Show $|f(z)|=1$ is constant

Question

Suppose that $\Omega$ is a domain that $f:\Omega\rightarrow\mathbb{C}$ is analytic in $\Omega$, and that $|f(z)|=1 \ \forall z\in\Omega$. By using the CR equations, show that $f$ is constant.

My attempt: \begin{align} |f(z)|&=1 \\ u^2(x,y)+v^2(x,y)&=1 \end{align} Differentiating w.r.t $x$: $$2uu_x+2vv_x=0 \ \ \ \ \ \ \ \ \ (1)$$ Differentiating w.r.t $y$: $$2uu_y+2vv_y=0 \ \ \ \ \ \ \ \ \ (2)$$ From $(1)$, $$2uv_y-2vu_y=0 \ \ \ \ \ \ \ \ \ (3)$$ Now equating $(2)$ and $(3)$: \begin{align} 2uu_y+2vv_y-2uv_y+2vu_y&=0 \\ u_y(2u+2v)-v_y(2v-2u)&=0 \end{align} But I am stuck at this point.

Answer 1

From $(1)$ we get

$u^2u_x+uvv_x=0 \ \ \ \ \ \ \ \ \ (a)$

and from $(2)$ we get

$uvu_y+v^2v_y=0 \ \ \ \ \ \ \ \ \ (b)$.

Since $u_y=-v_x$ it follows from $(b)$ that

$-uvv_x+v^2v_y=0 \ \ \ \ \ \ \ \ \ (c)$.

Addition of $(a)$ and $(c)$ gives

$u^2u_x+v^2v_y=0$.

Since $v_y=u_x$ it results that

$0=(u^2+v^2)u_x=u_x$.

In the same way we derive $0=v_y=u_y=v_x$.

Can you proceed ?

Answer 2

If you read $(2)$ and $(3)$ as dot products, you see that the vector $\pmatrix{u\\v}$ is orthogonal to both $\pmatrix{u_y\\v_y}$ and $\pmatrix{v_y\\-u_y}$. Now these last two vectors are orthogonal to each other. In dimension $2$, you can't have three non-zero vectors any two of which are orthogonal. Therefore either $\pmatrix{u\\v}=0$, or $\pmatrix{u_y\\v_y}=0$ and by the CR equations $\pmatrix{u_x\\v_x}=0$ as well.

In both cases $f$ is constant since a domain is connected.