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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 $\forall z(z\in aB_1(1) \Longleftrightarrow z \in B_{|a|}(a))$.2011-10-31
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    yes but I can take the $\phi$ for an interval for radii and circlii and then it will also give me a disc ??2011-10-31

5 Answers 5