- Let \(z_{1}\) and \(z_{2}\) be two complex numbers.
Show that \[ \left|1-\bar{z}_{1} z_{2}\right|^{2}-\left|z_{1}-z_{2}\right|^{2}=\left(1-\left|z_{1}\right|^{2}\right)\left(1-\left|z_{2}\right|^2\right) \]
Show that if \(\left|z_{1}\right|<1\) and \(\left|z_{2}\right|<1\), then \(\left|\frac{z_{1}-z_{2}}{1-\bar{z}_{1} z_{2}}\right|<1 .\)
Assume that \(z_{1} \neq z_{2}\). Show that \(\left|\frac{z_{1}-z_{2}}{1-\bar{z}_{1} z_{2}}\right|=1\) if only if \(\left|z_{1}\right|=1\) or \(\left|z_{2}\right|=1\).
The question as written on the original qual has several errors. What is below is the correct version of the inequality.
Solution
Part 1: For ease of notation, let \(z=z_1\) and \(w=z_2\) We want to show \[ \abs{1- z\bar{w}} - \abs{z-w}^2 = \qty{{1 - \abs{z}^2 }} \qty{{1 - \abs{w}^2}} .\]
So write \[ \abs{1- z\bar{w}} - \abs{z-w}^2 &= (1-z\bar w)(1-\bar z w) - (z-w)(\bar z - \bar w) \\ &= 1 - z\bar w - \bar z w - \abs{z}^2\abs{w}^2 - \abs{z}^2 - \abs{w}^2 + w\bar{z} + z\bar{w} \\ &= 1 - \abs{z}^2\abs{w}^2 - \abs{z}^2 - \abs{w}^2 \\ &=(1-\abs z^2)(1-\abs w^2) .\]
Part 2 and 3: \[ \abs{z-w \over 1 - \bar z w}^2 \leq 1 &\iff 0 \leq \abs{1 - \bar z w}^2 - \abs{z-w}^2 \\ &\iff 0 \leq (1 - \abs z^2 )(1 - \abs w^2) ,\] where we’ve used part 1. But this holds when \(\abs{z}, \abs{w} < 1\): then \(1 - \abs z^2 > 0\) and \(1 - \abs w^2 > 0\), so their product \((1 - \abs z^2)(1 - \abs w^2) > 0\), making the RHS positive. Moreover if \(\abs{z} = \abs{w} = 1\), the RHS is zero, yielding equalities at every step instead.