I was reading the reason why $S^1$ is Lie group
The circle $S^1 \subset \mathbb{C}^\star$ is a smooth manifold and a group under complex multiplication. With appropriate angle functions as local coordinates on open subsets of $S^1$ (see Problem 1-8), multiplication and inversion have the smooth coordinate expressions $(\theta_1 , \theta_2) \mapsto \theta_1 + \theta_2$ and $\theta \mapsto -\theta$ , and therefore $S^1$ is a Lie group, called the circle group.
Just to be sure, the atlas we are talking about here is $\{(U_p, \theta_p):p\in S^1 \}$ where $U_p=S^1\setminus\{p\}$ and $\theta_p : U_p \to \mathbb{R}, z=e^{it} \mapsto t$?
You have to be a little careful. As mentioned above in the comment, the map $e^{it}\mapsto t$ is not globally defined on $S^1$ (standard issues in defining a $\log$ function). However, you can define a pair of coordinate charts by $S^1\setminus \{1\}\to (0,2\pi)$ and $S^1\setminus\{-1\}\to (0,2\pi)$ by taking suitable branches of the complex logarithm for instance.
Actually, more generally by choosing appropriate branch cuts for $\log$, we can define a coordinate chart $S^1\setminus \{p\}\to (0,2\pi)$ by using $\log$. Now, if we take any pair of points $x,y\in S^1$, then we can find a chart containing $x,y,xy$. Then you can check that in coordinates the multiplication is given by addition of the angular coordinate.
Here is another way you might see that this is a Lie group structure: if $M$ is a manifold and $N$ is an embedded submanifold, then the restriction of a smooth function $f$ on $M$ to $N$ is smooth. Note that the multiplication on $S^1\subseteq \mathbb{C}^*$ is given by restricting the multiplication structure on $\mathbb{C}$. It is easy to see that $\mu:\mathbb{C}\times \mathbb{C}\to \mathbb{C}$ is smooth, hence its restriction to $S^1\times S^1\subseteq \mathbb{C}\times \mathbb{C}$ (valued in $S^1$) is smooth. The same reasoning shows that the inversion map $i$ is smooth.