For an analytic function $f(z)$, $|f(z)^2-1|<1$ implies $\Re f(z)>0$ or $\Re f(z)<0$? Doing a bit of self study, and I'm unsure about a problem. It says,
Suppose $f(z)$ (a complex valued function) is analytic and satisfies the condition $|f(z)^2-1|<1$ in a region $\Omega$. Show that either $\Re f(z)>0$ or $\Re f(z)<0$ throughout $\Omega$. 
I write $f=u+iv$ and suppose to the contrary that $\Re f(z)=0$ at some point $z_0$. Then $f(z_0)^2=-v(z_0)^2$. But $v$ is real valued, and so
$$
|f(z_0)^2-1|=|-v(z_0)^2-1|\geq 1
$$
a contradiction.
What makes me uneasy is I don't see if I used that fact that $f$ is analytic. Did I interpret the question correctly, or did it mean that $\Re f(z)>0$ on all of $\Omega$ or $\Re f(z)<0$ on all of $\Omega$, but doesn't take both positive and negative values? Thanks.
 A: 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...
A: 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.
