For an analytic function $f(z)$, $|f(z)^2-1|<1$ implies $\Re f(z)>0$ or $\Re f(z)<0$?
Solution 1:
I don't think $f$ even needs to be analytic - only continuous on $\Omega$. In any case, I think the latter interpretation you pose is the correct one: the problem wants an either-or on all of $\Omega$, i.e.
$$\left(\;\forall z\in\Omega:\operatorname{Re} f>0 \;\right)\text{ or }\left(\;\forall z\in\Omega:\operatorname{Re} f<0 \;\right).$$
This isn't too much more work than what you've already done. You've shown the real part can't be zero; now assume there are two arguments $z$ and $w$ in $\Omega$ with $\operatorname{Re} f(z)<0<\operatorname{Re}f(w)$. Since $\Omega$ is connected, there is a path going from $z$ to $w$ contained in $\Omega$. Consider how $\operatorname{Re}f$ looks on this path...
Solution 2:
Let $D$ be the open disk centered at $1$ and with radius $1$, and let $D'=\{w:w^2\in D\}$. Since $f(z)^2\in D$ for all $z\in\Omega$, $f(z)\in D'$. What do we know about $D'$? The following two facts are easy to prove:
- $D'$ is symmetric with respect to the imaginary axis;
- no point in the imaginary axis is in $D'$.
Thus, $\{w\in D':\Re z>0\}\ $ and $\{w\in D':\Re z<0\}\ $ are disconnected; since $f(\Omega)$ is connected, it must be contained in one of them.
In fact, $D'$ is the interior of a lemniscate.