Inequality Proof: Fun Problem | z,w <1 | Forum

[SweatingBear]

Here's a fun problem proof I came across. Show that

$$\left| \frac { z- w }{1 - \overline{z}w} \right| < 1$$

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

 

[Chris L T521]

Ironically, I thought it was a fun inequality problem too! XD

http://mathhelpboards.com/potw-graduate-students-45/problem-week-6-july-9th-2012-a-1408.html

 

[SweatingBear]

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

 

[alyafey22]

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]

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]

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.

 

[SweatingBear]

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 $