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It is to show for an $a\in \mathbb{C}^{\ast}$ that $aB_{1}(1)= B_{|a|}(a)$

where B denotes a disc

Okay, maybe this is correct:

$aB_{1}(1) = a(e^{i\phi}) = ae^{i\phi} = |a|e^{i\phi} = B_{|a|}(a)$

But this seems very wrong!

V

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It makes no sense to write $aB_1(1)=a(e^{i\phi})$ -- the former is a set of points, the latter is a single point that depend on a new variable $\phi$. I think you would be better off proving it as $\forall z(z\in aB_1(1) \Longleftrightarrow z \in B_{|a|}(a))$. –  Henning Makholm Oct 31 '11 at 1:12
    
yes but I can take the $\phi$ for an interval for radii and circlii and then it will also give me a disc ?? –  VVV Oct 31 '11 at 1:16
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5 Answers

let a be a complex number of the form: $a:= u+vi$ and $z:= x+yi$

$B_{1}(1)$ means that $ |z-1| < 1 $ and so $a|z-1| = (u+vi)|z-1| = (u+vi)(\sqrt{(x-1)^{2}+y^{2}} $ so we can write it as $|z-1|u+|z-1|vi< a|1| = |a| = \sqrt{u^{2}+v^{2}}$

This seems to be the wrong route also. ??

V

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Well, your idea is OK. But you should improve some things:

First: If B is a disk and you write $e^{i\phi}$, then it is a parametrization of a circle not the whole disk. But actually a disk is a sum of circles plus the middle point, so the strategy is OK.

Second: As @Adam wrote, you can't write $ae^{i\phi}=|a|e^{i\phi}$, because it is not equal if $a \neq |a|$. However it is true that $ae^{i\phi} \in B_0^{|a|}$. So in that way, you can prove that $aB_0^1 \subseteq B_0^{|a|}$ (1). So what remains is that it is not an inclusion but an equality. Since $a \cdot a^{-1}=1$ it is enough to show that $ a^{-1}B_0^{|a|} \subseteq B_0^1$ and proving that is actually the same as (1).

Third: You have to remember that the middlepoint of your circle is $1$ and when you write $e^{i\phi}$ it generates a circle around $0$, so you have to add $1$ to it.

P.S. There is no $a^{-1}$ for $a=0$, so it is a special case, but a very simple one.

Your second attempt also should lead to success, but you shouldn't multiply $a$ by distance between $z$ and $1$ (notice that it leaded you to write an inequality with complex numbers, which makes no sense!) - you should multiply $a$ by $z$ and look what is the distance between $az$ and $a$ assuming that the distance between $z$ and $1$ is smaller than $1$

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We look at: $|az-a|$ with $|z-1|<1$, how does that help??? –  VVV Oct 31 '11 at 17:45
    
$|az-a|=|a(z-1)|=|a|\cdot|z-1|<|a|$, so $az$ is in $B_{|a|}(a)$, right? –  savick01 Oct 31 '11 at 19:02
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First let me answer your specific question.

  1. Let $z \in B_{1}(1)$, that is to say, $|z - 1| \lt 1$. We want to show that $az \in B_{|a|}(a)$, that is, we want to show that $|az-a| \lt |a|$. But $$ |az - a| = |a| \cdot \underbrace{|z-1|}_{\lt 1} \lt |a|, $$ as we wanted, therefore $aB_1(1) \subset B_{|a|}(a)$.

  2. Conversely, let $w \in B_{|a|}(a)$, that is $|w -a| \lt |a|$. We want to show that $w \in aB_{1}(1)$. Since $a \neq 0$ we can write $$ |a| \gt |w-a| = |a| \cdot \left|\frac{w}{a}-1\right|, $$ so $\left|\frac{w}{a}-1\right| \lt 1$. But this means that $z = \frac{w}{a} \in B_{1}(1)$, so $w = az \in a B_{1}(1)$, hence $B_{|a|}(a) \subset aB_1(1)$.

Putting 1. and 2. together we have $aB_1(1) = B_{|a|}(a)$, as desired.


To make this a bit more useful, we generalize slightly:

Consider $B_{r}(p)$ with $r \gt 0$ and let $a \in \mathbb{C}^\ast$. Then $aB_{r}(p) = B_{|a|r}(ap)$.

Indeed, if $z \in B_{r}(p)$, so $|z-p| \lt r$, then $$ |az-ap| = |a|\cdot|z-p| \lt |a|r, $$ so $aB_{r}(p) \subset B_{|a|r}(ap)$.

Conversely, if $w \in B_{|a|r}(ap)$ then $$ |a|r \gt |w-ap| = |a|\cdot \left|\frac{w}{a} - p\right|, $$ so $\left|\frac{w}{a} - p\right| \lt r$, thus $z = \frac{w}{a} \in B_{r}(a)$ and therefore $w=az \in aB_{r}(p)$. The claimed equality $aB_{r}(p) = B_{|a|r}(ap)$ is proved.

To sum up: multiplying by a complex scalar $a \in \mathbb{C}^\ast$ scales all balls by a factor $|a|$ (i.e., multiplies the radii by $|a|$) and moves the centers from $p$ to $ap$.

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Since this looks like homework, I'll just give you a hint. You are on the right track, but it is not true that $a e^{i\phi} = |a| e^{i \phi}$. Indeed, remember that $a = |a| e^{i \theta}$ for some $\theta$ (the modulus of $a$). Geometrically, what is going on is that multiplication by $a = |a| e^{i \theta}$ scales things by $|a|$ and rotates things by $\theta$.

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I don't see how to describe $B(|a|)(a)$ or $B_{1}(1)$ without using $re^{i\phi}$ ? ? –  VVV Oct 31 '11 at 1:18
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Following the second attempt:

So we look at this system of inequalities:

$aB_{1}(1)$: $|az-a|$ and $|z-1|<1 $

what we want to show is that this equals $|z-a|<|a|$

Then stuck.

Following the first attempt:

for $a \in \mathbb{C}^{*} = \mathbb{C}\backslash \{0\}$

we look at $B_{1}(0) : |z|<1 $ so $|az|< |a|$ which is $B_{|a|}(0)$ this must be finished like this.

But I don't know where I have shown it.

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You don't know because it is quite easy:) Actually what is crucial is $|az|=|a|\cdot|z|$. But it's a ball with its centre in zero, and in your homework there is $1$ and $a$, so it is not so easy. Why you think you want to show that $|az-a|=|z-a|<|a|$? It is false, so showing that may be hard;) One more thing: do you know that $|x-y|$ is the distance between $x$ and $y$? –  savick01 Oct 31 '11 at 19:10
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