$p_n(z)=z^n$ is a smooth map on $S^1$

Question

I have to solve this one:

Consider local coordinates on $S^n$ given by the atlas $\{(S^n\setminus\{N\},\varphi_N),(S^n\setminus\{S\},\varphi_S)\}$, where $\varphi_N$ and $\varphi_S$ are the stereographic projection from the north and south pole respectively. Compute the expression in local coordinates of the function

\begin{equation} \begin{array}{crcl} p_n: & S^1 & \rightarrow & S^1\\ & z & \mapsto & z^n \end{array} \end{equation} with $n\in\mathbb{Z}$, and prove that is a smooth map.

Solution:

We can consider $z\in S^1$ as $e^{i\theta(z)}$, where $\theta$ is the angle function $\theta:U\subset S^1\rightarrow \mathbb{R}$ defined as $e^{i\theta(z)}=z$. According to this notation, we have $p_n(z)=e^{in\theta(z)}=\cos(n\theta)+i\sin(n\theta)$. Now we take $\varphi_N$ on $S^1\setminus\{N\}=S^1\setminus\{(0,i)\}$ and we get \begin{equation} \varphi_N\left(e^{in\theta(z)}\right)=\frac{\cos(n\theta)}{1-\sin(n\theta)}\ . \end{equation}

So this is the expression in local coordinates (stereographics' one) and from it we see that $p_n$ is smooth.

Is it all right?

Thanks a lot