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!

Here is another (perhaps interesting) proof: let $g(z)=(e^{z}+e^{-z})f(z)$. Then $|g(z)|\leq 2A $ because $|e^{z}+e^{-z}| \leq e^{x}+e^{-x} =2\cosh x$. By Louiville's Theorem $g$ is a constant, say $c$. Clearly, $c \neq 0$. We have $(e^{z}+e^{-z})f(z)=c$. You get a contradiction by taking $z=i\pi /2$
