How to prove that $S^{2n+1}/S^1$ is homeomorphic to $\mathbb CP^n$ under a given identification We represent an element $(x_1,y_1,...,x_n,y_n,x_{n+1},y_{n+1})\in S^{2n+1}$ as an element $(z_1,...,z_{n+1})\in \mathbb C^{n+1}$ where $z_k = x_k+iy_k$. Now I'm considering the action of $S^1$ on $S^{2n+1}$ as $e^{iθ}·(z_1,...,z_{n+1}) = (e^{iθ}z_1,...,e^{iθ}z_{n+1})$. Under this action we identify any two elements as $(z_1,...,z_{n+1}) ∼ e^{iθ} ·(z_1,...,z_{n+1})\forall (z_1,...,z_{n+1})\in S^{2n+1}\subseteq \mathbb C^{n+1}$ and $e^{iθ}\in S^1$. For the resulting quotient space $S^{2n+1}/S^1$, I'm trying to prove that $S^{2n+1}/S^1$ is homeomorphic to $\mathbb CP^n$. How do I prove this? Any solution will be highly appreciated.

What is your definition of $\mathbb{C}P^n$? Is it

the quotient space of $\mathbb{C}^{n+1}− {0} $ under the equivalence
relation $ v \sim \lambda v$ for $ \lambda \neq 0 $

? I assume it is defined as above. I used $\pi$ to denote the quotient space. So  $\pi: \mathbb{R}^{n+1} \setminus \{0\} \to \mathbb{R}P^n$.
Here we identify $\mathbb{C}^{n+1} = \mathbb{R}^{2n+2}$
You essentially just need to apply the universal property of quotients.
Consider the inclusion map $S^{
2n+1} \hookrightarrow \mathbb{R}^{2n+2}$. This is of course continuous, and thus the composition
$$\begin{array}{ccccccccc} S^{2n+1} & \xrightarrow{i} & \mathbb{R}^{2n+2} \backslash \{0\} &\\
 & \searrow{f} & \downarrow{\pi} \\
 & & \mathbb{C}P^n \end{array}$$
is continuous. Note that $f$ is defined on $S^{(2n+1)}$,  $f(x)=f(\lambda x)$, where $|\lambda| = 1$. So $f$ makes the same identifications as your map: $e^{i\theta}$.
Thus, based on passing to the quotient, $f$ induces a map $F$ in the quotient by
$$\begin{array}{ccccccccc} S^{2n+1} & \xrightarrow{i} & \mathbb{R}^{2n+2}\backslash \{0\} &\\
\downarrow{e^{i\theta}} & \searrow{f} & \downarrow{\pi} \\
S^{2n+1}/\sim & \xrightarrow{F} & \mathbb{C}P^n .\end{array}$$
We have that $F$ is continuous. Note that it is also a bijection. Since $S^{2n+1}/\sim$ is compact (it is the image by $e^{i\theta}$ of $S^{2n+1}$, which is compact) and $\mathbb{C}P^n$ is Hausdorff, it is a homeomorphism.
Actually, I just copied the proof from this answer, making some modifications.
