Let $C^*$ and $R^+$ denote the multiplicative group of non zero complex numbers and the subgroup of positive reals respectively. Identify the quotient group $\frac{C^*}{R^+}$.

To prove it regorusly, write $z_j = r_j e^{i\theta_j}$, $j = 1, 2$, we see from $z_1 = rz_2$ or $r_1 e^{i\theta_1} = rr_2e^{i\theta_2}$ that $r_1 = r r_2$ and $e^{i\theta_1} = e^{i\theta_2}$; restricting $\theta_1, \theta_2 \in [0, 2\pi)$ we have $\theta_1 = \theta_2$; each coset is thus represented by a unique $\theta \in [0, 2\pi)$ or, equivalently, a unique unimodular complex number $e^{i\theta}$.

Furthermore, it is easy to see that since $z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}$, the natural projection $C^\ast \to C^\ast/R^+ = S^1$ is a homomorphism taking $z_1z_2$ to $\theta_1 + \theta_2$, where we use the usual addition $\mod 2\pi$. $C^\ast/R^+$ is the circle group indeed.

.**Solution :****The elements of the group $C^\ast/R^+$ are the cosets of $R^+$ in $C^\ast$ and $z_1, z_2 \in C^\ast$ are in the same coset if and only if $z_1 z_2^{-1} \in R^+$, i.e., if and only if $z_1 = rz_2$ for some $r \in R^+$. This means that $z_1$ and $z_2$ lie on the same ray through the origin, so we must have $\arg z_1 = \arg z_2$. And so we guess that $C^\ast/R^+$ is the circle group $S^1$.**

To prove it regorusly, write $z_j = r_j e^{i\theta_j}$, $j = 1, 2$, we see from $z_1 = rz_2$ or $r_1 e^{i\theta_1} = rr_2e^{i\theta_2}$ that $r_1 = r r_2$ and $e^{i\theta_1} = e^{i\theta_2}$; restricting $\theta_1, \theta_2 \in [0, 2\pi)$ we have $\theta_1 = \theta_2$; each coset is thus represented by a unique $\theta \in [0, 2\pi)$ or, equivalently, a unique unimodular complex number $e^{i\theta}$.

Furthermore, it is easy to see that since $z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}$, the natural projection $C^\ast \to C^\ast/R^+ = S^1$ is a homomorphism taking $z_1z_2$ to $\theta_1 + \theta_2$, where we use the usual addition $\mod 2\pi$. $C^\ast/R^+$ is the circle group indeed.

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