$L_d = (\mathbb{C}^{n+1} - \{0\}) \times \mathbb{C}/\sim,$ where $(x_0, \dots, x_n, q) \sim (\lambda x_0, \dots \lambda x_n, \lambda^dq).$ I want to show that the following is a manfiold: $$L_d = (\mathbb{C}^{n+1} - \{0\}) \times \mathbb{C}/\sim, \text{ where } (x_0, \dots, x_n, q) \sim (\lambda x_0, \dots \lambda x_n, \lambda^dq).$$
I'm looking for a collection of open charts which cover $L_d.$ For $\mathbb{P}^n,$ we usually use $U_i,$ which is the set of points such that $x_i \neq 0.$ Can we do something similar here? So $U_0 = (1, \tfrac{x_1}{x_0}, \dots, \tfrac{x_n}{x_0}, \tfrac{q}{x_0}).$ Then each of these would be isomorphic to $\mathbb{C}^{n+1}.$ Is this anywhere near the right idea? I always get confused when thinking about coordinate charts. What would the transition functions be?
 A: Yes, I think you're in the right track. You can put
$$U_i = \{ [x_0:\ldots:x_n:q] \in L_d \mid i \neq 0 \}$$and define $\varphi\colon U_i \to \Bbb C^{n+1}$ by $$\varphi_i([x_0:\ldots: x_n: q]) = \left(\frac{x_0}{x_i}, \ldots,\frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i}, \ldots, \frac{x_n}{x_i}, \frac{q}{x_i^d}\right).$$ You have to divide the last component by $x_i^d$ instead of just $x_i$, because of the definition of $\sim$. Clearly we have $\bigcup_{i=0}^n U_i = L_d$, since at least one of the first $n+1$ entries of $[x_0,\ldots, x_n,q]$ has to be non-zero. The inverse is $\varphi^{-1}_i\colon \Bbb C^{n+1} \to U_i$ given by $$\varphi_i^{-1}(y_0,\ldots, y_{n+1}) = [y_0:\ldots: y_{i-1}:1:y_{i+1}:\ldots:y_n].$$If you don't divide $q$ by $x_i^d$ in the definition of $\varphi_i$, you'll see what goes wrong in checking that $\varphi_i^{-1}\circ \varphi_i$ is the identity on $U_i$. Checking that $\varphi_i\circ \varphi^{-1}_i$ is the identity goes just like what happens in $\Bbb C P^n$.
To see what the transition maps are, assume without loss of generality that $0=i<j=1$, and directly compute $\varphi_0\circ \varphi_1^{-1}$ to be $$\varphi_0\circ \varphi_1^{-1}(y_0,\ldots, y_{n+1}) = \left(\frac{1}{y_0}, \frac{y_1}{y_0}, \ldots, \frac{y_n}{y_0}, \frac{y_{n+1}}{y_0^d}\right).$$That is, the transitions are just like the transitions in $\Bbb C P^n$, but with the extra term divided by a $d$-th power. So you get a holomorphic atlas, all the same.
