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Silversonic

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## Homework Statement

For H [itex]\leq[/itex] G as specified, determine the left cosets of H in G.

(ii) G = [itex]\mathbb{C}[/itex]* H = [itex]\mathbb{R}[/itex]*

(iii) G = [itex]\mathbb{C}[/itex]* H = [itex]\mathbb{R}[/itex][itex]_{+}[/itex]

## The Attempt at a Solution

I have the answers, it's just a little inconsistency I don't understand.

For (ii) left cosets are

{r(cos∅ + isin∅); r [itex]\in[/itex] (0,∞)} ∅ [itex]\in[/itex] [0, 2[itex]\pi[/itex])

For (iii)

{r(cos∅ + isin∅); r [itex]\in[/itex] [itex]\mathbb{R}[/itex] \ {0} } ∅ [itex]\in[/itex] [0, [itex]\pi[/itex])

I'm told that the answers are different because the range of r and ∅ are different. It says in (ii) they are "half lines" coming out of the origin and in (iii) they are lines through the origin but excluding the origin itself.

What I don't get, though, is that surely the answer for (ii) should be the answer for (iii)? And vice versa? Basically in (ii) we have H is the set of all the real numbers, while G is the set of all the complex numbers. So when we multiply an element of H by an element of G (and the constant multiplying the euler's forumla is positive), surely r would then range over all the real numbers (excluding zero).

Yet in (iii) we have H is the set of all the positive real numbers, while G is still the set of all the complex numbers (and the constant multiplying the euler's forumla is positive). So when we multiply an element of H by an element of G, surely r would only range over the positive real numbers, as opposed to all the real numbers exluding zero like the answer says?

Does anyone understand my problem?

Thanks.

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