Simple complex question

Asked Sep 15 '15 at 15:32

Active Sep 15 '15 at 15:38

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Prove that $$\left|\frac{z-w}{1-\overline{z}w}\right|<1$$ for $|z|<1$, $|w|<1$ How I do it: First notice that:

$\left|\frac{z-w}{1-\overline{z}w}\right|<1 \leftrightarrow |z-w| < |1-\overline{z}w|$ Since $$|z-w|<|z-w|||\overline{z}| = |z^2-\overline{z}w| < |1-\overline{z}w|$$ we are done. The professor said this one might be a bit tricky though so I'm wondering if I did something wrong because this seemed pretty straightforward, no?

edited Sep 15 '15 at 15:38

asked Sep 15 '15 at 15:32

Lozansky

1,035

The first inequality only holds if $|z| >1$. – Hetebrij Sep 15 '15 at 15:34
@Hetebrij I added the conditions $|z|<1, |w|<1$. – Lozansky Sep 15 '15 at 15:38
Still, if we have $|z| = \frac{1}{2}$ and $w=0$, then we have $\frac{1}{2} < \frac{1}{4}$ in the first inequality. – Hetebrij Sep 15 '15 at 15:39
The inequality holds for all $(z, w) \in \mathbb{C}^2$ with $|z|, |w| < 1$ or $|z|, |w| > 1$, as is discussed here. This is not a duplicate of that question though. – Michael Albanese Sep 15 '15 at 15:40
@MichaelAlbanese Please note the added conditions. – Lozansky Sep 15 '15 at 15:40
@Lozansky Micheal states that your inequality is true, and in fact is also true if $|z|,|w|>1$. – Hetebrij Sep 15 '15 at 15:41
Your argument is not quite correct: on the bottom line, you mean $|z|^2$, not $z^2$, I believe. – user264781 Sep 15 '15 at 15:43
@user264781 Ah yes, that's what I meant. The next step should follow then? – Lozansky Sep 15 '15 at 15:47
@Lozansky Do you know Maximum Modulus theorem? – Dontknowanything Sep 15 '15 at 15:47
@MichaelAlbanese, yes, and that's "if and only if", as the inequality is equivalent to $(|z|^2-1)(|w|^2-1)>0$. – Oskar Limka Sep 15 '15 at 15:57
@Lozansky, it follows, but I think needs some more explanation. Consider what values the real part of $\overline{z}w$ can take. – user264781 Sep 15 '15 at 16:00
@Learner No don't think so. – Lozansky Sep 15 '15 at 16:04
@user264781 Hm how does that matter? It will be between -1 and 1 at least. – Lozansky Sep 15 '15 at 16:05
@Lozansky Suppose $|z|^2=0.5$ and $\overline{z}w=0.9$. Then the inequality would be false. This can't happen, but you need to show why, I don't think its clear that $\overline{z}w$ is closer to $|z|^2$ than to 1. – user264781 Sep 15 '15 at 16:48
@user264781 But I only use that $|z^2|$<1. – Lozansky Sep 15 '15 at 17:19
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possible duplicate of Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^| \neq 0$ and $| {{z-w} \over {1-zw^}}| < 1$ – Cameron Buie Sep 16 '15 at 17:48

Simple complex question

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