Describe the quotient groups $\mathbb{C}^{\times}/P$ and $\mathbb{C}^{\times}/U$ where $U$ is the subgroup of complex numbers of absolute value $1$ and $P$ denotes the positive reals.
$\mathbb{C}^{\times}/P=\{C_r: r>0\}$, where $C_r$ is the subgroup with if $z\in C_r$, $\mid z\mid =r$
But I could not write $\mathbb{C}^{\times}/U$ like $\mathbb{C}^{\times}/P$. What I can see that $U=\{r\in \mathbb{R}:r>0\}$ is somehow related to $\mathbb{C}^{\times}/P$. That's all.
For the quotient by $P$, observe that any two complex numbers are equivalent if either is a positive real multiple of the other. In other words, the class of equivalence of a complex number $re^{i\theta}$ is
$$\{te^{i\theta}\,|\,t\in(0,+\infty)\}.$$
A representative of each equivalence class may be then taken by setting $t=1$, so the set of representatives may be taken to be $\mathbb{S}^1=U$. You can check that indeed as a mulplicative group $U$ is homeomorphic to $\mathbb{C}^{\times}/P$.
For the quotient by $U$, observe that any two complex number are equivalent if either can be obtained from the other via multiplication by a complex number of modulus one. In other words, for a complex number $re^{i\theta}$, its equivalence class is
$$\{re^{i\alpha}\,|\,\alpha\in[0,2\pi]\}$$
A representative of each equivalence class may be then taken by setting $\alpha=0$, so the set of representatives may be taken to be $(0,+\infty)=P$. You can check that, indeed, as a mulplicative group $P$ is homeomorphic to $\mathbb{C}^{\times}/U$.
I guess the point of this exercise was to show that $\mathbb{C}^{\times}/P\simeq U$ and $\mathbb{C}^{\times}/U\simeq P$, which is kind of neat.
