $v_x$ are real numbers, so $u_x=v_x=0$ Dec 20, 2020 at 17:30
• 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?
– Nick
Dec 20, 2020 at 17:50
• Yes, that's right. Dec 20, 2020 at 19:35
• It is no wrong by evaluating $\Delta(\sqrt{u^2+v^2})$ directly. saulspatz is right. Dec 20, 2020 at 23:29
• @NikosBagis So do the computations above for $|f|$ rather than $|f|^2$?
– Nick
Dec 21, 2020 at 15:57

$$f(z)=f(x+iy)=u+iv$$. 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.

• You must also show first that f has no zeros in order to justify the differentiation. Dec 22, 2020 at 8:35
• I add the corrections in my answer. Thank you very much. Dec 22, 2020 at 10:09

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.

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