Assuming either $|z|=1$ or $|w|=1$ , prove that
|$\frac{z-w}{1-\bar zw}$| $=1$
When assuming that only $|z|=1$, we can easily deduce that $|{z-w}|=|{1-\bar zw}|$
But it's more complicated when we take the case where only $|w|=1$
Any hints please ?
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what is the question here? – Dr. Sonnhard Graubner Feb 04 '18 at 19:21
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In reality, we can easily deduce $z-w=z(1-\bar zw)$, if $|z|=1$. – Feb 04 '18 at 19:25
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It is "more complicated", you say? How so? The function is symmetrical! – Did Feb 04 '18 at 20:37
3 Answers
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If $|w|=1$, then$$\left|1-\overline zw\right|=\left|\overline ww-\overline zw\right|=\left|\overline w-\overline z\right|.\left|w\right|=|w-z|=|z-w|.$$
José Carlos Santos
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@Karen I made a mistake but I've already edited my answer. Is it clear now? – José Carlos Santos Feb 04 '18 at 19:39
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A series of algebraical manipulations show that $$ \frac{\lvert{w-z}\rvert}{\lvert{1 - \overline{w}z}\rvert} \leq 1 $$ if and only if \begin{align*} \lvert{w-z\rvert}^2 \leq \lvert{1 - \overline{w}z\rvert}^2 &\iff \lvert{w\rvert}^2 - \overline{w}z - w\overline{z} +\lvert{z\rvert}^2 \leq 1-\overline{w}z-w\overline{z} + \lvert{w}^2\rvert{z}^2 \\ &\iff 0 \leq (1-\lvert{w\rvert})(1-\lvert{z\rvert}) \end{align*} Indeed, the above inequality holds whenever $\lvert{w\rvert}, \lvert{z\rvert} \leq 1$. Furthermore, we have a strict inequality if $\lvert{w\rvert}, \lvert{z\rvert} < 1$ and equality if $\lvert{z\rvert} = 1$ or $\lvert{w\rvert}=1$.
Quoka
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If $|w|=1$, then $|\bar{w}|=1$, therefore $$|1-\bar{z}w|=|1-\bar{z}w|\cdot|\bar{w}|=|\bar{w}-|w|^2\bar{z}|=|\bar{w}-\bar{z}|=|w-z|$$
Arnaud Mortier
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