I was reading Dummit and Foote, where they have discussed that $\mathbb{Q}^+$ and $\mathbb{R}^+$ form group under usual multiplication while $\mathbb{Z}^+$ does not (as no inverse exists). I was wondering what is the behaviour of $\mathbb{C}^+$ under usual multiplication? Does it form a group?

edit: I was also wondering what would be the order of any element in the group of $\mathbb{C}-${0} under usual multiplication? (except identity, which is obviously 1) Thanks...

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    $\begingroup$ What is $\mathbb{C}^+$ for you? $\endgroup$ – Sigur Oct 17 '14 at 1:20
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    $\begingroup$ Probably not, since $(1+2i)^2 = -3 + 4i$. $\endgroup$ – Yuval Filmus Oct 17 '14 at 1:20
  • $\begingroup$ The unitary complex numbers makes a group under complex multiplication. This group is the unitary circle $\mathbb{S}^1$. $\endgroup$ – Sigur Oct 17 '14 at 1:21
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    $\begingroup$ This has been hinted at in one of the comments and answers, but just to be very explicit: there is no universal notion of a "positive complex number". $\endgroup$ – RghtHndSd Oct 17 '14 at 1:29

Think about this geometrically: multiplying a given point (represented by a complex number) by a complex number can be represented as scaling the point and rotating it about the origin (or vice versa). From here it's easy to see that arbitrary rotations can be built from powers of arbitrarily small ones, and that no half-plane of the complex numbers can be multiplicatively closed. In fact, it's easy to go further with a little analysis: since the principal branch of $z^{1/n}$ converges to $1$ as $n\to\infty$ (why?), the multiplicative closure of any open neighborhood of $1$ is the entire (non-zero) complex plane $\mathbb{C}\backslash\{0\}$.

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  • $\begingroup$ So, does it means that complex numbers are not multiplicatively closed except in the unitary circle, and thus don't form group? $\endgroup$ – Ritu Oct 17 '14 at 1:42
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    $\begingroup$ @Ritu Not at all - the full set of complex numbers is multiplicatively closed, and there are many discrete or even dense subsets (for instance, the set of all nonzero algebraic numbers) that are multiplicatively closed, and which even form a group. But none of those sets can include an open neighborhood of $1$ without being the whole complex plane. (In fact, they can't include an open neighborhood of any point on the complex plane; see if you can understand why!) $\endgroup$ – Steven Stadnicki Oct 17 '14 at 1:49
  • $\begingroup$ @hardmath Mea culpa, yes - I should (essentially) say 'punctured' here. $\endgroup$ – Steven Stadnicki Oct 17 '14 at 2:27
  • $\begingroup$ okay, now I get it.. Thanks:) $\endgroup$ – Ritu Oct 17 '14 at 2:45

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