An analytic function such that $|f^2(z)-1|=|f(z)-1|\,|f(z)+1|<1.$ Let $f$ be an analytic function such that $$|f^2(z)-1|=|f(z)-1|\,|f(z)+1|<1.$$ on a non empty connected set $U$. Then 
(A) $f$ is constant.
(B) $Im (f)>0$ on $U$.
(C) $Re(f)\not = 0$ on $U$.
(D) $Re(f)$ is of fixed sign on $U$.
I tried through taking $f(z)=u(x,y)+iv(x,y)$ and putting on the relation, but I could not conclude anything from there.
 A: Only the third one and the fourth one are correct in general, so I'll start with them.
$$\begin{align}
\Re(f(z))=0 & \implies |1-i\Im(f)|\cdot|1+i\Im(f)|<1 \\
 & \implies (1+\Im(f(x))^2)<1 \\
 & \implies \Im(f(x))^2 < 0
\end{align}$$
This can't be since $\Im(z)$ is a real number. So $(3)$ is true.
Since $f$ is analytic, it is also continuous. Therefore its real part is continuous. If it wasn't of fixed sign on $U$. You could find a continuous path between two points $z_1,z_2 \in U$ such that $\Re(f(z_1))<0$ and $\Re(f(z_2))>0$. You could deduce that $\exists z$ on that path such that $\Re(f(z))=0$, which can't be, as we've proven earlier. So $(4)$ is also true.
Let $f(z) := \displaystyle\frac{z}{2}+1$ on $U=U=D(0,1)\cap D(-4,4)$, which is connected (convex, even.)
$$\begin{align}
|f(z)^2-1| & =|\frac{z^2}{4}+z| \\
 & = \frac{1}{4}|z|\cdot|z+4|<\frac{1}{4}\cdot1\cdot4=1
\end{align}$$
So $f$ satisfies our condition. But $f$ is neither constant, nor is its imaginary part positive (or even of fixed sign) on all of $U$.
