Vtyjkyuk

Suppose that $Omega$ is a domain that $f:OmegarightarrowmathbbC$ is analytic in $Omega$, and that $|f(z)|=1 forall zinOmega$. By using the CR equations, show that $f$ is constant.

My attempt:
beginalign
|f(z)|&=1 \
u^2(x,y)+v^2(x,y)&=1
endalign
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 $(3)$:
beginalign
2uu_y+2vv_y-2uv_y+2vu_y&=0 \
u_y(2u+2v)-v_y(2v-2u)&=0
endalign
But I am stuck at this point.

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 ?

If you read $(2)$ and $(3)$ as dot products, you see that the vector $pmatrixu\v$ is orthogonal to both $pmatrixu_y\v_y$ and $pmatrixv_y\-u_y$. 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 $pmatrixu\v=0$, or $pmatrixu_y\v_y=0$ and by the CR equations $pmatrixu_x\v_x=0$ as well.

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