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Let $\mathbb C^*$ denote the multiplicative group of non-zero complex numbers and let $P$ denote the subgroup of positive real numbers. Identify the quotient group.

My thought $$\frac{\mathbb C^*}{P}=\{P,-P,iP,-iP\}.$$ Is it right?

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Which of those 4 cosets contains a complex cube root of 1? – Gerry Myerson Jan 25 '14 at 14:42
@GerryMyerson $P$ contains $1$.. – tattwamasi amrutam Jan 25 '14 at 14:45
@GerryMyerson : I do not understand your idea.... :O could you please explain a bit more... – Praphulla Koushik Jan 25 '14 at 14:55
@TattwamasiAmrutam : why do you think the quotient group is same as what you have written... it would be better if you can try writing that partially... – Praphulla Koushik Jan 25 '14 at 14:57
I guess Gerry is thinking that if none of these four cosets contains a complex cube root of $1$ then the conjectured solution cannot be corret. – Derek Holt Jan 25 '14 at 15:13
up vote 8 down vote accepted

I think you can set this map $$f:\mathbb C^*\to U,~~z=a+ib\mapsto \frac{a}{|z|}+i\frac{b}{|z|}$$ wherein $U=\{z\in\mathbb C^*\mid|z|=1\}$. Show this map is surjective with $\ker f=\mathbb R^*_{>0}$.

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@amWhy: Hello Amy. I am just online. :-) – Babak S. Jan 26 '14 at 14:17

I am not sure if this is in any way different than B.S's way. i am just omitting maps and trying to see just as a quotient.

We see $x+iy\in \mathbb{C}$ as $(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})$

Now, As $\sqrt{x^2+y^2}\in \mathbb{R}$ when we see $(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})$ modulo real numbers we would get :

$$(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})\equiv (\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}}) \text { mod }\mathbb{R}$$

And then we have $|(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})|=\sqrt{(\dfrac{x}{\sqrt{x^2+y^2}})^2+(\dfrac{y}{\sqrt{x^2+y^2}})^2}=1$

So,for any $z\in \mathbb{C}$ we have $z=|z|(z')$ for $|z|\in \mathbb{R}$ and $|z'|=1$.

Thus $\mathbb{C}^*/P=\{z\in \mathbb{C} : |z|=1\}$

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