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Unit Disc representation help

Show that $aB_{1}(1) = B_{|a|}(a)$ for an $a\in \mathbb{C}^{*}$

This is a remark in a book called theory of complex functions that I want to prove, but I am stuck at many things I tried:

  1. I tried to put $z=e^{i\phi}$ . That was not correct

  2. I tried to put $B_{1}(1) : |z-1|<1 $ and then multiply it with $a:= x+iy$ so: $a|z-1| < a$ but because no |a| exists in this this can not be right.

Does anybody see a way to show this equation. Tell me. Please.


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marked as duplicate by t.b., Gerry Myerson, Hans Lundmark, Henning Makholm, Did Oct 31 '11 at 13:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What's the difference to this question? –  t.b. Oct 31 '11 at 11:15

1 Answer 1

Let's think of the set of vectors that form $ B_1(1).$ Instead of naively drawing vectors straight to the points of the disc itself, pick an arbitrary point in the disc, draw a vector (call it $z_1$) from the origin to the center of the disc at $z=1$, then from the center of the disc out to the arbitrary point in the disc (call this one $z_2$).

As $a(z_1+z_2) = az_1 + az_2 $, the multiplication by $a$ changes $z_1$ into the vector from $0$ to $a.$ Since $z_2$ could have lengths in $[0,1)$, the vector $az_2$ could have lengths $[0,|a|)$. Also, $z_2$ could point in any direction, and thus $ az_2$ can do so as well. Hence, the set formed is the set spanned by the vectors whose tail is at $a$, which lengths $ [0, |a| ) $ and free to point in any direction, which is precisely $ B_{|a|} (a) .$

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