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.