Complex number condition on the modulus The set of all complex numbers $(z_1,z_2)$ which satisfy 
$$\frac{|z_1 -z_2|}{|1-\overline{z_1}z_2|} \lt 1 $$
is?
(Here $\overline{z_1}$ is $z_1$'s cojugate.)
I attempted to write $z_1$ a as $x_1 + iy_1$ and $z_2$ as $x_2 + iy_2 $ and tried to simplify. But then it didn't seem to work. How do I go about this?

The key to answering this question is knowing that for $a \in \mathbb{D} = \{z \in \mathbb{C} \mid |z| < 1\}$, $$T_a(z) = \frac{z - a}{1 - \overline{a}{z}}$$ is an automorphism of $\mathbb{D}$. In addition, if $|z| = 1$, then $|T_a(z)| = 1$, and if $|z| > 1, z \neq \frac{1}{\overline{a}}$, then $|T_a(z)| > 1$; note, we can make sense of $T_a(z)$ at $z = \frac{1}{\overline{a}}$ if we consider $T_a$ as an automorphism of the Riemann sphere but that isn't really necessary here.
First note that 
$$\frac{|z_1 - z_2|}{|1-\overline{z_1}{z_2}|} = \frac{|z_2 - z_1|}{|1-\overline{z_1}{z_2}|} = \left|\frac{z_2 - z_1}{1-\overline{z_1}{z_2}}\right| = |\,T_{z_1}(z_2)|.$$
So, if $\underline{|z_1| < 1}$, $(z_1, z_2)$ satisfies the desired inequality if and only if $|z_2| < 1$.
Now note that 
$$|\,T_{z_1}(z_2)| = \left|\frac{z_2 - z_1}{1-\overline{z_1}{z_2}}\right| = \left|\frac{\frac{z_2}{\overline{z_1}z_2} - \frac{z_1}{\overline{z_1}z_2}}{\frac{1}{\overline{z_1}z_2}-1}\right| = \left|\frac{\frac{z_1}{\overline{z_1}z_2} - \frac{1}{\overline{z_1}}}{1-\frac{1}{\overline{z_1}z_2}}\right| = \left|\frac{\frac{z_1}{\overline{z_1}z_2} - \frac{1}{\overline{z_1}}}{1-\frac{1}{z_1}\frac{z_1}{\overline{z_1}z_2}}\right| = \left|\,T_{\frac{1}{\overline{z_1}}}\left(\frac{z_1}{\overline{z_1}z_2}\right)\right|.$$
So, if $\underline{|z_1| > 1}$ (so that $\left|\frac{1}{\overline{z_1}}\right| < 1$), $(z_1, z_2)$ satisfies the desired inequality if and only if $\left|\frac{z_1}{\overline{z_1}z_2}\right| < 1$. As
$$\left|\frac{z_1}{\overline{z_1}z_2}\right| = \frac{|z_1|}{|\overline{z_1}||z_2|} = \frac{|z_1|}{|z_1||z_2|} = \frac{1}{|z_2|}$$
we see that if $|z_1| > 1$, $(z_1, z_2)$ satisfies the desired inequality if and only if $\frac{1}{|z_2|} < 1$, that is, $|z_2| > 1$.
Finally, if $\underline{|z_1| = 1}$, then 
\begin{align*}
\left|\frac{z_1 - z_2}{1-\overline{z_1}z_2}\right|^2 &= \frac{z_1 - z_2}{1-\overline{z_1}z_2}\overline{\left(\frac{z_1 - z_2}{1-\overline{z_1}z_2}\right)}\\
&\\
&= \frac{z_1 - z_2}{1-\overline{z_1}z_2}\frac{\overline{z_1} - \overline{z_2}}{1-z_1\overline{z_2}}\\ 
&\\
&= \frac{|z_1|^2 - z_1\overline{z_2} - \overline{z_1}z_2 + |z_2|^2}{1 - z_1\overline{z_2} - \overline{z_1}z_2 + |z_1|^2|z_2|^2}\\
&\\
&= \frac{1 - z_1\overline{z_2} - \overline{z_1}z_2 + |z_2|^2}{1 - z_1\overline{z_2} - \overline{z_1}z_2 + |z_2|^2}\\
&\\
&= 1.
\end{align*}
So, if $|z_1| = 1$, there is no $z_2$ such that $(z_1, z_2)$ satisfies the desired inequality.
Therefore, if 
$$S = \left\{(z_1, z_2) \in \mathbb{C}^2 : \frac{|z_1 - z_2|}{|1-\overline{z_1}z_2|} < 1\right\}$$ 
then $S = \{(z_1, z_2) \in \mathbb{C}^2 : |z_1| < 1, |z_2| < 1\} \cup \{(z_1, z_2) \in \mathbb{C}^2 : |z_1| > 1, |z_2| > 1\}$.
