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Inequality proof

sweatingbear

Member
May 3, 2013
91
Here's a fun problem proof I came across. Show that

\(\displaystyle \left| \frac { z- w }{1 - \overline{z}w} \right| < 1\)

given \(\displaystyle |z|<1\), \(\displaystyle |w|<1\). I attempted writing z and w in rectangular coordinates (a+bi) but to no avail. Any suggestions, forum?
 
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sweatingbear

Member
May 3, 2013
91
Would you look at that, it was already treated unbeknownst to me (which just nullifies this thread, I'll have it depreciated). Thanks!
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Would you look at that, it was already treated unbeknownst to me (which just nullifies this thread, I'll have it depreciated). Thanks!
Don't agree , others might have different approaches.
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Would you look at that, it was already treated unbeknownst to me (which just nullifies this thread, I'll have it depreciated). Thanks!
Nah, no need to worry about that. It's always good to revisit older problems. The thing I would be interested in is if there's another way to do it than the way I presented in that link.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I neglected to respond to POTW #6 in July 2012, so here is my solution to the problem. It relies on the fact that $\overline{z}z = |z|^2$.

Start with the fact that $(1-|z|^2)(1-|w|^2) > 0$. Then $$(1 - \overline{z}z)(1 - \overline{w}w) > 0,$$ $$1 - \overline{z}z - \overline{w}w + \overline{z}z\,\overline{w}w > 0,$$ $$\overline{z}z + \overline{w}w < 1 + \overline{z}z\,\overline{w}w.$$ Now subtract $\overline{z}w + \overline{w}z\ (=2\mathrm{Re}(\overline{z}w))$ from both sides: $$\overline{z}z - \overline{z}w - \overline{w}z + \overline{w}w < 1 - \overline{z}w - \overline{w}z + \overline{z}z\,\overline{w}w,$$ $$(\overline{z} - \overline{w})(z-w) < (1-\overline{z}w)(1-z\overline{w}),$$ $$|z-w|^2 < |1-\overline{z}w|^2$$ and finally, taking square roots, $$|z-w| < |1-\overline{z}w|,$$ $$\left|\frac{z-w}{1-\overline{z}w}\right|< 1.$$

Edit. Having looked at Chris's solution to POTW #6, I see that my solution is essentially the same as his.
 
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sweatingbear

Member
May 3, 2013
91
The thing I would be interested in is if there's another way to do it than the way I presented in that link.
The given statement can be written $ |z-w| < |1 - \overline{z}w| $, which equivalently is $ |z-w|^2 < |1 - \overline{z}w|^2 $.

Let $ z = a +bi $ and $ w = c + di $. Thus $a^2 + b^2 + c^2 + d^2 < (a^2+b^2)(c^2 + d^2) + 1$, or equivalently, $|z|^2 + |w|^2 < 1 + |z|^2 |w|^2 $. That statement is always true; we could write it as $ p + q < 1 + pq $ for $p<1$ and $q < 1$. This can be shown by expanding $ (1-p)(1-q) > 0 $.
$ \square $