I'm trying to show that if an analytic function $f$ maps a domain to the unit circle $\{z : |z| = 1\}$ then f must be constant. I want to use the Cauchy-Riemann equations.

What I have so far: Let $f(x+iy) = u(x+iy) + i v(x+iy)$, where $u$ and $v$ are real-valued functions. Since $|f(x+iy)| = 1$ for all $x+iy \in U$, then

$[(u(x+iy))^2 + (v(x+iy))^2]^{1/2} = 1$ so $(u(x+iy))^2 + (v(x+iy))^2 = 1$

So then

$2u(x+iy) u_x(x+iy) + 2v(x+iy) v_x(x+iy) = 0$ and $2u(x+iy) u_y(x+iy) + 2v(x+iy) v_y(x+iy) = 0$

Since f is analytic we have $u_x(x+iy) = v_y(x+iy)$ and $u_y(x+iy) = -v_x(x+iy)$.

I've tried plugging these into the above system in various ways and just can't seem to get anything meaningful out. I want to show somehow that $u_x = u_y = 0$ and $v_x = v_y = 0$ so that both $u$ and $v$ must be constant.

Thanks for any help!

  • $\begingroup$ Have you heard of the open mapping theorem ? It says the following : if $D$ is an open connected subset of the plane and $f$ is holomorphic on $D$ then the image of $f$ is either an open set or a singleton. $\endgroup$
    – M.G
    Commented May 9, 2016 at 1:04
  • $\begingroup$ An easy consequence of this theorem is that if any of $|f|, \text{Re}(f)$ and $\text{Im}(f)$ is constant than $f$ is a singleton. In your case $|f(z)| = 1$ therefore $f$ must be constant. $\endgroup$
    – M.G
    Commented May 9, 2016 at 1:08

1 Answer 1


So far so good. Differentiate $u^2 + v^2 = 1$ to get $uu_x + vv_x = 0$ and $uu_y + vv_y= 0$. You can write these in matrix form as $$\begin{bmatrix} u_x & v_x \\ u_y & v_y \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

Apply the Cauchy-Riemann equations to find that $$\begin{bmatrix} u_x & -u_y \\ u_y & u_x \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ at all points in the domain. Since at least one of $u$ and $v$ must be nonzero (the sum of their squares is one) the matrix is singular and thus has determinant zero. Thus $$u_x^2 + u_y^2 = 0$$ so that $u_x = u_y = 0$ throughout the domain making $u$ constant. Likewise $v_x = v_y = 0$ throughout the domain making $v$ constant too.


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