Given a non-empty connected subset of the complex plane $U$, and a holomorphic function $f : U \to \mathbb C$ satisfying $|f(z)^2 - 1| < 1$ for all $z \in U$, there were four options given, with at least one correct, and they were :
$f$ is constant.
$\mbox{Im} f(z) > 0$ on $U$.
$\mbox{Re} f(z) \neq 0$ on $U$.
$\mbox{Re} f(z)$ has fixed sign on $U$.
I am unable to completely conclude. Certainly the third point implies the fourth by continuity of the real part. Note that $0 < |f(z)| < \sqrt 2$ from above. Since this condition is invariant under changing $f$ to $-f$ I get the feeling that the second point is not true, and the conditions don't seem sufficient for $f$ to be constant.
Hence,my guess would be the third and fourth points, but I need proof for the same, along with counterexamples for the others.
Let $D_{1} = \{w: |w-1|< 1\}$ and $D_{2} = \{w: |w+1|< 1\}$ be the open discs centered at $1$ and $-1$ respectively with radius $1$.
$D_{1}$ lies to the right of the $y$-axis and $D_{2}$ to the left of the $y$-axis and no point on the $y$-axis lies on either $D_{1}$ or $D_{2}$. So $Re(w) > 0$ for all $w \in D_{1}$ and $Re(w) < 0$ for all $w \in D_{2}$. In particular $0$ is not in $D_{1}$ or $D_{2}$.
$|f(z)^{2}-1| < 1 \implies |f(z) - 1||f(z) + 1| < 1 \implies |f(z) - 1| < 1$ or $|f(z)+1| < 1$. Then $D_{1} \cup D_{2}$ is not connected and $f(U) \subseteq D_{1} \cup D_{2}$. Futhermore the connected components are $D_{1}$ and $D_{2}$. We are given that $f$ is holomorphic and $U$ connected, so $f(U)$ is connected. This implies that $f(U) \subseteq D_{1}$ or $f(U) \subseteq D_{2}$. Hence the third and fourth option hold.
The other three options do not hold. Simply taken $U = D_{1}$ and $f$ to be the identity function, i.e. $f(z) = z$.