Show that $\mathbb{R}/\mathbb{Z}$ is isomorphic to $\{e^{i\theta} : 0 \le \theta \le 2\pi \}$

This question is asking to prove that the quotient group $\mathbb{R}/\mathbb{Z}$ is isomorphic to the group of complex numbers with modulus 1 (under multiplication). It's hard for me to visualize the structure of $\mathbb{R}/\mathbb{Z}$, so I can't think of an isomorphism. I feel like the first isomorphism theorem can help? Help is appreciated.

The map $t\mapsto e^{2\pi it}$ is a homomorphism from $\mathbb{R}$ onto the circle group. Its kernel is $\mathbb{Z}$. Invoke the first isomorphism theorem.

Hint: Try to look at the obvious homomorphism (from addition in $\mathbb{R}$ to multiplication in $\mathbb{C}$) defined by $\phi: \mathbb{R} \to \mathbb{C}$ with $$\phi(x) = e^{2\pi ix}.$$

What is the image? What is the kernel? Then you should apply the first isomorphism theorem.

You can see the problem this way : $\mathbb{R}/\mathbb{Z}=[0,1)$. Then, at each point in this interval correspond an unique point in the circle $\{ e^{2\pi i \theta} : \theta \in [0,1) \}$. About the group operations, when you add two numbers in $[0,1)$, you multiply them in the set representing the circle (adding the argument).