$f$ is holomorphic in Ω such that $|f|^2$ is harmonic; we need to show that $f$ is constant.

Question

$f$ is holomorphic in Ω such that $|f|^2$ is harmonic; we need to show that $f$ is constant.

solution of the question In the solution attached, I don't really understand the transition between $∆|f(z)|^2 = 4|f'_z(z)|^2$. It would be great someone could answer this.

Answer 1

If $f(z)=u(x,y)+iv(x,y)$ then consider $h(x,y)=u^2(x,y)+v^2(x,y)=|f(z)|^2$

$h_{xx}=2(u_x)^2+2uu_{xx}+2(v_x)^2+2vv_{xx}$

$h_{yy}=2(u_y)^2+2uu_{yy}+2(v_y)^2+2vv_{yy}$

Then $∆|f(z)|^2=h_{xx}+h_{yy}=4((u_x)^2+(v_x)^2)=4|f'_z(z)|^2$ because of the Cauchy-Riemann equations and the fact that if $f$ is holomorphic then both $u$ and $v$ are harmonic

This gives us that $f'_z(z)=0$ (since $h=|f|^2$ is also harmonic from the hypothesis) and this means that $f$ is constant