Prove $|f|$ is constant implies $f$ is constant Let $f$ be an entire function (differentiable everywhere over $\mathbb{C})$. Suppose that $|f|$ is constant. Prove that $f$ is constant.
Hint: $|f|\equiv c$ implies that $u^2+v^2\equiv c^2$. Take partial derivatives and apply the Cauchy-Riemann equations.

Proof: Suppose $|f|=c$ for some $c \in \mathbb{C}$. Writing $f(z)=f(x+iy)=u(x,y)+iv(x,y)$, this means
\begin{equation}
u(x,y)^2+v(x,y)^2=c^2
\end{equation}
Please correct me if I'm not differentiating correctly, but taking the partial with respect to $x$ I find that
\begin{equation}
2u(x,y) u_x(x,y)+2v(x,y)v_x(x,y)=0
\end{equation}
Simiarly, taking the partial with respect to $y$ shows that
\begin{equation}
2u(x,y) u_y(x,y)+2v(x,y)v_y(x,y)=0
\end{equation}
Is this correct so far?
Edit: Thanks to AlexR for his tip. I continue the proof below.
Substituting the C-R equations ($u_x=v_y$ and $v_x=-u_y$), the first equation gives two possible equations:
\begin{equation}
2\left(u(x,y) v_y(x,y)+v(x,y)v_x(x,y)\right)=0
\end{equation}
or
\begin{equation}
2\left(u(x,y) u_x(x,y)-v(x,y)u_y(x,y)\right)=0
\end{equation}
Similarly, depending on the choice of substitution, the second equation gives two possible equations:
\begin{equation}
2\left(v(x,y)v_y(x,y)-u(x,y)v_x(x,y)\right)=0
\end{equation}
or
\begin{equation}
2\left(u(x,y) u_y(x,y)+v(x,y)u_x(x,y)\right)=0
\end{equation}
(does it matter which equations we choose to use?)

A short, but different proof can be given by using the fact that a non-constant holomorphic function is an open map: By assumption $f(\mathbb C)$ is contained in a circle in $\mathbb C$ with radius $c$. Any non-empty subset of such circle is not open in $\mathbb C$, hence $f$ is constant, q.e.d.
