<s>[INST] Prove that $|z_{1}+z_{2}| < |1+\overline{z_{1}}\cdot z_{2}|\;$ if $|z_{1}| < 1$ and $|z_{2}| < 1$. Given $z_{1},z_{2} \in \mathbb{C}$
Prove that $|z_{1}+z_{2}| < |1+\overline{z_{1}}\cdot z_{2}|\;$, if $|z_{1}| < 1$ and $|z_{2}| < 1$.
My Professor gave my classmates and I, the following answer:
$|z_{1} + z_{2}| \leq |z_{1} + z_{2}|\cdot |\overline{z_{1}}|\;$ $\Leftarrow$ The part I'm having trouble with
$|z_{1} + z_{2}|\cdot |\overline{z_{1}}| = |z_{1}\overline{z_{1}} + z_{2}\overline{z_{1}}| = ||z_{1}|^{2} + z_{2}\overline{z_{1}}| < |1+\overline{z_{1}}z_{2}|$
$\therefore |z_{1}+z_{2}| < |1+\overline{z_{1}}z_{2}|$
I don't understand why the line highlighted is correct.
Since $|\overline{z_{1}}| = |z_{1}| < 1$, $\;|\overline{z_{1}}|\cdot  |z_{1} + z_{2}|< 1\cdot |z_{1} + z_{2}|$ which contradicts the proof above.
If my reasoning is correct, How could I prove this statement?
 [/INST] Square both sides:
\begin{align*}
|z + w| < |1 + \overline{z}w| & \Longleftrightarrow |z + w|^{2} < |1 + \overline{z}w|^{2}\\\\
& \Longleftrightarrow (z + w)(\overline{z} + \overline{w}) < (1 + \overline{z}w)(1 + z\overline{w})\\\\
& \Longleftrightarrow z\overline{z} + z\overline{w} + \overline{z}w + w\overline{w} < 1 + z\overline{w} + \overline{z}w + z\overline{z}w\overline{w}\\\\
& \Longleftrightarrow |z|^{2} + |w|^{2} < 1 + |z|^{2}|w|^{2}\\\\
& \Longleftrightarrow |z|^{2}(|w|^{2} - 1) + 1 - |w|^{2} > 0\\\\
& \Longleftrightarrow (1 - |z|^{2})(1 - |w|^{2}) > 0
\end{align*}
which clearly holds based on the assumption that $|z| < 1$ and $|w| < 1$.
Hopefully this helps!
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