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