Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Which of those 4 cosets contains a complex cube root of 1? –  Gerry Myerson Jan 25 at 14:42
    
@GerryMyerson $P$ contains $1$.. –  tattwamasi amrutam Jan 25 at 14:45
    
@GerryMyerson : I do not understand your idea.... :O could you please explain a bit more... –  Praphulla Koushik Jan 25 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 at 14:57
2  
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 at 15:13

2 Answers 2

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 $\ker f=\mathbb R^*_{>0}$.

share|improve this answer
    
@amWhy: Hello Amy. I am just online. :-) –  Babak S. Jan 26 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 $\mathbb{C}^*/P=\{z\in \mathbb{C} : |z|=1\}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.