Edited: Key first step at the bottom, link to full solution provided in first comment.
I'm struggling to show that complex numbers inequality: $$ |z - w| < |1 - z\cdot \overline{w}| \quad z, w \in \mathbb{C}, \quad \overline{w}=Re(w) - i\cdot Im(w) $$ is equivalent to condition below: $$ (|z| < 1 \land |w| < 1) \lor (|z| > 1 \land |w| > 1) $$ So far I used $|z-w|=|w-z| \rightarrow |z-w|+|w-z|<2\cdot |1 - z\cdot \overline{w}|$ representation to get: $$ \sqrt {(z-w)\cdot \overline{(z-w)}}+\sqrt {(w-z)\cdot \overline{(w-z)}}< 2\cdot \sqrt {(1 - z\cdot \overline{w})\cdot \overline{(1-z\cdot \overline{w})}} $$ which didn't really lead anywhere, but: $|z-w|+|w-z|<2\cdot |1 - z\cdot \overline{w}|$ with triangle inequality leads to: $$ 0=|z-w+w-z|\le ||z-w|+|w-z|<2\cdot |1 - z\cdot \overline{w}| \rightarrow 0<|1 - z\cdot \overline{w}| $$ So, if I'm correct, I got to: $$ 0<|1 - z\cdot \overline{w}| $$ Now, that $|x|=\sqrt {x\cdot \overline{x}}$, $$ 0< \sqrt{(1 - z\cdot \overline{w})\cdot \overline{(1-z \cdot \overline{w})}} = \sqrt{(1 - z\cdot \overline{w})\cdot (\overline{1}-\overline{z} \cdot \overline{\overline{w}})} $$ Ultimately, I receive: $$ 0<\sqrt{(1 - z\cdot \overline{w})\cdot (1-\overline{z} \cdot w)} $$ Which after squaring both sides takes form: $$ 0<(1 - z\cdot \overline{w})\cdot (1-\overline{z} \cdot w) $$ Not sure how to proceed though or even if it's the right direction.
Edited: Disregard the above, the key is to notice that: $$ |z - w|= (z-w)(\overline{z}-\overline{w}) $$ $$ |1-z \cdot \overline{w}|=(1-z \cdot \overline{w}) \cdot (1-\overline{z} \cdot w) $$ as seen in: Inequalty with complex numbers provided in first comment.
