Normal Families Suppose $\mathcal{F}$ is a famliy of analytic functions of the unit disc.   Suppose also that
$( Re(f(z)) )^2 \ne ( Im(f(z)) ) $for all $|z|<1$ and all $f \in \mathcal{F}$.   
It follows from the Fundamental Normality test   that $\mathcal{F}$ is a normal family. 
Is there a for elementary way of showing $\mathcal{F}$ is a normal family without invoking the Fundamental Normality Test?
 A: Your condition means that  if $\gamma$ is the parabola $x^2-y=0$ then for any $f\in\mathcal{F}$ we have $f(\mathbb{D})\cap\gamma=\emptyset$. Thus 
$\mathcal{F}=\mathcal{G}\cup \mathcal{H}$ where each $g$ in $\mathcal{G}$ maps the disk in the open set $x^2-y>$0 and each $h$ in $\mathcal{H}$ maps the disk in the open set $x^2-y<0$. 
Now just note that these two open sets are biholomorphic to the unit disk by uniformization theorem, thus $\mathcal{G}$ and $\mathcal{H}$ are normal families.  If you don't want to use uniformization theorem try to prove by hand that the open sets are biholomorphic to the disk.
A: Actually, you can avoid uniformisation, and just use the following simple normality criterion:
Theorem (Montel). If there is a nonempty open set $U$ such that all functions in $\mathcal{F}$ omit all points of $U$, then $\mathcal{F}$ is normal.
(By postcomposing with a M\"obius transformation, we can assume that the family $\mathcal{F}$ is uniformly bounded, and the claim follows e.g. from Marty's theorem.)
Since each of your functions omits one of the two complementary domains of the mentioned parabola, your family is the union of two normal families, and hence itself normal.
