Submanifolds of $\mathbb{C}P_1$ I've recently started to study some differential geometry, but I am having some trouble with this problem:
Let $\displaystyle{\mathbb{C}P_1:= \frac{\mathbb{C}^2 - \{0\}}{\sim}} $ be the projective space, with $\sim$ the equivalence relation given by
$$(z_1,z_2) \sim (w_1,w_2) \iff (z_1,z_2) = \lambda (w_1,w_2), \text{ for some $\lambda \in \mathbb{C}$. }$$
And denote by $[z_1 : z_2]$ the equivalence class of $(z_1,z_2)$.
I am trying to prove that
\begin{align}
N_1 &= \{[z_1 : z_2] \in \mathbb{C}P_1 : \left |\frac{z_1}{z_2} \right| = 1, z_2 \neq 0\},\\
N_2 &= \{[z_1: z_2] \in \mathbb{C}P_1 : e^{i\frac{\pi}{3}}z_1 \overline{z_2} = e^{- i\frac{\pi}{3}}\overline{z_1}z_2 \}
\end{align}
are rank one submanifolds of $\mathbb{C}P_1$.
With the knowledge I already have, I've tried to find smooth functions $F_i: \mathbb{C}P_1 \to \mathbb{C}$, $i = 1,2$, such that for some regular value $q_i \in \mathbb{C}$ one has $F_i^{-1}(q) = N_i$, which would, if I am not wrong, solve the problem. However, I do not know how to find functions that do the job. Could someone point out what I could do?
 A: The functions are more or less there: for $N_1$, consider
$$ U_1 = \{ [z_1, z_2] | z_2\neq 0\} = \mathbb {CP}^1\setminus \{[0,1]\}.$$
and on $U_1$ define
$$F: U_1 \to \mathbb R, \ \ \ F([z_1, z_2]) = \left| \frac{z_1}{z_2}\right|^2$$
(We take square so that $F$ is smooth). Note that $F$ is well defined since $F([\lambda z_1, \lambda z_2]) = F([z_1, z_2])$ .
It is clear that $F$ is regular except when $z_1 = 0$, that is, at $[0,1]$. Indeed, in the local coordinate $\phi: U_1 \to \mathbb C$, $\phi ([z_1, z_2]) = z_1/z_2$, we have $F\circ \phi^{-1} (u) = |u|^2$.
Thus $F^{-1}(1) = N_1$ is a 1-dimensional submanifold of $N_1$, and thus $\mathbb{CP}^1$.
For $N_2$ it is similar: note that $e^{i\frac{\pi}{3}}z_1 \overline{z_2} = e^{- i\frac{\pi}{3}}\overline{z_1}z_2 $ if and only if $e^{i\frac{\pi}{3}}z_1 \overline{z_2}$ is real. So define
$$G: \mathbb {CP}^1 \to \mathbb R, \ \ \ G([z_1, z_2]) = \frac{\operatorname{Im}(e^{i\frac{\pi}{3}}z_1 \overline{z_2})}{|z_1|^2 + |z_2|^2}.$$
(We put $|z_1|^2 + |z_2|^2$ at the bottom, so that $G$ is well-defined: i.e. $G([\lambda z_1, \lambda z_2]) = G([z_1, z_2])$).
Now we show that $G^{-1}(0) = N_2$ is a regular level set. Let $[z_1, z_2]\in G^{-1}(0)$. Then Write $r = e^{i\frac{\pi}{3}}z_1 \overline{z_2}$, which is real by assumption. Now we split into two cases:
Cases 1 When $r \neq 0$: let $\gamma (t) = [e^{it}z_1, z_2]$ be a curve in $\mathbb {CP}^1$. Hence
$$ DG_{[z_1, z_2]} \gamma'(0) = \frac{d}{dt} G(\gamma(t)) \bigg|_{t=0} = \frac{d}{dt}\frac{\operatorname{Im} (re^{it})}{|z_1|^2 + |z_2|^2}\bigg|_{t=0} =\frac{r}{|z_1|^2 + |z_2|^2} \neq 0, $$
thus $DG$ is surjective there.
Cases 2 When $r=0$: then either $z_1 = 0$ or $z_2 = 0$. That is, either $[z_1, z_2] = [1,0]$ or $[0,1]$. In either case, one could use $\gamma (t) = [1,t]$, $[t, 1]$ respectively to show that $DG$ is surjective.
Thus $G^{-1}(0)$ is a regular level set and $N_2$ is a 1-submanifold.
Remark If your question you stated that you are looking for functions $\mathbb{CP}^1 \to \mathbb C$, but since $\mathbb C$ is two real dimensional, any such regular level sets would be zero dimensional submanifolds, thus will not be $N_i$.
