Identifying a quotient group (NBHM-$2014$) 
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?
 A: 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 $\ker f=\mathbb R^*_{>0}$.
A: One another way,
The map $f:\mathbb{C}^* \to \mathbb{C}^*$ defined by $f(z=r.e^{i\theta})=1.e^{i\theta}$ is a homomorphism with kernel being the set of positive reals.
By the Fundamental theorem of homomorphism we can say that $\mathbb{C}^*/P$ is isomorphic to $f(\mathbb{C}^*)=S^1$. 
where $S^1=\{z\in\mathbb{C} | |z|=1\}$.
A: 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 $\mathbb{C}^*/P=\{z\in \mathbb{C} : |z|=1\}$
