No analytic function $f$ has modulus $|f(z)|=1/\cosh(\Re z)$ An analytic function $f(z) = f(x+iy)$ in $\mathbb{C}$ cannot have modulus $\frac{A}{\cosh(x)}$ for some constant $A \neq 0$.
Can we do so simply using the Cauchy-Riemann Equations?
I tried working by contradiction:
Say there is such a $f(z)$. Given $f(z) = u(x,y) + i v(x,y)$ is analytic, it satisfies: $u_x = v_y$ and $u_y = -v_x$
We also see that: $$|f(z)|^2 = u(x,y)^2 + v(x,y)^2 = (\frac{A}{\cosh x})^2$$
I tried reaching a contradiction, but seem to be getting lost in a mess of reformulations. 
My start point is: 
$$u_x = v_y \text{ and } u_y = -v_x$$
along with
$$u(x,y)u_x + v(x,y)v_x = -A^2 \frac{sinh x}{(cosh x )^3}$$
and 
$$u(x,y)u_y + v(x,y)v_y = 0$$
I would appreciate some hints!

So
$$
\begin{align*}
uu_x+vv_x&=-A^2\frac{\sinh x}{\cosh^3 x}\\
-uv_x+vu_x&=0\\
\end{align*}
$$
So
$$
u_x=-u\frac{\sinh x}{\cosh x}
$$
(remember $u^2+v^2=\dfrac{A^2}{\cosh^2 x}$) and similarly $v_x=-v\dfrac{\sinh x}{\cosh x}$.  So
$$
u=\frac{1}{\cosh x}\cdot f(y)\text{ and }v=\frac{1}{\cosh x}\cdot g(y)
$$
and there are no $g$ that gives $u_x=v_y$.
