Show that a function $f(z)$ holomorphic on an open set $\Omega$ cannot have $|f(a + bi)|^2$ = $\frac{K}{cosha}$ How can we show that a function $f(z)$ holomorphic on an open set $\Omega$ cannot have $|f(a + bi)|^2$ = $\frac{K}{cosha}$ for all $a + bi \in \Omega$, where $K \neq 0$ is a real constant? Please try to keep answers elementary, without using integration. 
 A: Let's consider the more general problem of determining all differentiable functions $g(x)$ that are such that there exist a homomorphic function $f(z)$ satisfying
$$|f(x+iy)| = g(x)$$
Before we start I should warn that this solution cannot be said to be "elementary, without using integration". Lets start by writing $f(z) = u(x,y) + i v(x,y)$. We then have 
$$|f(z)|^2 = g^2(x) \implies u^2(x,y) + v^2(x,y)  = g^2(x)$$
Taking partial derivatives of this equation with respect to $x$ and $y$ gives us
$$u u_y + vv_y = 0~~~~\text{and}~~~uu_x + vv_x = g(x)g'(x)$$
Now if $f$ is holomorphic then the Cauchy–Riemann equations can be used to simplify the equations above. The first equation gives us
$$\frac{u_x}{u} = \frac{v_x}{v} \implies v(x,y) = u(x,y) F(y)$$
for some differentiable function $F(y)$. The second equation gives us
$$u^2\frac{\partial}{\partial y}\left(\frac{v}{u}\right) = \frac{g'(x)}{g(x)}(u^2+v^2) \implies F'(y) = \frac{g'(x)}{g(x)}\left(1+F^2(y)\right)$$
and by integrating we get
$$F(y) = \tan\left(\frac{g'(x)}{g(x)}y + C\right)$$
for some constant $C$. Since $F(y)$ is a function of $y$ only we must have $\frac{g'(x)}{g(x)} = b$ for some real constant $b$ which gives us that $g(x)$ has to be on the form $g(x) = e^{a+bx}$ with $a,b\in\mathbb{R}$.
