We want to show that the function $f$, holomorphic on some domain $D$, is constant in the following cases:

  1. $z \mapsto \overline{f(z)}$ is holomorphic
  2. $z \mapsto f(\overline{z})$ is holomorphic
  3. $|f(z)|$ is constant



Suppose $g(z)= \overline{f(z)}$ is holomorphic. $f$ is holomorphic so $f$ satisfies the Cauchy-Riemann equations in $D$.

So letting $u=\Re(f), v=\Im(f)$, we have that $u_x=v_y$ and $u_y=-v_x$.

Now $g(z)= \overline{f(z)}$, so $g=u-iv$. Let $w=-v$. Then $u_x=w_y=-v_y$ and $u_y=-w_x=v_x$.

So, $-v_x=v_x$ and so $v$ is a function of $y$ only. However, $v_y=-v_y$ and so $v$ is constant. A similar results holds for $u$ and we are done.

So I think 1. is prove, but I can't see how 2 & 3 work. Help would be appreciated.


Suppose $f = u + iv$.

  1. In this case $u = Re f$ is holomorphic, but it is real valued, so $\frac{\partial u}{\partial y} = \frac{\partial u}{\partial x} = 0$, i.e. it is a constant function.

  2. Cauchy-Riemann equation have the following form $\frac{\partial f}{\partial \bar z} = 0$, so then if $f(\bar z)$ is holomorphic, then $\frac{\partial f}{\partial z} = 0$ as well, which means that $f$ is a constant function.

  3. Cauchy-Riemann equation tells you that $u_x = v_y, u_y = -v_x$. $|f|^2 = u^2 + v^2$ is a constant tells you that $uu_x + vv_x = uu_y + vv_y = 0$, combining them, you have a system of equations $uu_x - vu_y = 0, uu_y + vu_x = 0$, which has determinant $u^2 + v^2$. If $|f| = 0$, then $f = 0$ a constant. If $f \neq 0$, then the system of equations have a unique trivial solution, so again $f$ is a constant.


Use the Cauchy-Riemann equations also for 2) and 3). For 2) note that $ g $ has $ g (z)=f (x-iy) $. For 3) that 3) implies that $ u^2+v^2 $ is constant.


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