We are told to show that the quotient group $\mathbb{R}/\mathbb{Z}$ is isomorphic to the group $U$ = $\{z \in \mathbb{C}\mid \vert z \vert = 1\}$ endowed with multiplication as operation.
My answer:
The group $U$ symbolizes the unit circle. Euler's formula gives an isomorphism from an previous assignment as $z = e^{i2\pi r}$. We also showed that $\phi:\mathbb{R} \longmapsto \mathbb{C}$ is a homomorphism. The equation for that was $\phi(r)=\cos (2\pi r) + i \sin (2\pi r)$. Plotting Euler's formula on an imaginary-real axis gives us a circle. Here $r = 1$ (as seen on the kernel since $\epsilon = 1$). This indicates that it is isomorphic to $U$. $\mathbb{Z}$ is just any set of integers and will only make the values smaller and thereby into the unit circle.
What do you think of my answer?
So what you want to say is something like that $\psi:\Bbb R\to U$ by $\psi(r)=e^{2\pi i r}$ induces an isomorphism between $\Bbb R/\Bbb Z$ and $U$. Here $U$ is the circle group. This is true by the first isomorphism theorem, since $\operatorname{ker}\psi=\Bbb Z$.
This is true.