# Coordinate representation of the map $e^{2\pi it}$ using an angle coordinate map

This is Example 2.13 (b) from Lee's Introduction to Smooth Manifolds.

If the circle $$\mathbb{S}^1$$ is given its standard smooth structure, the map $$\epsilon: \mathbb{R} \to \mathbb{S}^1$$ defined by $$\epsilon(t)=e^{2\pi it}$$ is s mooth, because with respect to any angle coordinate $$\theta$$ for $$\mathbb{S}^1$$ it has a coordinate representation of the form $$\hat{\epsilon}(t)=2\pi t+c$$ for some constant $$c$$, as you can check.

Here $$\theta$$ is a function from some open subset $$U \subset \mathbb{S}^1$$ to $$\mathbb{R}$$ such that $$e^{i\theta (z)}=z$$ for all $$z \in U$$. So we have $$\hat{\epsilon} = \theta \circ \epsilon$$. I can't see why we have $$\hat{\epsilon}(t)=2\pi t+c$$ for some constant $$c$$ here. I would appreciate any help.

Well, $$e^{i\theta (z)}=z\implies \theta(z)=\ln z/i$$.
Note, your $$S^1$$ sits in $$\Bbb C$$.
Now, we need a branch of log, because the complex logarithm is not a single valued function. The principal branch can be assumed, for instance, in which case, we have $$\ln z=\ln r +i\theta$$, where $$z=re^{i\theta}$$.
In this case your $$c$$ will be $$0$$. For a different branch of log, it will be $$\ln z=\ln r+i\theta+2\pi ki$$. See https://math.stackexchange.com/a/1475592/403337.
So let's compose and see what we get: $$\hat\epsilon(t)=\theta\circ\epsilon(t)=\theta(e^{2\pi i t})=\ln(e^{2\pi i t})/i=2\pi t+2\pi k =2\pi t+ c$$, where $$c=2\pi k$$.