Given that $f:D\to \mathbb{C}$ is holomorphic on $D$ and $\left| f \right|$ is constant on $D$, then $f$ is constant on $D$. Where $D$ is a connected open set.

My approach:

Write $f=u(x,y)+iv(x,y)$. Since $f$ is holomorphic on $D$, it is complex-differentiable on $D$, which implies that the $\mathbb{R}^2$-Jacobian of $f$ is defined on all of $D$. That is,

$$Df=\begin{bmatrix} u_x & v_x \\ u_y & v_y \end{bmatrix}$$

Now, since $\left| f \right|$ is constant, it must be true that

$$\left| Df \right|=\det\left(\begin{bmatrix} u_x & v_x \\ u_y & v_y \end{bmatrix}\right)=u_xv_y-v_xu_y=0 \text{ (*)}$$

Since $f$ is $\mathbb{C}$-differentiable on $D$, the Cauchy-Riemann equations hold for $f$. Thus,

$$ u_xv_y-v_xu_y=u_x^2+u_y^2=0 \iff u_x^2=-u_y^2$$

Thus $u_x=u_y\equiv 0 \implies u(x,y)$ is constant. Similarly, we can make the substitutions in (*) to obtain $v_y^2=-v_x^2$ and $v_x=v_y\equiv 0$. This implies that both $u$ and $v$ are constant, and thus $f$ is constant.

Please let me know if my approach is correct. I'm just a little concerned about my treatment of the Jacobian of $\left|f\right|$ as the determinant of the Jacobian of $f$, but I think this should be correct.

  • 1
    $\begingroup$ Another approach is to use the open mapping theorem which states that a non constant analytic function is an open map. $\endgroup$
    – copper.hat
    Commented Jan 31, 2017 at 6:07

2 Answers 2


The implication "$u_x=u_y\equiv 0 \implies u(x,y)$ is constant" only works if $D$ is connected !

Example: let $D_1$ and $D_2$ open sets with $D_1\cap D_2 = \emptyset$ and $D:= D_1 \cup D_2$. Then define $f:D \to \mathbb C$ by

$f(z)=1$ if $z \in D_1$ and $f(z)=-1$ if $z \in D_2$.

Then $f$ is holomorphic on $D$, $|f|$ is constant on $D$, but $f$ is not constant on $D$.

So, your proof is correct, if $D$ is a region (open and connected).

  • $\begingroup$ Indeed, I forgot to mention that $D$ is connected. Let me edit that. Sorry. $\endgroup$
    – sequence
    Commented Jan 31, 2017 at 5:29

Here is a way elaborating @copper.hat's comment:

$f(D)$ is a subset of the circle of radius $|f|$ centered at the origin, hence $f(D)$ is not open, it follows by the open mapping theorem, that $f(D)$ must be a singleton set. In other words, $f$ must be a constant map.


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