Help for complex numbers (1 Viewer)

asianese · Mar 9, 2013

Hint: uses the triangle inequality.

HeroicPandas · Mar 9, 2013

asianese said:
Hint: uses the triangle inequality.

I have tried but it seems dodgy and can you attempt it and show me? (without lettting using cartesian or polar substitutions (eg. z = x+iy or rcis theta)

asianese · Mar 9, 2013

Actually I'm not really sure anymore. The inequality flips on itself if you try to do that.

Hmm

Sy123 · Mar 9, 2013

HeroicPandas said:
$$ Suppose that \alpha $ $ is a fixed complex number and $|\alpha| <1.$ Show that $|z| \leq 1 $ if and only if $\left | \frac{z - \alpha}{1 - \bar\alpha z} \right | \leq 1$

$|z-\alpha|^2 \leq |1-\bar{\alpha}z|^2$

$(z-\alpha)\overline{(z-\alpha)} \leq (1-\bar{\alpha}z) \overline{(1-\bar{\alpha}z)}$

Distribute the conjugates, then expand, we are left with:

$|z|^2 + |\alpha|^2 \leq 1 + |\alpha z|^2$

$(1-|\alpha|^2)|z|^2 + |\alpha|^2 - 1 \leq 0$

Common factor

$(1-|\alpha|^2)(|z|^2-1) \leq 0$

$|\alpha| \neq 1$

$|z|^2 \leq 1$

$|z| \leq 1$

$\square$

RealiseNothing · Mar 9, 2013

Sy123 said:
$|z-\alpha|^2 \leq |1-\bar{\alpha}z|^2$

$(z-\alpha)\overline{(z-\alpha)} \leq (1-\bar{\alpha}z) \overline{(1-\bar{\alpha}z)}$

Distribute the conjugates, then expand, we are left with:

$|z|^2 + |\alpha|^2 \leq 1 + |\alpha z|^2$

$(1-|\alpha|^2)|z|^2 + |\alpha|^2 - 1 \leq 0$

Common factor

$(1-|\alpha|^2)(|z|^2-1) \leq 0$

$|\alpha| \neq 1$

$|z|^2 \leq 1$

$|z| \leq 1$

$\square$

Does that constitute an iff proof though? You'd have to go the other way as well.

Sy123 · Mar 10, 2013

RealiseNothing said:
Does that constitute an iff proof though? You'd have to go the other way as well.

Hmm, yes you are right I think.

So simply going backwards is sufficient to fulfil the iff condition?

Help for complex numbers (1 Viewer)

HeroicPandas

Heroic!

asianese

Σ

HeroicPandas

Heroic!

asianese

Σ

Sy123

This too shall pass

RealiseNothing

what is that?It is Cowpea

Sy123

This too shall pass

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